Cap 03

effective

1. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Chapter 3, Solution 1.













Resolve 90 N force into vector components P and Q
where
( )90 N sin 40Q= °
57.851 N=
Then
/BAB
M rQ=−
( )( )N851.57m225.0−=
13.0165 N m=−⋅

13.02 N m
B
=⋅M

W

2. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 2.

















( )90 N cos 25
x
F= °
81.568 N =
( )90 N sin 25
y
F= °
38.036 N =
( )0.225 m cos65x= °
0.095089 m=
( ) °= 65sinm225.0y
0.20392 m =
B yx
M xF yF=−
( )( )( )( )0.095089 m 38.036 N 0.20392 m 81.568 N=−
13.0165 N m=−⋅
13.02 N m
B
=⋅M
W

3. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 3.
















()3lb sin30
x
P= °
1.5 lb=
()3lb cos30
y
P= °
2.5981 lb=

//ABAyBAx
M xP yP=+
( )( )( )( )3.4 in. 2.5981 lb 4.8 in. 1.5 lb=+
16.0335 lb in.= ⋅
16.03 lb in.
A
=⋅M
W

4. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 4.
















For
P to be a minimum, it must be perpendicular to the line joining points
A and B
with
()()
22
3.4 in. 4.8 in.
AB
r=+
5.8822 in.=

1
tan
y
x
αθ
−
==




14.8 in.
tan
3.4 in.
−

= 



54.689
= °
Then
minAAB
M rP=
or
min
19.5 lb in.
5.8822 in.
A
AB
M
P
r ⋅
==
3.3151 lb
=

min
3.32 lbP∴=

54.7°


or
min
3.32 lbP=

35.3° W

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 5.













By definition
/
sin
ABA
MrP
θ=
where ( )90θφα=+ °−
and
14.8 in.
tan
3.4 in.
φ
−

= 


54.689
= °
Also
()()
22
/
3.4 in. 4.8 in.
BA
r=+
5.8822 in.
=

Then ( )( )( )( )17 lb in. 5.8822 in. 2.9 lb sin 54.689 90α⋅= °+°−
or
(
)sin 144.689 0.99658α°− =
or 144.689 85.260 ; 94.740α°− = ° °
49.9 , 59.4
α∴= ° ° W

6. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 6.


(a)


(
b)










(
a)
/ABABF
=×Mr T

y x
ABF BF
M xT yT=+
()( ) ( )( )2 m 200 N sin 60 0.4 m 200 N cos60= °+ °
386.41 N m= ⋅
or 386 N m
A
= ⋅M

W

(b) For
C
F to be a minimum, it must be perpendicular to the line
joining
A and C.

()
min
AC
MdF∴=
with
()( )
22
2 m 1.35 md=+
2.4130 m =
Then ( )()
min
386.41 N m 2.4130 m
C
F⋅=
()
min
160.137 N
C
F =
and
11.35 m
tan 34.019
2m
φ
−

= =°

90 90 34.019 55.981
θ φ=−=°− °= °

(
)
min
160.1 N
C
∴=F 56.0°W

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

=
Chapter 3, Solution 7.


(a)






(b)





(
c)






yx
A BF BF
MxT yT=+

()( ) (
)()2 m 200 N sin 60 0.4 m 200 N cos60=°+ °

386.41 N m=⋅
or
386 N m
A
=⋅M

=



Have
A C
MxF=
or
386.41 N m
2m
A
C
M
F
x
⋅
==

193.205 N=

193.2 N
C
∴=F
=
For
B
F to be minimum, it must be perpendicular to the line joining A
and B

()
minAB
MdF∴=
with
()( )
22
2 m 0.40 m 2.0396 md=+ =
Then
() ()
min
386.41 N m 2.0396 m
C
F⋅=

()
min
189.454 N
C
F =
and
12m
tan 78.690
0.4 m
θ
−

==°


or
()
min
189.5 N
C
=F
78.7°=

8. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

(
Chapter 3, Solution 8.


(a)






(b)






(c)







()/
cos15
BAB
Mr W=°

()( )()14 in. cos15 5 lb=°

67.615 lb in.=⋅
or
67.6 lb in.
B
=⋅M
(



/
sin85
BDB
MrP=°
()67.615 lb in. 3.2 in. sin85P ⋅= °
or 21.2 lbP= (


For
()
min
,FF must be perpendicular to BC.
Then,
/BCB
MrF=

()67.615 lb in. 18 in.F ⋅=
or
3.76 lb=F

75.0°(

9. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

=
Chapter 3, Solution 9.









(a) Slope of line
35 in. 5
76 in. 8 in. 12
EC==
+

Then
()
12
13
ABx AB
TT=
()
12
260 lb 240 lb
13
==

and
()
5
260 lb 100 lb
13
ABy
T==
Then
() ()35 in. 8 in.
D ABx ABy
MT T=−

()()()()240 lb 35 in. 100 lb 8 in.=−

7600 lb in.=⋅
or
7600 lb in.
D
=⋅M

=+
(b) Have
() ()
D ABx ABy
MTyTx=+

(
)() ( )( )240 lb 0 100 lb 76 in.=+

7600 lb in.=⋅
or
7600 lb in.
D
=⋅M

=

10. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

=
Chapter 3, Solution 10.








Slope of line
35 in. 7
112 in. 8 in. 24
EC==
+

Then
24
25
ABx AB
TT=
and
7
25
ABy AB
TT=
Have
() ()
D ABx ABy
MT
yTx=+
() ()
24 7
7840 lb in. 0 112 in.
25 25
AB AB
TT∴⋅= +
250 lb
AB
T=
or
250 lb
AB
T= =

11. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 11.



The minimum value of d can be found based on the equation relating the moment of the force
AB
T about D:
( )()
maxDAB
y
M Td=
where 1152 N m
D
M
= ⋅
( ) ( )
max max
sin 2880 N sin
AB AB
y
TT θ θ==
Now
()()
22
1.05 m
sin
0.24 1.05 md
θ=
++



()()
()
22
1.05
1152 N m 2880 N
0.24 1.05
d
d


∴⋅=

++


or
()()
22
0.24 1.05 2.625dd++ =
or
()()
22 2
0.24 1.05 6.8906dd++ =
or
2
5.8906 0.48 1.1601 0dd−− =
Using the quadratic equation, the minimum values of d are 0.48639 m and −0.40490 m.
Since only the positive value applies here, 0.48639 m
d
=
or 486 mm
d
= W

12. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 12.

















with
()( )
22
42 mm 144 mm
AB
d=+
150 mm=

42 mm
sin
150 mm
θ=

144 mm
cos
150 mm
θ=
and sin cos
AB AB AB
FF
θ θ=− −Fij
()()
2.5 kN
42 mm 144 mm
150 mm
 =−−
 
ij
( )( )700 N 2400 N=− −ij
Also ( )( )
/
0.042 m 0.056 m
BC
=− +rij
Now
/CBCAB
=×Mr F

( )( )0.042 0.056 700 2400 N m=−+ ×−− ⋅ij ij
( )140.0 N m=⋅ k
or 140.0 N m
C
=⋅M
W

13. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 13.

















with
()( )
22
42 mm 144 mm
AB
d=+
150 mm=

42 mm
sin
150 mm
θ=

144 mm
cos
150 mm
θ=
sin cos
AB AB AB
FF
θ θ=− −Fij
()()
2.5 kN
42 mm 144 mm
150 mm
 =−−
 
ij
( )( )700 N 2400 N=− −ij
Also ( )( )
/
0.042 m 0.056 m
BC
=− −rij
Now
/CBCAB
=×Mr F

( )( )0.042 0.056 700 2400 N m=−− ×−− ⋅ij ij
( )61.6 N m=⋅ k
or 61.6 N m
C
= ⋅M W

14. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 14.









() ()
88 105
: 0.090 m 80 N 0.280 m 80 N
137 137
DD
MM
 
Σ= ×− ×
 
 

12.5431 N m=− ⋅
or 12.54 N m
D
=⋅M

W

15. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 15.


Note: (
)cos sinBβ β=+Bij

(
)cos sinBβ β′=−Bij

(
)cos sinCα α=+Cij
By definition: ( )sinBCαβ×= −BC (1)
( )sinCBC αβ′×= +B (2)
Now ...
(
)( )cos sin cos sinBCβ βαα×= + × +BC i j i j

(
)cos sin sin cosBCβαβα=− k (3)
and
(
)( )cos sin cos sinBCβ βαα′×= − × +BC i j i j

(
)cos sin sin cosBCβα βα=+ k (4)
Equating the magnitudes of ×
BC from equations (1) and (3) yields:

(
) ( )sin cos sin sin cosBC BCαββαβα−= − (5)
Similarly, equating the magnitudes of ′×BC from equations (2) and (4) yields:

(
) ( )sin cos sin sin cosBC BCαββαβα+= + (6)
Adding equations (5) and (6) gives:

(
) ( )sin sin 2cos sinαβαβ βα−+ +=
or () ()
11
sin cos sin sin
22
αβ αβ αβ=++− W

16. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Chapter 3, Solution 16.


Have
/AB O A
d=× rλ
where
/
/
BA
AB
BA
r
=
r
λ

and
()
/
210 mm 630 mm
BA
=− −ri

()()270 mm 225 mm+−− j

()()840 mm 495 mm=− + ij

()()
22
/
840 mm 495 mm
BA
r=− +
975 mm =
Then
()()
840 mm + 495 mm
975 mm
AB
λ
−
=
ij


()
1
56 33
65
=−+ ij
Also
()( )
/
0 630 0 ( 225)
OA
=− +−−rij

()()630 mm 225 mm=− + ij

( )()()
1
56 33 630 mm 225 mm
65
d  ∴= − + ×− +
 
ij i j
126.0 mm =

126.0 mmd= W

17. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

×
Chapter 3, Solution 17.


(a)
×
=
×
AB
AB
λλλλ

where 12 6 9=−+Ai
jk

397.5=− + −Bi
j k
Then
12 6 9
3 9 7.5
×= −
−−
ijk
AB


()( )( )45 81 + 108 18=− −27+90+ −i
j k

()94 7 10=−++i
jk
And
22 2
9 ( 4) (7) (10) 9 165×= −+ + =AB

()
94 7 10
9165
−+ +
∴=
ij k
λλλλ

or
()
1
4710
165
=−++ ij kλλλλ ×
(b)
×
=
×AB
AB
λλλλ
where
14 2 8=− − +Ai
jk

31.5=+ −Bi
jk
Then
14 2 8
3 1.5 1
×=− −
−
ijk
AB


()( )( )212 2414 216=− + − +−+i
j k

()52 2 3=−+ −i
jk
and
22 2
5(2) (2) (3) 517×= −+ +− =AB

()
5223
517
−+ −
∴=
ijk
λλλλ

or
()
1
223
17
=−+− ijk ×λλλλ
×

18. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 18.


(a) Have
A=×PQ

2
37 2in.
51 3
×= −
−
ijk
PQ

[
]
2
(21 2) (10 9) (3 35) in.=++−++ijk

()
( )( )
22 2
23 in. 1 in. 38 in.=++ ij k

22 2 2
(23) (1) (38) 44.430 in.A∴= + + =
or
2
44.4 in.A= W
(
b)
A=×PQ

2
243in.
615
×= −
−
ijk
PQ

[
]
2
(20 3) (18 10) (2 24) in.=− − +− − +−+ijk

(
)( )( )
22 2
23 in. 28 in. 22 in.=− − + ijk

222 2
(23) (28) (22) 42.391in.A∴=− +− + =
or
2
42.4 in.A= W

19. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 19.


(a) Have
O
=×MrF

63 1.5 Nm
7.5 3 4.5
=− ⋅
−
ijk


[
]( 13.5 4.5) (11.25 27) ( 18 22.5) N m=− − + − +− − ⋅ijk

(
)18.00 15.75 40.5 N m=− − − ⋅ijk
or ( ) ( )18.00 N m m 40.5N m
O
=− ⋅ (15.75Ν⋅ ) − ⋅Mijk W−
(
b) Have
O
=×MrF

20.751Nm
7.5 3 4.5
=− −⋅
−
ijk


[
](3.375 3) ( 7.5 9) (6 5.625) N m=++−+++ ⋅ij k

(
)6.375 1.500 + 11.625 N m=+ ⋅ij k
or ( ) ( )6.38 N m m 11.63 m
O
=⋅+(1.500Ν⋅)+Ν⋅Mijk W
(
c) Have
O
=×MrF

2.5 1 1.5 N m
7.5 3 4.5=− − ⋅
ijk

[
](4.5 4.5) (11.25 11.25) ( 7.5 7.5) N m=−+ − +−+ ⋅ijk
or 0
O
=MW
This answer is expected since
r and F are proportional
( )3.=−Fr Therefore, vector F has a line of action
passing through the origin at
O.

20. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 20.


(a) Have
O
=×MrF

7.5 3 6 lb ft
364=− − ⋅
−
ijk

[
](12 36) ( 18 30) (45 9) lb ft=−+−++− ⋅ijk
or ( )( )( )24.0 lb ft lb ft lb ft
O
=− ⋅ +12.00⋅ +36.0⋅Mijk W
(
b) Have
O
=×MrF

7.5 1.5 1 lb ft
364=− − ⋅
−
ijk

[
](6 6) ( 3 3) (4.5 4.5) lb ft=−+−++ − ⋅ij k
or 0
O
=MW
(
c) Have
O
=×MrF

82 14lbft
364=− − ⋅
−
ijk

[
](8 84) ( 42 32) (48 6) lb ft=− +−+ + − ⋅ijk
or ( )( )( )76.0 lb ft lb ft lb ft
O
= − ⋅ − 10.00 ⋅ + 42.0 ⋅Mijk W

21. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 21.









With
( )369 N
AB
=−Tj
()
( )( )( )
()( )( )
22 2
2.4 m 3.1 m 1.2 m
369 N
2.4 m 3.1 m 1.2 m
AB AD
AD
T
AD −−
==
+− +−
ijk
T
JJJG

( )( )( )216 N 279 N 108 N
AD
=−−Tijk
Then 2
AABAD
=+RTT

( )( )( )216 N 1017 N 108 N=− − ijk
Also ( )( )
/
3.1 m 1.2 m
AC
=+rik
Have
/CAC A
=×Mr R

03.11.2Nm
216 1017 108
= ⋅
−−
ijk

( )( )( )885.6 N m 259.2 N m 669.6 N m=⋅+⋅−⋅ ijk
( )( )( )886 N m 259 N m 670 N m
C
=⋅+⋅−⋅Mijk W

22. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 22.
















Have
/ACA
= ×Mr F
where ( )( )( )
/
215 mm 50 mm 140 mm
CA
=−+rijk
( )36 N cos45 sin12
x
F=−°°
( )36 N sin 45
y
F=−°
( )36 N cos 45 cos12
z
F=−°°
( )( )( )5.2926 N 25.456 N 24.900 N∴=− − −Fijk
and 0.215 0.050 0.140 N m
5.2926 25.456 24.900
A
= −⋅
−−−
ijk
M
( )( )( )4.8088 N m 4.6125 N m 5.7377 N m=⋅+⋅−⋅ ijk
( )( )( )4.81 N m 4.61 N m 5.74 N m
A
=⋅+⋅−⋅Mijk W

23. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Chapter 3, Solution 23.
















Have
/OAO
= ×Mr R
where ()()
/
30 ft 3 ft
AD
=+rjk
( ) ( )
1
62 lb cos10 62 lb sin10  =−°−°
  
Tij


( )( )61.058 lb 10.766 lb=− − ij

22
AB
T
AB
=T
JJJG


()
()()()
()( )()
222
5ft 30ft 6ft
62 lb
5ft 30ft 6ft−+
=
+− +
ijk


()( )()10 lb 60 lb 12 lb=−+ ijk
( )( )()51.058 lb 70.766 lb 12 lb∴=− − +Rijk
0303lbft
51.058 70.766 12
O
= ⋅
−−
ijk
M
( )( )( )572.30 lb ft 153.17 lb ft 1531.74 lb ft=⋅−⋅+ ⋅ ij k
( )( )( )572 lb ft 153.2 lb ft 1532 lb ft
O
=⋅− ⋅+ ⋅Mijk W

24. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 24.








(a) Have
/OBOBD
=×Mr T
where
(
)()2.5 m 2 m
B/O
=+rij

BD BD
BD
T
BD
=T
JJJG


()
()()()
()()()
222
1m 2m 2m
900 N
1m 2m 2m
 −− +
 
=
−+−+
ijk


( )( )( )300 N 600 N 600 N=− − + ijk
Then
2.5 2 0 N m
300 600 600
O
=⋅
−−
ijk
M

( )( )( )1200 N m 1500 N m 900 N m
O
=⋅−⋅−⋅Mijk W

continued

25. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



(b) Have
/OBOBE
=×Mr T
where
(
)()2.5 m 2 m
B/O
=+rij

BE BE
BE
T
BE
=T
JJJG


()
( )()()
()()()
22 2
0.5 m 2 m 4 m
675 N
0.5 m 2 m 4 m
 −−−
 
=
+− +−
ijk


( )( )( )75 N 300 N 600 N=− − −ijk
Then
2.5 2 0 N m
75 300 600
O
=⋅
−− −
ijk
M

( )( )( )1200 N m 1500 N m 600 N m
O
=− ⋅ + ⋅ − ⋅Mijk W

26. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 25.



Have
/CAC
= ×Mr P
where

///AC BC AB
=+rrr

( )( )= 16 in. cos80 cos15 sin80 cos80 sin15− °°−°− °°ij k
( )( )15.2 in. sin 20 cos15 cos20 sin 20 sin15+ −°°+ °− °°ij k
( )( )( )7.7053 in. 1.47360 in. 2.0646 in.=− − − ijk
and
(
)( )150 lb cos5 cos70 sin 5 cos5 sin 70=°°+°−°°Pijk

(
)( )( )51.108 lb 13.0734 lb 140.418 lb=+ − ijk
Then
7.7053 1.47360 2.0646 lb in.
51.108 13.0734 140.418
C
=− − − ⋅
−
ijk
M

(
)( )( )233.91 lb in. 1187.48 lb in. 25.422 lb in.=⋅− ⋅−⋅ ijk
or ( )( )( )19.49lbft 99.0lbft 2.12lbft
C
=⋅−⋅−⋅Mijk W

27. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 26.






Have
/CACBA
=Mr F ×
where
( )( )( )
/
0.96 m 0.12 m 0.72 m
AC
=−+rijk
and
BABABA
F=F λ

()()()
() () ()
()
22 2
0.1 m 1.8 m 0.6 m
228 N
0.1 1.8 0.6 m

−+ −

=

++

ijk

( )( )( )12.0 N 216 N 72 N=− + − ijk
0.96 0.12 0.72 N m
12.0 216 72
C
∴ =− ⋅
−−
ijk
M
( )( )( )146.88 N m 60.480 N m 205.92 N m=− ⋅ + ⋅ + ⋅ijk
or ( )( )( )146.9 N m 60.5 N m 206 N m
C
=− ⋅ + ⋅ + ⋅Mijk W

28. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 27.













Have
CAD
Td=M
where
d
= Perpendicular distance from C to line AD
JJJG

with
/CACAD
=MrT
and
( )( )
/
3.1 m 1.2 m
AC
=+rjk

AD AD
AD
T
AD
=
T
JJJG


()
( )( )( )
()( )( )
222
2.4 m 3.1 m 1.2 m
369 N
2.4 m 3.1 m 1.2 m
AD
 −−
 
=
+− +−
ijk
T


( )( )( )216 N 279 108 N=−− iNj k
Then
03.11.2Nm
216 279 108
C
=⋅
−−
ijk
M


(
)( )259.2 N m 669.6 N m=⋅−⋅ jk
and ()( )
22
259.2 N m 669.6 N m
C
=⋅+−⋅M
718.02 N m
=⋅

(
)718.02 N m = 369 Nd∴⋅
or 1.946 m
d= W

29. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 28.













Have
OAC
d=MT
where
d
= Perpendicular distance from O to rope AC
with
/OAOAC
=×Mr T
and
()()
/
30 ft 3 ft
AO
=+rjk
( ) ( )62 lb cos10 62 lb sin10
AC
  =−°−°
  
Tij


( )( )61.058 lb 10.766 lb=− − ij
Then
0303lbft
61.058 10.766 0
O
=⋅
−−
ijk
M


(
)( )( )32.298 lb ft 183.174 lb ft lb ft= ⋅ − ⋅ + 1831.74 ⋅ijk
and ()( )( )
222
32.298 lb ft 183.174 lb ft 1831.74 lb ft
O
=⋅+−⋅+⋅M
1841.16 lb ft
=⋅

( )1841.16 lb ft = 62 lbd∴⋅
or 29.7 ft
d= W

30. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 29.













Have
OAB
Td=M
where
d
= Perpendicular distance from O to rope AB
with
/OAOAB
=×Mr T
and
()()
/
30 ft 3 ft
AO
=+rjk

AB AB
AB
T
AB
=
T
JJJG


()
()()()
()( )()
222
5ft 30ft 6ft
62 lb
5ft 30ft 6ft
 −+
 
=
+− +
ijk


()( )()10 lb 60 lb 12 lb=−+ ijk
Then
0303lbft
10 60 12
O
=⋅
−
ijk
M


(
)( )( )540lbft 30lbft 300lbft=⋅+⋅−⋅ ij k
and ()()( )
22 2
540lbft 30lbft 300lbft
O
=⋅+⋅+−⋅M
618.47 lb ft
=⋅

( )618.47 lb ft = 62 lbd∴⋅
or
9.98 ftd= W

31. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 30.













Have
CBD
Td=M
where
d
= Perpendicular distance from C to cable BD
with
//CBCBD
=×Mr T
and
()
/
2m
BC
=rj

BD BD
BD
T
BD
=
T
JJJG


()
()()()
()( )()
222
1m 2m 2m
900 N
1m 2m 2m
 −− +
 
=
−+− +
ijk


( )( )( )300 N 600 N 600 N=− − +ijk
Then
020Nm
300 600 600
C
=⋅
−−
ijk
M

(
)( )1200 N m 600 N m=⋅+⋅ ik
and ()()
22
1200 N m 600 N m
C
=⋅+⋅M
1341.64 N m
=⋅

( )1341.64 = 900 Nd∴
or 1.491 m
d= W

32. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Chapter 3, Solution 31.














Have
C
M Pd=
From the solution of problem 3.25
( )( )( )233.91 lb in. 1187.48 lb in. 25.422 lb in.
C
=⋅− ⋅−⋅Mijk

Then

()( )( )
222
233.91 1187.48 25.422
C
M=+−+−

1210.57 lb in.= ⋅

and
1210.57 lb.in.
150 lb
C
M
d
P
==

or
8.07 in.d= W

33. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



Chapter 3, Solution 32.







Have
||
DBA
Fd=M
where perpendicular distance from to line .dDAB
=

/DADBA
=Mr F ×

( )( )
/
0.12 m 0.72 m
AD
=− +rjk
( )( )( )( )
() () ()
()
22 2
0.1 m 1.8 m 0.6 m
228 N
0.1 1.8 0.6 m
BA BA BA
F
−+ −
==
++
ijk
F λ

( )( )()12.0 N 216 N 72 N=− + − ijk
0 0.12 0.72 N m
12.0 216 72
D
∴ =− ⋅
−−
ijk
M
( )( )( )146.88 N m 8.64 N m 1.44 N m=− ⋅ − ⋅ − ⋅ijk
and
()()()
222
| | 146.88 8.64 1.44 147.141 N m
D
= ++= ⋅M
( )147.141 N m 228 Nd∴⋅=
0.64536 m
d
=
or 0.645 m
d
= W

34. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 33.







Have | |
CBA
Fd=M
where perpendicular distance from to line .
dCAB
=

/CACBA
=Mr F ×

( )( )( )
/
0.96 m 0.12 m 0.72 m
AC
=−+rijk
( )( )()( )
() () ()
()
22 2
0.1 m 1.8 m 0.6
228 N
0.1 1.8 0.6 m
BA BA BA
F
−+ −
==
++
ijk
F λ

( )( )()12.0 N 216 N 72 N=− + − ijk
0.96 0.12 0.72 N m
12.0 216 72
C
∴ =− ⋅
−−
ijk
M
( )( )( )146.88 N m 60.48 N m 205.92 N m=− ⋅ − ⋅ + ⋅ij k
and
()()()
22 2
| | 146.88 60.48 205.92 260.07 N m
C
= ++ = ⋅M
( )260.07 N m 228 Nd∴⋅=
1.14064 m
d
=
or 1.141 m
d
= W

35. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 34.










(
a) Have
//
sin
CA AB CA
dr θ==× rλ
where
d
=Perpendicular distance from C to pipe AB
with
() () ( )
22 2
7432
74 32
AB
AB
+−
==
++−
AB i j k
λ
()
1
7432
33
=+−ij k
and ()()( )
/
14 ft 5 ft 22 ft
CA
L =− + + −
 
rij k

Then
/
1
74 32ft
33
14 5 22
AB C A
L
×= −
−−
ij k
rλ

( ) () ( ) ( ) () ( ){}
1
4 22 32 5 32 14 7 22 7 5 4 14 ft
33
LL =−++−−++
 
ijk
()( )
1
472 7602 91ft
33
LL=++−++

ijk
and ()( )()
2221
472 7602 91
33
dL L=++−++
For ()( ) ( )( )
2
min 2
1
() , 24 4 72 2 7 7 602 0
d 33
dd
dLL
L
=++−−+=


or
65 3926 0L−=
or 60.400 ftL=
But
greenhouse
LL> so 30.0 ft L= W
(b) with
()( )()
2221
30 ft, 4 30 72 7 30 602 91
33
Ld==×++−×++
or 13.51 ftd=
W
Note: with 60.4 ft,L=

()( )()
2221
4 60.4 72 7 60.4 602 91 11.29 ft
33
d=×++−×++=

36. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 35.



( )( )483 9 7⋅=−+ − ⋅ −−PQ i j k i j k
()()()()()()49 8 1 3 7=−+−+−−
23 =−
or 23⋅=−
PQ W

( )( )483 562⋅=− + − ⋅ − +PS i j k i j k
()()()()()()45 8 6 32=− + − +−
74 =−
or 74⋅=−
PS W

( )( )97562⋅= −− ⋅ − +QS i j k i j k
()()()()()()95 1 6 72=+−−+−

37
=
or 37⋅=
QS W

37. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



Chapter 3, Solution 36.





By definition
( )cosBCαβ⋅= −BC
where ( )( )cos sinBββ =+
 
Bij


( )()cos sinCαα =+
 
Cij

( )( )( )( ) ( )cos cos sin sin cosBC BC BCβ αβα αβ∴+=−
or
(
)cos cos sin sin cosβαβα αβ+=− (1)
By definition
( )cosBCαβ′⋅= +BC
where ( )( )cos sinββ′ =−
 
Bij

( )( )( )( ) ( )cos cos sin sin cosBC BC BCβ αβα αβ∴+−=+
or
(
)cos cos sin sin cosβαβα αβ−=+ (2)
Adding Equations (1) and (2),
( ) ( )2cos cos cos cosβααβ αβ=−++
or () ()
11
cos cos cos cos
22
αβαβ αβ=++− W

38. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 37.












First note:

( )( )
/
0.56 m 0.9 m
BA
=+rij
( )( )
/
0.9 m 0.48 m
CA
=−rjk
( )( )( )
/
0.52 m 0.9 m 0.36 m
DA
=− + +rijk

()()
22
/
0.56 m 0.9 m 1.06 m
BA
=+=r

()( )
22
/
0.9 m 0.48 m 1.02 m
CA
=+−=r

()()()
22 2
/
0.52 m 0.9 m 0.36 m 1.10 m
DA
=− + + =r
By definition
// //
cos
BA DA BA DA
θ⋅=rr rr
or
(
)( )()()0.56 0.9 0.52 0.9 0.36 1.06 1.10 cos θ+⋅−++ =ij ij k
()( )()()()()0.56 0.52 0.9 0.9 0 0.36 1.166cos θ−+ + =

cos 0.44494
θ=

63.6θ
= °W

39. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 38.










From the solution to problem 3.37
( )( )
//
1.02 m with 0.9 m 0.48 m
CA CA
==−rrij
( )( )( )
//
1.10 m with 0.52 m 0.9 m m
DA DA
==−++0.36rrijk
Now by definition

// //
cos
CA DA CA DA
θ⋅=rr rr
or
(
)( )()()0.9 0.48 0.52 0.9 0.36 1.02 1.10 cos θ−⋅−++ =jk ijk
( )()()( )()0 0.52 0.9 0.9 0.48 0.36 1.122cos θ−+ +− =

cos 0.56791
θ=
or 55.4
θ
= °W

40. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 39.






(
a) By definition
(1) (1) cos
BC EF
θ+=λλ
where

()()()
() ( ) ( )
22 2
32ft 9ft 24ft
32 9 24 ft
BC
−−
=
+− +−ij k
λ

()
1
32 9 24
41
=−−
ij k

()()()
()()()
222
14 ft 12 ft 12 ft
14 12 12 ft
EF
−− +
=
−+−+ijk
λ

()
1
766
11
=−−+
ijk
Therefore
( )( )32924 766
cos
41 11
θ
−− −−+
⋅=ij k ijk


(
)()()()()()()()32 7 9 6 24 6 41 11 cos θ−+− −+− =
cos 0.69623 θ=−
or 134.1
θ
= °W
(
b) By definition ()
()cos
EG EF
BC
TT θ=
( )( )110 lb 0.69623=−
76.585 lb =−
or
(
) 76.6 lb
EFBC
T =− W

41. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 40.






(
a) By definition
(1) (1) cos
BC EG
θ⋅=λλ
where

()()()
() ( ) ( )
22 2
32ft 9ft 24ft
32 9 24 ft
BC
−−
=
+− +−ij k
λ

()
1
32 9 24
41
=−−
ij k

()()()
() ( ) ( )
22 2
16 ft 12 ft 9.75
16 12 9.75 ft
EG
−+
=
+− +ijk
λ

()
1
16 12 9.75
22.25
=−+
ij k
Therefore
( )( )32 9 24 16 12 9.75
cos
41 22.25
θ
−− − +
⋅=ij k i j k


(
)()()()()()()( )32 16 9 12 24 9.75 41 22.25 cos θ+− − +− =

cos 0.42313
θ=
or 65.0
θ
= °W
(
b) By definition ()
()cos
EG EG
BC
TT θ=
( )( )178 lb 0.42313=
75.317 lb =
or
() 75.3 lb
EG
BC
T = W

42. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 41.













First locate point
B:

3.5
22 14d
=
or 5.5 m
d
=
(
a)
()()()
222
5.5 0.5 22 3 23 m
BA
d= + +− +− =

Locate point
D:

( )( )3.5 7.5sin 45 cos15 , 14 7.5cos 45 ,−−°°+ °



( )0 7.5sin 45 sin15 m+°°

or ( )8.6226 m, 19.3033 m, 1.37260 m−
Then

()()()
222
8.6226 5.5 19.3033 22 1.37260 0 m
BD
d=− + + − + −


4.3482 m
=
and
( )( )
()( )
6 22 3 3.1226 2.6967 1.37260
cos
23 4.3482
BA BD
ABD
BA BD
dd
dd
θ
−−⋅− − +⋅
==ijk i j k

0.36471
=
or 68.6
ABD
θ
= °W
(
b) (
) cos
BABAABDBD
TT θ=

(
)( )230 N 0.36471=
or
(
) 83.9 N
BABD
T = W

43. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 42.













First locate point B:

3.5
22 14d
=
or 5.5 m
d
=
(
a) Locate point D:

( )( )3.5 7.5sin 45 cos10 , 14 7.5cos45 ,−−°°+ °



( )0 7.5sin 45 sin10 m+°°


or ( )8.7227 m, 19.3033 m, 0.92091 m−

Then
( )( )( )5.2227 m 5.3033 m 0.92091 m
DC
=−−dijk
and

()()()
222
5.5 8.7227 22 19.3033 0 0.92091 m
DB
d=−+ + − +−


4.3019 m
=
and
( )( )
()()
3.2227 2.6967 0.92091 5.2227 5.3033 0.92091
cos
4.3019 7.5
DB DC
BDC
DB DC
dd
dd
θ
+− ⋅ −−⋅
==ij k ij k

0.104694 =
or 84.0
BDC
θ
= °W
(b) () ( )( )cos 250 N 0.104694
BD BD BDC
DC
TT θ==
or
(
) 26.2 N
BDDC
T = W

44. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 43.


Volume of parallelopiped is found using the mixed triple product
(
a)
( )Vol=⋅×PQ S

3
341
76 8in.
923
−
=− −
−−

( )
3
54 288 14 48 84 54 in.=− + + − + −

3
230 in.=
or
3
Volume 230 in.= W

(
b)
( )Vol=⋅×PQ S

3
574
625in.
48 9
−−
=−
−−

( )
3
90 140 192 200 378 32 in.=− + + + − −

3
32 in.=
or
3
Volume 32 in.= W

45. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 44.


For the vectors to all be in the same plane, the mixed triple product is zero.

( )0⋅×=PQ S

375
0214
86
y
S
−−
∴=− −
−

10 12 84 40
yy
SS0 = 18 + 224 − − + −
So that 22 286
y
S=

13
y
S
=
or 13.00
y
S= W

46. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 45.











Have
( )2.25 m
C
=rk

CE CE
CE
T
CE
=T
JJJG


()
( )( )( )
()()( )
22 2
0.90 m 1.50 m 2.25 m
1349 N
0.90 1.50 2.25 m
CE
 +−
 
=
++−
ijk
T


( )( )( )426 N 710 N 1065 N=+−ij k
Now
OCCE
=×MrT
002.25Nm
426 710 1065
=⋅
−
ij k


( )( )1597.5 N m 958.5 N m=− ⋅ + ⋅ij
1598 N m, 959 N m, 0
xyz
MMM∴=− ⋅ = ⋅ = W

47. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Chapter 3, Solution 46.










Have (
)( )0.90 m 1.50 m
E
=+rij

DE DE
DE
T
DE
=T
JJJG


()
( )( )( )
()()()
22 2
2.30 m 1.50 m 2.25 m
1349 N
2.30 1.50 2.25 m
−+ −

=
−+ +−
ijk


(
)( )( )874 N 570 N 855 N=− + −ijk
Now
OEDE
=×MrT

0.90 1.50 0 N m
874 570 855
= ⋅
−−
ijk


(
)( )( )1282.5 N m 769.5 N m 1824 N m=− ⋅ + ⋅ + ⋅ijk
1283 N m, 770 N m, 1824 N m
xyz
MMM∴=− ⋅ = ⋅ = ⋅ W

48. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 47.







Have
() ()
zBBACCD y y
 
=⋅ × + ×
  
Mkr T kr T

where
(
)48 lb ft
z
=− ⋅Mk

(
)() ()3ft
BC
y y
==rr j
()
( )()()
() () ()
222
4.5 ft 3 ft 9 ft
14 lb
4.5 3 9 ft
BA BA
BA
T
BA
 −+
 
==
+− +
ijk
T
JJJG


(
)()()6lb 4lb 12lb=−+ij k

()()()
() ( ) ( )
222
6ft 3ft 6ft
636ft
CD CD CD
CD
TT
CD
 −−
 
==
+− +−
ijk
T
JJJG


()22
3
CD
T
=−− ij k
Then
()()()()(){
}48lbft 3ft 6lb 4lb 12lb −⋅=⋅ × − +
 
kj i j k

()
()3ft 2 2
3
CD
T
 
+⋅ × − −  
 
kj ijk
or
48 18 2
CD
T−=−−

15.00 lb
CD
T= W

49. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 48.








Have
() ()
yBBACCD z z
 =⋅ × ×⋅ ×
  
Mjr T jr T
where 156 lb ft
y
=⋅M

(
)() ()()24 ft ; 6 ft
BC
z z
==rkrk

( )()()
() () ()
222
4.5 ft 3 ft 9 ft
4.5 3 9 ft
BA BA BA
BA
TT
BA
 −+
 
==
+− +
ijk
T
JJJG


()4.5 3 9
10.5
BA
T
=−+ ijk

()
()()()
() ( ) ()
222
6ft 3ft 9ft
7.5 lb
639ft
CD CD
CD
T
CD
 −+
 
==
+− +
ijk
T
JJJG


(
)( )()5lb 2.5lb 5lb=− −ijk
Then
()()
()156 lb ft 24 ft 4.5 3 9
10.5
BA
T
⋅=⋅ × −+

jk ijk

() ()( )(){
}6ft 5lb 2.5lb 5lb +⋅ × − −
 
jk i j k
or
108
156 30
10.5
BA
T=+
12.25 lb
BA
T= W

50. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 49.








Based on
(
)( ) ( )( )cos 0.225 m sin sin 0.225 m cos
x
MP P φ θφ θ =−
  (1)

(
)( )cos 0.125 m
y
MP φ=− (2)

(
)( )sin 0.125 m
z
MP φ=− (3)
By
(
)
()
( )( )
()()
Equation 3 sin 0.125
:
Equation 2 cos 0.125
z
y
PM
MPφ
φ−
=
−

or
4
tan 9.8658
23
φφ
−
=∴= °
−

or 9.87
φ
= °W
From Equation (2)

(
)( )23 N m cos9.8658 0.125 mP−⋅=− °
186.762 NP=
or 186.8 NP=
W
From Equation (1)

(
) ( )26 N m 186.726 N cos9.8658 0.225 m sinθ  ⋅= °
  


(
) ( )186.726 N sin 9.8658 0.225 m cosθ−°

or
0.98521sin 0.171341cos 0.61885
θ θ−=
Solving numerically,
48.1
θ
= °W

51. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 50.









Based on
(
)( ) ( )( )cos 0.225 m sin sin 0.225 m cos
x
MP P φ θφ θ =−
  (1)

(
)( )cos 0.125 m
y
MP φ=− (2)

(
)( )sin 0.125 m
z
MP φ=−
By
(
)
()
( )( )
()()
Equation 3 sin 0.125
:
Equation 2 cos 0.125
z
y
PM
MPφ
φ−
=
−

or
3.5
tan ; 9.9262
20
φφ
−
==°
−

From Equation (3):

(
)( )3.5 N m sin 9.9262 0.125 mP−⋅=− °
162.432 N P=
From Equation (1):

(
)( )( )162.432 N 0.225 m cos9.9262 sin 60 sin 9.9262 cos60
x
M=°°−°°
28.027 N m=⋅
or 28.0 N m
x
M=⋅ W

52. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 51.


First note:

BA BA
BA
T
BA
=
T
JJJG


()
() ( )()
() () ()
222
41.5 1 6
70 lb
41.5 1 6
BC
BC
L
L
+− ++−

=
+− ++−

ijk


()
( )
()
2
40.5 6
70 lb
52 0.5
BC
BC
L
L
+− −
=
+−ijk


()( )()4 ft 1.5 ft 12 ft
A
=+ −rijk
Have
OABA
=×MrT

()
()
2
70 lb
4ft 1.5ft 12ft
52 + 0.5
40.5 6
BC
BC
L
L
=−
−
−−
ij k

For the
i components:
()
() ()
2
70
763 lb ft 1.5 6 12 0.5 lb ft
52 + 0.5
BC
BC
L
L
−⋅= −+ − ⋅

−

or
()
2
10.9 52 0.5 3 12
BCBC
LL+− =+
or
() ()
22 2
10.9 52 0.5 9 72 144
BCBCBC
LLL

+− =+ +


or
2
25.19 190.81 6198.8225 0
BC BC
LL+−=
Then
()()( )
()
2
190.81 190.81 4 25.19 6198.8225
2 25.19
BC
L
−± − −
=
Taking the positive root 12.35 ft
BC
L= W

53. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 52.


First note:

BA BA
BA
T
BA
=
T
JJJG


()
() ( )()
() () ()
222
41.5 1 6
70 lb
41.5 1 6
BC
BC
L
L
+− ++−

=
+− ++−

ijk


()
( )
()
2
40.5 6
70 lb
52 0.5
BC
BC
L
L
+− −
=
+−ijk


()( )()4ft 1.5ft 12ft
A
=+ −rijk
Have
OABA
=×MrT

()
()
2
4ft 1.5ft 12ft
52 + 0.5
40.5 6
BA
BC
BC
T
L
L
=−
−
−−
ij k

For the
i components:
()
() ()
2
900 lb ft 1.5 6 12 0.5 lb ft
52 + 0.5
BA
BC
BC
T
L
L
−⋅= −+ − ⋅

−

or
()
()
2
300 1 4
52 + 0.5
BA
BC
BC
T
L
L
=+
−
(1)
For the
k components:
()
() ()
2
315lbft 40.5 1.54 lbft
52 + 0.5
BA
BC
BC
T
L
L
−⋅= −− ⋅

−

or
()
()
2
4
315 1
52 + 0.5
BA
BC
BC
T
L
L
=+
−
(2)
Then,
(
)
() ()
1 300 1 4
231541
BC
BC
L
L
+
⇒=
+

or
59
ft
4
BC
L=
14.75 ft
BC
L= W

54. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 53.











Have
()/AD AD B A BH
M=⋅×λrT
where
(
)( )
()( )
22
0.8 m 0.6 m
0.8 0.6
0.8 m 0.6 m
AD
−
==−
+−ik
λ ik


(
)
/
0.4 m
BA
=ri
()
( )( )( )
() () ( )
22 2
0.3 m 0.6 m 0.6 m
1125 N
0.3 0.6 0.6 m
BH BH
BH
T
BH
 +−
 
==
++−
ijk
T
JJJJG

Then

0.8 0 0.6
0.4 0 0 180 N m
375 750 750
AD
M
−
==−⋅
−

or 180.0 N m
AD
M=− ⋅ W

55. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 54.











Have
()/AD AD B A BG
M=⋅×λrT
where
(
)( )0.8 m 0.6 m
AD
=−λ ik

(
)
/
0.4 m
BA
=ri
()
( )()( )
()()( )
22 2
0.4 m 0.74 0.32 m
1125 N
0.4 m 0.74 m 0.32 m
BG BG
BG
T
BG
 −+−
 
==
−+ +−
ij k
T
JJJG


( )( )( )500 N 925 N 400 N=− + −ijk
Then

0.8 0 0.6
0.4 0 0
500 925 400
AD
M
−
=
−−

or 222 N m
AD
M=− ⋅ W

56. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 55.


Have

( )/AD AD E A EF
M=⋅ ×λrF
where
AD
AD
AD
=λ
JJJG


( )( )
()()
22
7.2 m 0.9 m
7.2 m 0.9 m
AD
+
=
+ij
λ

0.99228 0.124035
= +ij
( )( )
/
2.1 m 0.9 m
EA
=−rij
()
( )( )( )
()()()
22 2
0.3 m 1.2 m 2.4 m
24.3 kN
0.3 m 1.2 m 2.4 m
EF EF
EF
F
EF
 ++
 
==
++
ijk
F
JJJG


(
)( )( )2.7 kN 10.8 kN 21.6 kN=+ + ijk
Then

0.99228 0.124035 0
2.1 0.9 0 kN m
2.7 10.8 21.6
AD
M=− ⋅
19.2899 5.6262=− −
24.916 kN m=− ⋅
or 24.9 kN m
AD
M=− ⋅ W

57. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 56.


Have
( )/AD AD G A EF
M=⋅ ×λrE
Where
( )( )
()()
22
7.2 m 0.9 m
7.2 m 0.9 m
AD
+
=
+ij
λ

0.99228 0.124035
= +ij
()( )
/
6m 1.8m
GA
=−rij
()
( )( )( )
( )()()
222
1.2m 2.4m 2.4m
21.3 kN
1.2m 2.4m 2.4m
GH GH
GH
F
GH
 −+ +
 
==
−+ +
ijk
F
JJJJG


(
)( )( )7.1 kN 14.2 kN 14.2 kN=− + +ijk
Then

0.99228 0.124035 0
61.80kNm
7.1 14.2 14.2
AD
M=− ⋅
−

25.363 10.5678=− −
35.931 kN m=− ⋅
or 35.9 kN m
AD
M=− ⋅ W

58. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 57.





Have
( )/OA OA C O
M=⋅ ×rPλλλλ
where
From triangle OBC

()
2
x
a
OA=

() ()
1
tan30
2323
zx
aa
OA OA

=°==


Since
() () () ( )
222 2
zxy
OA OA OA OA=++
or
()
22
22
2 23
y
aa
aOA

=+ +
 
 


()
22
2
2

412 3
y
aa
OA a a∴=−−=

Then
/
2
23 23
AO
aa
a=+ +
rijk
and
121
23 23
OA
=+ +ij kλλλλ

BC
P=Pλλλλ

()( )
()
sin30 cos30aa
P
a
°− °
=
ik

()3
2
P=−ik

/CO
a=ri

()
121
23 23

10 0 2
10 3
OA
P
Ma

∴=


−

()()
2
13
23
aP
=− −




2
aP
=
2
OA
aP
M=

59. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 58.






(a) For edge OA to be perpendicular to edge BC,
0OA BC=
uuur uuur
⋅⋅⋅⋅
where
From triangle OBC

()
2
x
a
OA=

() ()
1
tan30
2323
zx
aa
OA OA

=°==




()
2 23
y
aa
OA OA

∴= + +

 
i
j k
uuur

and
(
)()sin30 cos30BC a a=°−° ik
uuur


3
22
aa
=−ik

()3
2
a=−
ik
Then () ()30
22 23
y
aaa
OA
 
++ −= 

ij kik ⋅⋅⋅⋅

or
()()
22
00
44
y
aa
OA
+−=
0OA BC∴=
uuur uuur
⋅⋅⋅⋅

so that
OA
uuur
is perpendicular to.
BC
uuur


(b) Have
,
OA
MPd= with P acting along BC and d the
perpendicular distance from
to .OA BC
uuur uuur

From the results of Problem 3.57,
2
OA
Pa
M=

2
Pa
Pd∴=
or
2
a
d=



60. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 59.


Have
( )/DI DI F I EF
M=⋅ ×λrT
where
( )( )
()( )
22
4.8 ft 1.2 ft
4.8 ft 1.2 ft
DI
DI
DI
−
==
+−
ij
λ
JJJG

0.97014 0.24254 = −ij
( )
/
16.2 ft
FI
=rk


EF EF
EF
T
EF
=T
JJJG
()
( )( )( )
()( )( )
222
3.6 ft 10.8 ft 16.2 ft
29.7 lb
3.6 ft 10.8 ft 16.2 ft
 −+
 
=
+− +
ijk


( )( )( )5.4 lb 16.2 lb 24.3 lb=− +ijk
Then

0.97014 0.24254 0
0016.2lbft
5.4 16.2 24.3
DI
M
−
=⋅
−

21.217 254.60=− +
233.39 lb ft=⋅
or 233 lb ft
DI
M=⋅ W

61. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 60.


Have
( )/DI DI G I EG
M=⋅ ×λrT
where
( )( )
()( )
22
4.8 ft 1.2 ft
4.8 ft 1.2 ft
DI
DI
DI
−
==
+−
ij
λ
JJJG

0.97014 0.24254 = −ij
( )
/
35.1 ft
GI
=−rk

EG EG
EG
T
EG
=T
JJJG
()
( )( )( )
()( )( )
222
3.6 ft 10.8 ft 35.1 ft
24.6 lb
3.6 ft 10.8 ft 35.1 ft
 −−
 
=
+− +−
ijk


( )( )( )2.4lb 7.2lb 23.4lb=−− ij k
Then

0.97014 0.24254 0
0035.1lbft
2.4 7.2 23.4
DI
M
−
=−⋅
−−


20.432 245.17=−
224.74 lb ft
=− ⋅
or
225 lb ft
DI
M=− ⋅ W

62. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Chapter 3, Solution 61.



First note that
111
F=F
λ and
222
F=F λ
Let
12
moment of M= Fabout the line of action of
1
M
and
21
moment of M= Fabout the line of action of
2
M
Now, by definition
( ) ( )11/ 2 1/ 22BA BA
M F=⋅ =⋅rF rλ×λ×λ
( ) ( )22/ 1 2/ 11AB AB
M F=⋅ =⋅rF rλ×λ×λ
Since
12
FFF
== and
//AB BA
=−rr
( )11/ 2 BA
M F=⋅rλ×λ
( )22 / 1 BA
M F=⋅−rλ×λ
Using Equation (3.39)
( ) ( )1/ 2 2 / 1BA BA
⋅=⋅−rrλ ×λ λ ×λ
so that ( )21/ 2 BA
M F=⋅rλ×λ

12 21

M M∴= W

63. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 62.


From the solution of Problem 3.53:
0.8 0.6
AD
=−ikλ

( )( )( )375 N 750 N 750 N ; 1125 N
BH BH
T=+− =Tijk
180 N m
AD
M
=−⋅
Only the perpendicular component of
BH
T contributes to the moment of
BH
T about line AD. The
parallel component of
BH
T will be used to find the perpendicular component.
Have
(
)
ParallelBHADBH
=⋅TT λ

[
]( )( )( )0.8 0.6 375 N 750 N 750 N =− ⋅ + −
 
ik i j k


(
)300 450 N=+
750 N
=
Since
(
) ()
Perpendicular Parallel
BH BH BH
=+TT T
Then
() ()()
22
Perpendicular ParallelBH BH BH
T =−TT

()()
22
1125 N 750 N=−
838.53 N
=
and
()
Perpendicular
AD BH
d=MT

(
)180 N m 838.53 Nd⋅=
0.21466 m
d=
or 215 mm
d= W

64. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 63.


From the solution of Problem 3.54:
0.8 0.6
AD
=−ikλ

( )( )( )500 N 925 N 400 N
1125 N
BG
BG
T
=− + −
=
Tijk

222 N m
AD
M
=−⋅
Only the perpendicular component of
BG
T contributes to the moment of
BG
T about line AD. The
parallel component of
BG
T will be used to find the perpendicular component.
Have
()
Parallel
BGADBG
=⋅TT λ

[
]( )( )( )0.8 0.6 500 N 925 N 400 N =− ⋅− + −
 
ik i j k


(
)400 240 N=− +
160 N
=−
Since
()
()
Perpendicular Parallel
BG BG BG
=+TT T
Then
() ()()
22
Perpendicular Parallel
BG BG BG
TTT =−

()( )
22
1125 N 160 N=−−
1113.56 N
=
and
()
Perpendicular
AD BG
M Td=

(
)222 N m 1113.56 Nd⋅=
0.199361 m
d=
or 199.4 mm
d= W

65. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 64.


From the solution of Problem 3.59:
0.97014 0.24254
DI
=− ijλ

( )( )( )5.4lb 16.2lb 24.3lb
29.7 lb
EF
EF
T
=− +
=
Tijk

233.39 lb ft
DI
M
= ⋅
Only the perpendicular component of
EF
T contributes to the moment of
EF
T about line DI. The
parallel component of
EF
T will be used to find the perpendicular component.
Have
(
)
ParallelEFDIEF
=⋅TT λ

[
]( )( )( )0.97014 0.24254 5.4 lb 16.2 lb 24.3 lb =−⋅−+
 
iji j k


(
)5.2388 3.9291=+
9.1679 lb
=
Since
(
) ()
Perpendicular Parallel
EF EF EF
=+TT T
Then
() ()()
22
Perpendicular Parallel
EF EF EF
T =−TT

()( )
22
29.7 9.1679=−
28.250 lb
=
and
()
Perpendicular
DI EF
d=MT

(
)233.39 lb ft 28.250 lbd⋅=
8.2616 ft
d=
or 8.26 ft
d= W

66. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 65.


From the solution of Problem 3.60:
0.97014 0.24254
DI
=− ijλ

( )( )( )2.4 lb 7.2 lb 23.4 lb
24.6 lb
EG
EG
T
=−−
=
Tijk

224.74 lb ft
DI
M
=−⋅
Only the perpendicular component of
EG
T contributes to the moment of
EG
T about line DI. The
parallel component of
EG
T will be used to find the perpendicular component.
Have
()
Parallel
EGDIEG
=⋅TT λ

[
]( )( )( )0.97014 0.24254 2.4 lb 7.2 lb 23.4 lb =−⋅−−
 
ij ij k


(
)2.3283 1.74629=+
4.0746 lb
=
Since
()
()
Perpendicular Parallel
EG EG EG
T=+TT
Then
() ()()
22
Perpendicular Parallel
EG EG EG
TTT =−

()( )
22
24.6 4.0746=−
24.260 lb
=
and
()
Perpendicular
DI EG
M Td=

(
)224.74 lb ft 24.260 lbd⋅=
9.2638 ft
d=
or 9.26 ft
d= W

67. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 66.


From the solution of Prob. 3.55:
0.99228 0.124035
AD
=+ ijλ

( )( )( )2.7 kN 10.8 kN 21.6 kN
EF
=+ +Fijk
24.3 kN
EF
F=
24.916 kN m
AD
M
=−⋅
Only the perpendicular component of
EF
F contributes to the moment of
EF
F about edge AD. The
parallel component of
EF
F will be used to find the perpendicular component.
Have
(
)
ParallelEFADEF
=⋅FF λ

[
]( )( )( )0.99228 0.124035 2.7 kN 10.8 kN 21.6 kN =+ ⋅ + +
 
iji j k
4.0187 kN
=
Since
(
) ()
Perpendicular Parallel
EF EF EF
=+FF F
Then
() ()()
22
Perpendicular Parallel
EF EF EF
FFF =−

()( )
22
24.3 4.0187=−
23.965 kN
=
and
()
Perpendicular
AD EF
M Fd=

(
)24.916 kN m 23.965 kNd⋅=
1.039683m
d= or 1.040 m d= W

68. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 67.


From the solution of Prob. 3.56:
0.99228 0.124035
AD
=+ ijλ

(
)( )( )7.1 kN 14.2 kN 14.2 kN
GH
=− + +Fijk
21.3 kN
GH
F=
35.931 kN m
AD
M=− ⋅
Only the perpendicular component of
GH
F contributes to the moment of
GH
F about edge AD. The
parallel component of
GH
F will be used to find the perpendicular component.
Have
()
Parallel
GH AD GH
F =⋅Fλ

(
)( )( )( )0.99228 0.124035 7.1 kN 14.2 kN 14.2 kN =+ ⋅−+ +
 
ij i jk
5.2839 kN
=−
Since
()
()
Perpendicular Parallel
GH GH GH
=+FF F
Then
() ()()
22
Perpendicular Parallel
GH GH GH
FFF =−

()( )
22
21.3 5.2839=−
20.634 kN
=
and
()
Perpendicular
AD GH
M Fd=

(
)35.931 kN m 20.634 kNd⋅=
1.741349m
d= or 1.741 m d= W

69. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 68.


(a) Have
111
M dF=
Where
11
0.6 m and 40 NdF==

(
)( )
1
0.6 m 40 NM∴=
or
1
24.0 N mM
= ⋅W
(
b) Have
Total 1 2
=+MMM

( )( )( )8 N m 24.0 N m 0.820 m cos 24 Nα⋅= ⋅−
cos 0.81301α∴=
or 35.6
α
= °W
(
c) Have
12
0+=MM

(
)
2
24 N m 24 N 0d⋅− =
or
2
1.000 md= W

70. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 69.


( a)
MFd=
( )12 N m 0.45 mF⋅=
or 26.7 N
F= W

(
b)
MFd=
( )12 N m 0.24 mF⋅=
or 50.0 N
F= W

(
c)
MFd=

()()
22
Where 0.45 m 0.24md=+
0.51 m =

( )12 N m 0.51 m=F⋅
or 23.5 N
F= W

71. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 70.


(a) Note when 8 in.,a
=
/CF
r is perpendicular to the inclined 10 lb forces.
Have
d
M F=Σ()

() ()()
()10 lb 8 in. + 2 1 in. 10 lb 2 2 2 1 in.aa  =− + − +
  

For 8 in.,
a=

(
)( )10 lb 18 in. 24.627 in.M=− +
426.27 lb in.
=− ⋅
or 426 lb in.
M
= ⋅ W
(
b) Have 480 lb in.
= ⋅M
Also ( )
d
M MF=Σ + ()
= Moment of couple due to horizontal forces at
A and D +
Moment of force-couple systems at
C and F about C.

Then
() ( ) ()480 lb in. 10 lb 8 in. 2 1 in. 8 in. 2
CFX y
aMMFaFa  −⋅=− ++ +++++
  

Where
(
)()10 lb 1 in. 10 lb in.
C
M=− =− ⋅
10 lb in.
FC
MM==−⋅

10
lb
2
x
F
−
=

10
lb
2
y
F
−
=
( )480 lb 10 in. 10 lb in. 10 lb in.in. 10 lba∴− + − ⋅ − ⋅⋅=−
() ()
10 lb 10 lb
8in. 2
22
aa−+−

303.43 = 31.213
a
or 9.72in. a= W

72. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 71.


(a) Have
d
M F=Σ()

(
)( )( )( )9 lb 13.8 in. 2.5 lb 15.2 in.=−

(
)86.2 lb in.=⋅
86.2 lb in.= ⋅M W
(
b) Have 86.2 lb in. MTd
==⋅
For
T to be a minimum, d must be maximum.

min
T∴ must be perpendicular to line AC.

15.2 in.
tan
11.4 in.
θ=
53.130
θ= °
or 53.1θ= ° W
(
c) Have
min max
MTd= Where 86.2 lb in.M= ⋅

()()
()
22
max
15.2 in. 11.4 in. 2 1.2 in.d=++
21.4 in.
=

(
)
min
86.2 lb in. 21.4 in.T∴⋅=

min
4.0280 lbT=
or
min
4.03 lbT= W

73. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 72.


Based on
12
=+MM M

( )
1
18 N m=⋅Mk
( )
2
7.5 N m=⋅Mi
( )( )7.5 N m 18 N mi∴= ⋅ + ⋅Mk
and
()()
22
7.5Nm 18NmM=⋅+⋅
19.5 N m = ⋅
or 19.50 N m
M=⋅ W
With
( )( )7.5 N m 18 N m
19.5 N mM
⋅+ ⋅
==
⋅
ikM
λ

512
13 13
=+ik
Then
5
cos 67.380
13
xx
θθ
= ∴= °
cos 0 90
yy
θθ
= ∴=°

12
cos 22.620
13
zz
θθ
= ∴= °
or 67.4 , 90.0 , 22.6
xyz
θθθ
= °=°=° W

74. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 73.


Have
12
=+MM M
Where
1/CB IC
=×Mr P

( )( )
/
38.4 in. 16 in.
CB
=−rij
( )25 lb
IC
=−Pk

1
38.4 16 0 lb in.
0025
∴ =− ⋅
−
ijk
M
( )( )400 lb in. 960 lb in.=⋅+⋅ ij
and
2/DA ZE
=×Mr P

(
)( )
/
8in. 22in.
DA
=−rjk
()
( )( )
()()
22
19.2 in. 22 in.
36.5 lb
19.2 in. 22 in.
ZE ZE
ED
P
ED
 −+
 
==
−+
ik
P
JJJG


(
)( )24 lb 27.5 lb=− + ik

2
08 22lbin.
24 0 27.5∴= − ⋅
−
ijk
M

(
)( )( )
2
220 lb in. 528 lb in. 192 lb in.=⋅+⋅+⋅Mijk
and
12
=+MM M

(
)( ) ( )( )( )400 lb in. 960 lb in. 220 lb in. 528 lb in. 192 lb in. =⋅+⋅+⋅+⋅+⋅
 
ij ijk

(
)( )( )620 lb in. 1488 lb in. 192 lb in.=⋅+ ⋅+⋅ ijk

continued

75. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



()( )()
222
620 1488 192 lb in.M

=++⋅



1623.39 lb in.
=⋅
or 1.623 kip in.M=⋅ W

( )( )( )620 lb in. 1488 lb in. 192 lb in.
1623.39 lb in.M
⋅+ ⋅+ ⋅
==
⋅
ijkM
λ

0.38192 0.91660 0.118271
=++ ij k
cos 0.38192
x
θ= or 67.5
x
θ
= °W
cos 0.91660
y
θ= or 23.6
y
θ
= °W
cos 0.118271
z
θ= or
z
83.2θ
= °W

76. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 74.


Have
12
=+MM M
Where
1/
EDD
=×Mr F

(
)( )0.7 m 80 N=− × kj

(
)56.0 N m=⋅ i
And
2/GF B
=×Mr F
Now
( )()()
222
0.300 m 0.540 m 0.350 m
BF
d=− + +
0.710
= m
Then
B BF B
Fλ=F

(
)( )( )
()
0.300 m 0.540 m 0.350 m
71 N
0.710 m
−+ +
=
ijk


(
)( )( )30 N 54 N 35 N=− + +ijk

∴ (
) ( )( )( )
2
0.54 m 30 N 54 N 35 N=×−++

Mjijk

(
)( )18.90 N m 16.20 N m=⋅+⋅ ik
Finally
(
)( )( )56.0 N m 18.90 N m 16.20 N m =⋅+ ⋅+ ⋅
 
Mi i k


(
)( )74.9 N m 16.20 N m=⋅+ ⋅ ik
and
()( )
22
74.9 N m 16.20 N mM=⋅+⋅
76.632 N m
=⋅ or 76.6 N m =⋅M W

74.9 0 16.20
cos cos cos
76.632 76.632 76.632
xyz
θθθ===
or 12.20 90.0 77.8
xyz
θθθ
= °=° =° W

77. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 75.


Have (
)
12 P
=+ +MMM M
From the solution to Problem 3.74

(
)( )( )
12
74.9 N m 16.20 Nm+= ⋅+ ⋅MM i k
Now
/PDEE
=×Mr P

(
)( ) ( )0.54 m 0.70 m 90=+ ×

jkNi

(
)( )63.0 N m 48.6 N m=⋅−⋅ jk

∴ (
)( )74.9 16.20 63.0 48.6=+ + −Mik jk

(
)( )( )74.9 N m 63.0 N m 32.4 N m=⋅+⋅−⋅ ijk
and
()()( )
22 2
74.9 N m 63.0 N m 32.4 N mM=⋅+⋅+−⋅
103.096 N m
=⋅
or 103.1 N mM
=⋅ W
and
74.9 63.0 32.4
cos cos cos
103.096 103.096 103.096
xyz
θθθ
−
===
or 43.4 52.3 108.3
xyz
θθθ
= °=°=° W

78. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Chapter 3, Solution 76.


Have
12 P
=+ +MM M M
From Problem 3.73 solution:

()
( )
1
400 lb 960 lbin. in.=+ ⋅⋅Mij

()
( )( )
2
220 lb 528 lb 192 lbin. in. in.=++ ⋅⋅⋅Mijk
Now
/PEAE
=×Mr P

(
)()( )
/
19.2 in. 8 in. 44 in.
EA
=+−rijk

(
)52.5 lb
E
=Pj
Therefore
19.2 8 44
052.50
P
=−
ijk
M

(
)( )2310 lb. in. 1008 lb. in.=+ ik
and
12 P
=++MM M M

()
( )( )[] in.lb10081925289602310220400 ⋅++++++= kji

(
)( )( )2930 lb in. 1488 lb in. 1200 lb in.=⋅+⋅+⋅ ijk
()()()
222
2930 1488 1200=++M
3498.4 lb in. =⋅
or
3.50 kip in.=⋅M W
continued

79. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.




2930 1488 1200
3498.4
+ +
==
Mijk
M
λ

0.83753 0.42534 0.34301
=++ ijk
cos 0.83753
x
θ=
or 33.1
x
θ
= °W
cos 0.42534
y
θ=
or 64.8
y
θ
= °W
cos 0.34301
z
θ=
or 69.9
z
θ
= °W

80. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 77.








Have
123
=++MMM M
Where
(
) ( )
1
1.2 lb ft cos25 1.2 lb ft sin 25=− ⋅ ° − ⋅ °Mjk

(
)
2
1.3 lb ft=− ⋅Mj

(
) ( )
3
1.4 lb ft cos20 1.4 lb ft sin 20=− ⋅ ° + ⋅ °Mjk
∴ ( )( )1.08757 1.3 1.31557 0.507142 0.478828=− − − +− +Mjk

(
)( )3.7031 lb ft 0.028314 lb ft=− ⋅ − ⋅ jk
and ()( )
22
3.7031 0.028314 3.7032 lb ft=− +− = ⋅M
or 3.70 lb ft
=⋅M W

3.7031 0.028314
3.7032
− −
==
Mjk
M
λ

0.99997 0.0076458
=−− jk
cos 0
x
θ=
or 90
x
θ
=°W
cos 0.99997
y
θ=−
or 179.6
y
θ
= °W
cos 0.0076458
z
θ=−
or 90.4
z
θ
= °W

81. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 78.











(
a) :
B
=FP 160.0 N
B
F∴=
50.0°W
cos10
BBA
MrP=− °

(
)( )0.355 m 160 N cos10=− °
55.937 N m
=− ⋅
or 55.9 N m
B
M
= ⋅

W
(b) :
C
=FP 160.0 N
C
F∴=
50.0°W

(
)
CBCBB
MMrF
⊥
=−
sin 55
BCBB
MrF=− °

(
)( )55.937 N m m 160 N sin 55=− ⋅ − 0.305 °
or 95.9 N m
C
=⋅M

W

82. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 79.








(
a) : 135 N
B
Σ=FF
or 135 N
B
=F

W
:
B B
M PdΣ=M

(
)( )135 N 0.125 m=
16.875 N m
=⋅
or 16.88 N m
B
=⋅M

W
(
b) :
BBC
M FdΣ=M

(
)16.875 N m = 0.075 m
C
F⋅
225 N
C
F=
or 225 N
C
=F

W
:0
B C
FFΣ=−+F
225 N
BC
FF==
or 225 N
B
=F

W

83. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 80.






(a) Based on
:700 N
C
FP PΣ==

or 700 N
C
=P
60°W
:
CC xCyyCx
M MPdPdΣ=−+
where
()700 N cos60 350 N
x
P=°=
()700 N sin60 606.22 N
y
P=°=
1.6 m
Cx
d=
1.1 m
Cy
d=
()()( )() 350 N 1.1 m 606.22 N 1.6 m
C
M∴=− +
385 N m 969.95 N m=− ⋅ + ⋅
584.95 N m =⋅
or 585 N m
C
=⋅M

W

(b) Based on : cos60
xDx
FP PΣ=°

()700 N cos60=°

350 N=
()()(): cos60
DDABDB
MP d PdΣ°=
() ()()700 N cos60 0.6 m 2.4 m
B
P °=

87.5 N
B
P=
or 87.5 N
B
=P

W

84. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.







:sin60
yBDy
FP PPΣ°=+
()700 N sin 60 87.5 N
Dy
P°= +
518.72 N
Dy
P=

() ()
22
DDxDy
PP P=+

()( )
22
350 518.72 625.76 N=+ =
11 518.72
tan tan 55.991
350Dy
Dx
P
P
θ
−−
 
== =° 


or 626 N
D
P=
56.0°W

85. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 81.







: 2.8cos65 cos cos
xAC
FFF
θ θΣ°=+
( )cos
AC
FF θ=+ (1)
:2.8sin65 sin sin
yAC
FFF
θ θΣ°=+
( )sin
AC
FF θ=+ (2)
Then
(2)
tan 65 tan
(1)
θ⇒°=
or
65.0
θ= °
( )( ) ( )(): 27 m 2.8 kN sin 65 72 m sin 65
AC
MFΣ°=°
or 1.050 kN
C
F=
From Equation (1): 2.8 kN 1.050 kN
A
F=+
or 1.750 kN
A
F=
1.750 kN
A
∴=F
65.0°W
1.050 kN
C
=F 65.0°W

86. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 82.


Based on

( ): 54 lb cos30 cos cos
xBC
FFF α αΣ− °=− −
( ) ( )cos 54 lb cos30
BC
FF α+ =° (1)
( ): 54 lb sin 30 sin sin
yBC
FFF α αΣ°=+
or ( ) ( )sin 54 lb sin 30
BC
FF α+ =° (2)
From
()
()
2
:tan tan30
1
Eq
Eq
α
= °
30 α∴=°
Based on ( )( )( )( )( ): 54 lb cos 30 20 10 in. cos10 24 in.
CB
MF Σ°−°=°

22.5 lb
B
F∴=
or 22.5 lb
B
=F
30°W
From Eq. (1),
()
()22.5 cos30 54 cos30
C
F+°= °
31.5 lb
C
F=
or 31.5 lb
C
=F
30°W

87. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 83.


( a) Based on

( ):54lbcos30 cos30
xC
FFΣ −°=−°
54 lb
C
F∴=
or 54.0 lb
C
=F
30°W
( ) ( ): 54 lb cos10 10 in.
CC
M MΣ°=

531.80 lb in.
C
M
∴ =⋅
or 532 lb in.
C
= ⋅M W
( b) Based on
( ):54lbsin30 sin
yB
FF αΣ°=
or sin 27
B
Fα= (1)

( ) ( ): 531.80 lb in. lb cos10 24 in.
B
M Σ⋅−54°

( )24 in. cos20
C
F =−°
 

33.012 lb
C
F=
or 33.0 lb
C
=F
W
And
(
): 54 lb cos30 33.012 lb cos
xB
FF αΣ− °=− −
cos 13.7534
B
Fα= (2)
From
(
)
()
1 27
: tan 63.006
2 13.7534
Eq
Eq
αα
= ∴= °
From Eq. (1),
()
27
30.301 lb
sin 63.006
B
F==
°

or 30.3 lb
B
=F
63.0°W

88. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Σ
Chapter 3, Solution 84.









(a) Have :
yCDE
FFFFFΣ++=
200lb 150 lb 150 lbF=− + −
200 lbF=−
or
200 lb=F
Σ
Have ()():4.5ft6ft0
GC D
MFd FΣ−−=

()( )()()200 lb 4.5 ft 150 lb 6 ft 0d−− =

9ftd=
or
9.00 ftd= Σ
(b) Changing directions of the two 150-lb forces only changes the sign of
the couple.

200 lbF∴=−
or
200 lb=F
Σ
And ()():4.5ft6ft0
GC D
MFd FΣ−+=

()( )()()200 lb 4.5 ft 150 lb 6 ft 0d −+ =

0d=
or
0d=Σ

89. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 85.


(a)






(b)









(c)





Based on :
z
FΣ
200N 200N 240N
A
F−+ + =
240 N
A
F=
or
( )240 N
A
=Fk W
Based on :
A
MΣ
( )( )( )( )200 N 0.7 m 200 N 0.2 m
A
M−=
100 N m
A
M
= ⋅
or ( )100.0 N m
A
=⋅Mj W
Based on :
z
FΣ
200N 200N 240N F
− ++=
240 N F=
or ( )240 N=Fk W
Based on :
A
MΣ
( )()100 N m 240 N x⋅=
0.41667 mx=
or 0.417 m x= From A along AB
W
Based on :
B
MΣ
( )( )( )( )()()200 N 0.3 m 200 N 0.8 m 1 m 0PR−+ −=
100 N P=
or 100.0 N P=
W

90. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 86.




Let
R be the single equivalent force...
:
AC
Σ=+FRFF

(
)( )( )( )260 N cos10 sin10 320 N cos8 sin8=°−°+−°−°ik ik

(
)( )60.836 N 89.684 N=− − ik
or
(
)( )60.8 N 89.7 N=− −Rik W
:cos8
AADxACC
MrRrFΣ=°

(
)( )( )60.836 N 0.690 m 320 N cos8
AD
r = °
3.5941 m
AD
r=

∴R Would have to be applied 3.59 m to the right of A W
on an extension of handle ABC.

91. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 87.



(a) Have :
B CD A
Σ++=FFFFF
Since
B D
=−FF
22 lb
AC
∴== FF
20°
or 22.0 lb
A
=F
20°W
Have () () ():
ABTCTDTA
M Fr Fr Fr MΣ−−+=

(
) ()( ) ()28 lb sin15 8 in. 22 lb sin 25 8 in.−°− °


(
) ()28 lb sin 45 8 in.
A
M+°=

26.036 lb in.
A
M
= ⋅
or 26.0 lb in.
A
=⋅M
W
(b) Have :
AE
Σ=FFF
or 22.0 lb
E
=F
20°W

[ ](): cos 20
AE
M FaΣ=°M
( ) ()26.036 lb in. 22 lb cos 20a∴⋅=°

1.25941 in. a
=
or 1.259 in. BelowaA=
W

92. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 88.



(a) Let
R be the single equivalent force. Then

(
)120 N=Rk 120 N R= W
( )( )( ) ( )( ): 120 N 0.165 m 90 N cos15 0.201 m 90 N sin15
B
MaΣ− =− °+ °
0.080516 m a=
∴The line of action is
201
mm 80.516 mm 19.984 mm
2
y=− =
or 19.98 mmy=
W
(b)
( )( )( )( ) ( )( ): 0.201 0.040 m 120 N 0.165 m 90 N cos 0.201 m 90 N sin
B
M θ θΣ−− =− +

or cos 1.21818sin 1.30101
θ θ−=
or
()
22
cos 1.30101 1.21818sin
θ θ=+
or
22
1 sin 1.69263 3.1697sin 1.48396sin
θ θθ−= + +
or
2
2.48396sin 3.1697sin 0.69263 0θθ
+ +=
Then
()( )( )
()
2
3.1697 3.1697 4 2.48396 0.69263
sin
2 2.48396
θ
−± −
=
or 16.26 and 85.0
θθ
=−° =−° W

93. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



Chapter 3, Solution 89.


(a) First note that F = P and that F must be equivalent to (P, MD) at point D,
Where 57.6 N m
D
= ⋅M
For
(
)
min
FF= F must act as far from D as possible
Point of application is at pointB∴ W
(b) For
(
)
min
F F must be perpendicular to BD
Now
()( )
22
630 mm 160 mm
DB
d=+−
650 mm =

63
tan
16
α=
75.7
α= °
Then
DDB
M dF=
( )57.6 N m = 0.650 mF⋅
88.6 N
F
=
or 88.6 N=
F
75.7°W

94. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 90.




Have
(
):250kNΣ− =FjF
or ( )250 kN=−Fj W
Also have :
GP
Σ×=MrPM

0.030 0 0.060 kN m =
02500
−⋅
−
ijk
M


( )( )15 kN m 7.5 kN m∴= ⋅ + ⋅Mi k
or ( )( )15.00 kN m 7.50 kN m=⋅+⋅Mik W

95. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Chapter 3, Solution 91.













Have
:
AB
Σ =FT F
where
AB AB
AB
T
AB
=
T
JJJG


()
()()()
222
2.25 18 9
54 lb
2.25 18 9
−+
=
+− +
ijk

()( )( )6lb 48lb 24lb=− +ijk
So that ( )( )( )6.00 lb 48.0 lb 24.0 lb=−+Fijk W
Have
/
:
EAEAB
Σ×=MrTM

022.5 0lbft
64824
⋅=
−
ijk
M
( )( )540 lb ft 135 lb ft∴= ⋅ − ⋅Mik
or
(
)( )540 lb ft 135.0 lb ft=⋅− ⋅Mi k W

96. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 92.













Have
:
CD
Σ =FT F
where
CD CD
CD
T
CD
=T
JJJG


()
()( )()
222
0.9 16.8 7.2
61 lb
0.9 16.8 7.2
− −+
=
−+− +
ijk

()( )( )3lb 56lb 24lb=− − +ijk
So that ( )( )( )3.00 lb 56.0 lb 24.0 lb=− − +Fijk W
Have
/OCDCD
Σ= × =Mr T M
022.50lbft
35624
⋅=
−−
ijk
M
( )( )540 lb ft 67.5 lb ft∴= ⋅ + ⋅Mik

(
)( )540 lb ft 67.5 lb ft=⋅+ ⋅Mik W

97. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 93.













Have :
AB
Σ =FT F
where
AB AB
AB
T
AB
=T
JJJG


()
() ( ) ()
222
4.75 2
10.5 kN
14.752
−−+
=
−+− +
ijk

( )( )( )2kN 9.5kN 4kN=− − +ijk
So that ( )( )( )2.00 kN 9.50 kN 4.00 kN=− − +Fijk W
Have :
OAAB
Σ×=MrTM

34.750kNm
29.54
⋅=
−−
ijk
M
( )( )( )19 kN m 12 kN m 19 kN m∴= ⋅ − ⋅ − ⋅Mijk
( )( )( )19.00 kN m 12.00 kN m 19.00 kN m=⋅−⋅−⋅Mijk W

98. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 94.



Let
( ),
O
RM be the equivalent force-couple system
Then ( )( )220 N sin 60 cos60= −°− °Rjk

()
( )110 N 3=−− jk
or
(
)( )190.5 N 110 N=− −Rjk W
Now :
OOOC
Σ=×MMrR
Where
(
)( ) ( )0.2 m 0.1 0.4sin 20 m 0.4cos 20 m
OC
=+−°+ °

ri j k
Then
()( ) ( )
()0.1 110 N 2 1 4sin 20 4cos20 m
03 1
O
=− − ° °
ij k
M

()( )()( )
(){ }
11 N m 1 4sin 20 1 4cos 20 3 2 2 3

=− ⋅ − ° − ° − +

ij k
or ( )( )( )75.7 N m 22.0 N m 38.1 N m
O
=⋅+⋅−⋅Mijk W

99. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 95.













Have :
D
Σ=FFF
where
AI
F
AI
=F
JJG


()
()( )()
222
14.4 4.8 7.2
63 lb
14.4 4.8 7.2
− +
=
+− +
ijk

So that ( )( )( )54.0 lb 18.00 lb 27.0 lb=− +Fi jk W
Have
/
:
DIOD
Σ+×=MMrFM
where
AC
M
AC
=M
JJJG


()
() ( )
22
9.6 7.2
560 lb in.
9.6 7.2
−
=⋅
+−
ik

( )( )448 lb in. 336 lb in.=⋅−⋅ ik
Then
()()
448 lb in. 336 lb in. 0 0 14.4 lb in.
54 18 27
D
= ⋅− ⋅ + ⋅
−
ijk
Mik
( )( ) ( )( )448 lb in. 336 lb in. 259.2 lb in. 777.6 lb in. =⋅−⋅+ ⋅+ ⋅
 
ik i j

or
( )( )( )707 lb in. 778 lb in. 336 lb in.
D
=⋅+⋅−⋅Mijk W

100. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 96.






First assume that the given force W and couples
1
Mand
2
Mact at the
origin.
Now W=−
j
W
and
()( )
12 2 12
cos 25 sin 25MMM=+ =− °+ − °MM M i k
Note that since
W and M are perpendicular, it follows that they can be
replaced with a single equivalent force.
(a) Have
() or 2.4 NFW==−=−
j
WF j

()or 2.40 N=−Fj W
(b) Assume that the line of action of F passes through point P (x, 0, z).
Then for equivalence
/PO
=×Mr F
where
/PO
xz=+rik

()( )
212
cos 25 sin 25MMM∴− °+ − ° ik

() ()0
00
x zWz Wx
W
==−
−
ijk
ik


Equating the i and k coefficients,

12
cos 25 sin 25
and
z
MMM
zx
WW
−° − ° 
==−



(b) For
12
2.4 N, 0.068 N m, 0.065 N mWM M==⋅=⋅
0.068 0.065sin 25
0.0168874 m
2.4
x
−°
==−
−

or 16.89 mm x=−
W

0.065cos 25
0.024546 m
2.4
z
−°
==−
or 24.5 mm z=−
W

101. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 97.



(a) Have
2
:0
Bz z
MMΣ =
( )/1 1
0
HB z
M⋅ +=kr F× (1)
where ( )()
/
31 in. 2 in.
HB
=−rij
11 EH
F=Fλ

()()()
()
6 in. 6 in. 7 in.
20 lb
11.0 in.
+−
=
ijk

()
20 lb
667
11.0
=+−
ijk
11z
M
=⋅kM

11 EJ
M=M λ

()()
()
2
3 in. 7 in.
480 lb in.
58 in.
d
d
−+ −
= ⋅
+
ijk

Then from Equation (1),
()( )
2
001
7480 lbin.20 lb in.
31 2 0 0
11.0
58
66 7
d
−⋅⋅
− +=
+
−
continued

102. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Solving for d, Equation (1) reduces to
()
2
20 lb in. 3360 lb in.
186 12 0
11.0
58d
⋅ ⋅
+−=
+
From which
5.3955 in.d
=

or 5.40 in.d
= W
(b) ()
21
20 lb
667
11.0
== +−FF ijk
( )10.9091 10.9091 12.7273 lb=+− ijk
( )( )( )
2
or 10.91 lb 10.91 lb 12.73 lb=+−Fijk W

2/11HB
=+Mr FM ×

()
()
5.3955 3 720 lb in.
31 2 0 480 lb in.
11.0 9.3333
66 7
−+−⋅
=− + ⋅
−
ijk
ijk


(
)25.455 394.55 360 lb in.=++⋅ ijk

(
)277.48 154.285 360 lb in.+− + − ⋅ijk

(
)( )
2
252.03 lb in. 548.84 lb in.=− ⋅ + ⋅Mij
( )( )
2
or 21.0 lb ft 45.7 lb ft=−⋅+ ⋅Mij W

103. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 98.


(a) : : 400 N 600 N
ya
aF RΣ=−−
or 1000
a
N=R

W

(
)( )()( ): 2 kN m 2 kN m 5 m 400 N
Ba
MMΣ=⋅+⋅+
or 6.00 kN m
a
=⋅M

W
: : 1200 N 200 N
yb
bF RΣ=−+
or 1000 N
b
=R

W


(
)()( ):0.6kNm5m1200N
Bb
MMΣ=⋅+
or 6.60 kN m
b
=⋅M

W

: : 200 N 1200 N
yc
cF RΣ=−
or 1000 N
c
=R

W


(
)( )()( ): 4 kN m 1.6 kN m 5 m 200 N
Bc
MMΣ=−⋅−⋅−
or 6.60 kN m
c
=⋅M

W

: : 800 N 200 N
yd
dF RΣ=−−
or 1000 N
d
=R

W


(
)( )()( ): 1.6 kN m 4.2 kN m 5 m 800 N
Bd
MMΣ=−⋅+⋅+
or 6.60 kN m
d
=⋅M

W


continued

104. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


: : 500 N 400 N
ye
eF RΣ=−−
or 900 N
e
=R

W


(
)( )()( ): 3.8 kN m 0.3 kN m 5 m 500 N
Be
MMΣ=⋅+⋅+
or 6.60 kN m
e
=⋅M

W

: : 400 N 1400 N
yf
fF RΣ=−
or 1000 N
f
=R

W


(
)( )()( ): 8.6 kN m 0.8 kN m 5 m 400 N
Bf
MMΣ=⋅−⋅−
or 5.80 kN m
f
= ⋅M

W

: : 1200 N + 300 N
yg
gF RΣ=−
or 900 N
g
=R

W


(
)( )()( ): 0.3 kN m 0.3 kN m 5 m 1200 N
Bg
MMΣ=⋅+⋅+
or 6.60 kN m
g
= ⋅M

W

: : 250 N 750 N
yh
hF RΣ=−−
or 1000 N
h
=R

W


(
)( )()( ): 0.65kNm 6kNm 5m 250N
Bh
MMΣ=−⋅+⋅+
or 6.60 kN m
h
= ⋅M

W

(
b) The equivalent loadings are (b), (d), (h) W

105. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 99.











The equivalent force-couple system at
B is...
: 650 N 350 N
y
FRΣ=−−
or 1000 N
=R
( )( )( )()( ): 1.6 m 800 N 1.27 kN m 5 m 650 N
B
MMΣ= +⋅+
or 5.80 kN m= ⋅M

∴ The equivalent loading of Problem 3.98 is (f) W

106. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 100.





Equivalent force system...
(
a) : 400 N 200 N
y
FRΣ=−−
or 600 N
=R

W


(
)( )( )()( ): 600 N 200 N m 100 N m 4 m 200 N
A
MdΣ− =−⋅+⋅−


or 1.500 md= W

(
b) : 400 N + 100 N
y
FRΣ=−
or 300 N
=R

W


(
)( )( )()( ): 300 N 200 N m 600 N m 4 m 100 N
A
MdΣ − =− ⋅− ⋅+


or 1.333 md= W

(
c) : 400 N 100 N
y
FRΣ=−−
or 500 N
=R

W


(
)( )( )()( ): 500 N 200 N m 200 N m 4 m 100 N
A
MdΣ− =−⋅−⋅−


or 1.600 md= W

107. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 101.


The equivalent force-couple system at A for each of the five force-couple systems will be determined and
compared to

() ( )( )2 lb 48 lb in. 32 lb in.==⋅+⋅FjM i k
To determine if they are equivalent
Force-couple system at
B:
Have
():2lbΣ=FF j
and ( )/
:
A BBAB
Σ=Σ+×MMMrF
( )( )32 lb in. 16 lb in.=⋅+⋅Mik ()()8in. 2lb +×
 
ij
( )( )32 lb in. 32 lb in.=⋅+⋅ ik

∴ is not equivalent W
Force-couple system at C:
Have
():2lbΣ=FF j
And ( )/
:
A CCAC
Σ=+×MMMrF
( )()( ) ()68 lb in. 8 in. 10 in. 2 lb=⋅+ + ×

M iikj
( )( )48 lb in. 16 lb in.=⋅+⋅ ik

∴ is not equivalent Wcontinued

108. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Force-couple system at E:
Have ():2lbΣ=FF j
and ( )/
:
A EEAE
Σ=+×MMMrF
( )( )( )()48 lb in. 16 in. 3.2 in. 2 lb=⋅+ − ×

Miijj
( )( )48 lb in. 32 lb in.=⋅+⋅ ik
∴ is equivalent W
Force-couple system at
G:
Have
()():2lb2lbΣ=+FF i j
F has two force components
∴ is not equivalent W


Force-couple system at
I:
Have
():2lbΣ=FF j
and ( )/
:
A IIAI
ΣΣ+×MMrF
( )( )80 lb in. 16 in.=⋅−Mik
( )()( ) ()16 in. 8 in. 16 in. 2 lb+−+ ×

ij k j
( )( )48 lb in. 16 lb in.=⋅+⋅Mik
∴ is not equivalent W

109. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 102.






First
(
)38 kg
AA
Wmg g==

(
)29 kg
BB
Wmg g==
(
a) (
)27 kg
CC
Wmg g==
For resultant weight to act at
C, 0
C
MΣ=
Then
( )()( )()( )()38 kg 2 m 27 kg 29 kg 2 m 0ggdg − −=

76 58
0.66667 m
27
d
−
∴= =

or 0.667 md
= W
(
b) (
)24 kg
CC
Wmg g==
For resultant weight to act at
C, 0
C
MΣ=
Then
( )()( )()( )()38 kg 2 m 24 kg 29 kg 2 m 0ggdg − −=

76 58
0.75 m
24
d
−
∴= =
or 0.750 m
d
= W

110. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Σ
Chapter 3, Solution 103.











(a) Have
:
CDE
F WWW RΣ−−−=
200 lb 175 lb 135 lbR∴=− − −

510 lb=−
or
510 lb=R
Σ
Have
:
A
MΣ

()()()()()( )(
)200 lb 4.5 ft 175 lb 7.8 ft 135 lb 12.75 ft Rd−−− =−

()3986.3 lb ft 510 lbd∴− ⋅ = −
or
7.82 ftd= Σ
(b) For equal reactions at A and B,
The resultant R must act at midspan.
From
2
A
L
MR

Σ=−



()()()( )(
)()200 lb 4.5 ft 175 lb 4.5 ft + 135 lb 4.5 ft + 2.5aa∴− − −

()()510 lb 9 ft=−
or
2295 512.5 4590a+=
and
4.48 fta= Σ

111. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 104.











Have : 12kN 18kN 40kN 40kN
L
WΣ− −− =− −F
50 kN
L
W=
or 50.0 kN
L
W= W

( )()( )( )(): 12kN 5m 50kN 40kN 5m
B
dΣ+=M
2.8 m
d
=
or heaviest load
(
)50 kN is located W
2.80 m from front axle

112. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 105.


(a) (
)( )( ):80N40N60NΣ=− −FR i j j ( )( )90 N sin 50 cos50+ −°− °ij

(
)( )11.0560 N 157.851 N=− ij

()( )
22
11.0560 N 157.851 NR=+−
158.2 N
=

157.851
tan
11.0560
θ
−
=


86.0
θ=°
or 158.2 N
=R

86.0°W

(b)






(
)( )( )( )( )( )( ): 157.851 N 0.32 m 80 N 0.15 m 40 N 0.35 m 60 N
F
MdΣ− = − −


(
)( ) ( )( )0.61 m 90 N cos50 0.16 m 90 N sin 50−°−°


or 302 mm
d= to the right of F W

113. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Chapter 3, Solution 106.








(a)
()()()()()()()(): 0 0.32 m 80 N 0.1 m 40 N 0.1 m 60 N 0.36 m 90 N cos
I
M αΣ= + − −

()()0.16 m 90 N sinα−
or
4sin 9cos 6.5556αα+=

()( )
22
9cos 6.5556 4sinαα=−

()
22
81 1 sin 42.976 52.445sin 16sinααα−=− +

2
97sin 52.445sin 38.024 0αα−−=
Solving by the quadratic formula gives for the positive root

sin 0.95230α=

72.233α=°
or
72.2α
=° (
Note: The second root
() 24.3α=− ° is rejected since
090.α<<°
(b)
()()(): 80N 40N 60NΣ=−−FR i
jj

()( )90 N sin 72.233 cos72.233+− °− ° i
j

()( )5.7075 N 127.463 N=− − i
j

()( )
22
5.7075 N 127.463 NR=− +−

127.6 N=

127.463
tan
5.7075
θ
−
=
−


87.4θ=°
or
127.6 N=R
87.4°(

(

114. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 107.


(a) Have
( )( )( )( ): 0 0.8 in. 40 lb 2.9 in. 20 lb cos30
D
MMΣ=− − ° ( )( )3.3 in. 20 lb sin 30− °
or 115.229 lb in.M=⋅
or 115.2 lb in. =⋅
M

W





Now,
R is oriented at 45° as shown (since its line of action
passes through
B and D).
Have
( ) ( ): 0 40 lb cos 45 20 lb cos15
x
F
′Σ =°−°
( )( )90 lb cos 45α− +°
or
39.283
α= °
or 39.3
α=° W
(
b) : 40 20sin 30 90cos39.283
xx
FRΣ=−°− °
39.663 lb=−
Now
2
x
RR= or 56.1 lb = R 45.0° W

115. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 108.


(a) Reduce system to a force and couple at B:





Have
()( ) ( ) ( )10 lb 25 lb cos60 25 lb sin 60 40 lb=Σ =− + ° + ° −RF j i j i
( )( )27.5 lb 11.6506 lb=− +ij
or
()( )
22
27.5 lb 11.6506 lb 29.866 lbR=− + =

111.6506
tan 22.960
27.5
θ
−
==°



or 29.9 lb =
R
23.0°W
Also
(
)( )( )()( )80 lb in. 12 in. 10 lb 8 in. 40 lb
BB
=Σ = ⋅ − × − − × −MM k i j j i

(
)120 lb in.=− ⋅ k
(
b)




Have
(
) ()( )120 lb in. 11.6506 lb
B
u=− ⋅ =− ×Mkij

(
) ( )()120 lb in. 11.6506 lbu−⋅=− kk
10.2999 in.
u= and 12 in. 10.2999 in.x=−
1.7001 in.
=

116. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.









Have
(
) ()( )120 lb in. 27.5 lb
B
v=− ⋅ =− × −Mkji

(
) ( )()120 lb in. 27.5 lbv−⋅=− kk
4.3636 in.
v=
and 8 in. 4.3636 in. 3.6364 in.
y=− =
or 1.700 in. to the right of
A and 3.64 in. above C W

117. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 109.

(a) Position origin along centerline of sheet metal at the intersection with
line EF.
( a) Have
Σ=FR

()0.52 1.05 2.1 sin 45 cos45 0.64 kips=− − − °+ ° −

Rjj iji

()(
)2.1249 kips 3.0549 kips=− −Ri j

()()
22
2.1249 3.0549R=− +−

3.7212 kips=

13.0549
tan 55.179
2.1249
θ
−
−
==°
−

or
3.72 kips=R
55.2°
Have
EF EF
MM=Σ
Where
()()( )()0.52 kip 3.6 in. 1.05 kips 1.6 in.
EF
M=+

()
( )( )( )2.1 kips 0.8 in. 0.64 kip 1.6 in. sin 45 1.6 in.−− °+



0.123923 kip in.=⋅
To obtain distance d left of EF,
Have
()3.0549 kips
EF y
MdRd==−

0.123923 kip in.
0.040565 in.
3.0549 kips
d
⋅
==−
−

or
0.0406 in. left ofdEF= 

(b)
Have
0
EF EF
MM
=Σ =

()()( )()0.52 kip 3.6 in. 1.05 kips 1.6 in.
EF
M=+

()()2.1 kips 0.8 in.−

()(
)0.64 kip 1.6 in. sin 1.6 in.α−+



()1.024 kip in. sin 0.848 kip in.α∴⋅=⋅
or
55.9α=° 



118. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 110.



(a) Have
Σ=FR

( )0.52 1.05 2.1 sin cos 0.64 kipsαα=− − − + −

Rjj iji

(
)() ( )0.64 kip 2.1 kips sin 1.57 kips 2.1 kips cosα α=− + − +
 ij
Then
0.64 2.1sin
tan
1.57 2.1cos
x
y
R
R
α
α
α
+
==
+

1.57 tan 2.1sin 0.64 2.1sin
α αα+=+

0.64
tan
1.57
α=

22.178
α= °
or
22.2α
= °W
(b) From
22.178
α=°

(
)0.64 kip 2.1 kips sin 22.178
x
R=− − °

1.43272 kips=−

(
)1.57 kips 2.1 kips cos22.178
y
R=− − °

3.5146 kips=−

119. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.





()()
22
1.43272 3.5146R=− +−
3.7954 kips=

or
3.80 kips=R
67.8°W
Then
EFEF
MM=Σ
Where
(
)( )( )( )( )( )0.52 kip 3.6 in. 1.05 kips 1.6 in. 2.1 kips 0.8 in.
EF
M=+ −

(
)( )0.64 kip 1.6 in. sin 22.178 1.6 in. −°+
 


0.46146 kip in.=⋅
To obtain distance d left of EF,
Have
EFy
M dR=

(
)3.5146 kipsd=−

0.46146 kip in.
3.5146 kips
d
⋅
=
−

0.131298 in.=−
or
0.1313 in. left ofdEF= W

120. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 111.









Equivalent force-couple at
A due to belts on pulley A
Have :120 N160 N
A
RΣ− − =F
280 N
A
∴=R

Have ( ): 40 N 0.02 m
AA
MΣ− =M
0.8 N m
A
∴ =⋅M
Equivalent force-couple at
B due to belts on pulley B
Have
( ):210 N150 NΣ+F 25
B
°=R
360 N
B
∴=R
25°
Have ( ):60 N0.015 m
B B
MΣ− =M
0.9 N m
B
∴ =⋅M
Equivalent force-couple at
F
Have
( )( )( ): 280 N 360 N cos 25 sin 25
F
Σ =− + °+ °FR j i j
( )( )326.27 N 127.857 N=− ij
()( )
2222
326.27 127.857 350.43 N
FFxFy
RR R R== + = + =
11 127.857
tan tan 21.399
326.27Fy
Fx
R
R
θ
−−
 −
= ==−° 

or 350 N
F
==RR
21.4°W

121. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.








Have

( )( ): 280 N 0.06 m 0.80 N m
FF
MΣ =− − ⋅M
( ) ( )360 N cos25 0.010 m−°

( ) ( )360 N sin 25 0.120 m 0.90 N m+ °−⋅

( )3.5056 N m
F
=− ⋅Mk
To determine where a single resultant force will intersect line
FE,
Fy
M dR=
3.5056 N m
0.027418 m 27.418 mm
127.857 N
F
y
M
d
R
− ⋅
∴= = = =
−

or 27.4 mm
d
= W

122. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 112.

(a) Have =ΣRF

()( )()()25 N cos40 sin 40 15 N 10 N=°+°−−Ri
j i j

()()4.1511 N 6.0696 N=+ i
j

()()
22
4.1511 6.0696=+R

7.3533 N=

16.0696
tan
4.1511
θ
−
=




55.631=°
or
7.35 N=R
55.6°
( a) and (b)
(b) From
B By
MMdR=Σ =
where
() ( )25 N cos 40 0.375 m sin 50
B
M  =− ° °
 

() ( )25 N sin 40 0.375 m cos50 −° °
 

()( ) ()( )15 N 0.150 m sin 50 10 N 0.150 m+°−


6.25 N m+⋅

2.9014 N m
B
M∴=− ⋅
and
B
y
M
d
R
=

2.9014 N m
6.0696 N
−⋅
=


0.47802 m=
or
478 mm to the left of
dB= 
(c) From
/BDB
=×Mr R
()( )
11
2.9014 N m cos50 sin50dd ⋅=− °+ °ki j−−−−

()()4.1511 N 6.096 N×+

ij


()()
1
2.9014 N m 7.0814 d−⋅=− kk

1
0.40972 md∴=
or
1
410 mmd= from B along line AB
or 34.7 mm above and to left of A 


123. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 113.








Based on 0
x
F
Σ=
cos 15 N 0Pα− =

cos 15 NP
α∴ = (1)
and 0
y
F
Σ=
sin 10 N 0Pα− =
sin 10 N
P
α∴ = (2)
Dividing Equation (2) by Equation (1),
10
tan
15
α=
33.690
α∴ =°
Substituting into Equation (1),
15 N
18.0278 N
cos33.690
P==
°

or 18.03 N =P 33.7°
(a) Based on 0
B
MΣ=
( ) ( )18.0278 N cos33.690 0.150 m sin 50d  − °+ °
  

( ) ( )18.0278 N sin 33.690 0.150 m cos50d  − °+ °
  

()( ) ()( )15 N 0.150 m sin 50 10 N 0.150 m 6.25 N m 0+ °− + ⋅ =

17.9186 3.7858d
− =−
0.21128 md∴=
or 211 mm d= W

124. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.








(b) Based on 0
D
M
Σ =
( ) ( )18.0278 N cos33.690 0.150 m sin 50d  − °+ °
  

( ) ( )18.0278 N sin 33.690 0.150 m cos50 0.150 md  −°+°+
  

()( )15 N 0.150 m sin 50 6.25 N m 0+ °+ ⋅ =

17.9186 3.7858d
− =−
0.21128 md∴=
or 211 mm d= W
This result is expected, since
0=R and 0
R
B
=M for

211 mmd
= implies that 0 and 0= =RM at any other point for
the value of d found in part a.

125. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 114.


(a) Let (),
D
RM be the equivalent force-couple system at D.
First note... At
;xb
yh==
For
2
ykx=
We have
2
hkb=
or
2
h
k
b
=

2
2hy x
b

∴=


For any contact point c alone the surface
2
2h
yx
b

=




2
2
dy h
x
dx b
=


F=R

2
1
tan
2
b
hx
−





and
() ( ): sin cos
DD
MxF hyFΜθθΣ=−+−

2
422
4
b
xF
bhx

=−

+

2
2
422 2
4
hhx
hxF
b bhx

+−

 +


or
()
22
2
422
2
4
D
h
xb h x hx
b
F
bhx
 
−+−


=
+
M


23
22
2
422
2
2
4
hx
xb h x
b
F
bhx

−+ −

=
+

or
()
23
22
2
422
2
2
4
D
hx
hbx
b
F
bhx

−−
=
 +


M



126. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


(b) With 1ft, 2ftbh==

3
2
78
116
D
xx
MF
x
−
=

+


For
D
M to be a maximum
Then
() () ()()
()
1
22 3 2
2
2
1
7 24 1 16 7 8 32 1 16
2
0
116
D
xxxxxx
dM
F
dx x
−

−+−− +


==

+



For the non-trivial solution:

()()()
22 3
0724116 1678xxxxx=− + − −

42
0256 24 7xx=+−
Solving by the quadratic formula gives for the positive root.

0.354 ftx= 

127. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 115.













For equivalence
:
B CD A
Σ++=FF F F R

( )( )( )( )240 N 125 N 300 N 150 N
A
=− − − +Rjkik

( )( )( )300 N 240 N 25 N
A
∴=− − +Rijk W


Also for equivalence

// /
:
BABCACDAD AΑ
Σ×+×+×=rFrFrFMΜ
or 0 0.12m 0 0.06m 0.03m 0.075m 0.06m 0.08m 0.75m
0 240 N 125 N 300 N 0 0 0 0 150 N
A
M=+−+−
−− −
ij k i j k i j k


(
) ( )( ) ( )( )15 N m 22.5 N m 9 N m 12 N m 9 N m  =− ⋅ + ⋅ + ⋅ + ⋅ − ⋅
  ijkij
or ( )( )( )3N m 13.5N m 9N m
A
=− ⋅ + ⋅ + ⋅Mijk W

128. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 116.


Let (
),
D
RM be the equivalent force-couple system at O.
Now
:Σ=ΣFR F

(
)( )()( )()( )1.8lbsin40 cos40 11lbsin12 cos12 18lbsin15 cos15=−°−°+−°−°+−°−°i k jk jk
or ( )( )( )1.157 lb 6.95 lb 29.5 lb=− − − ijk WR
Note that each belt force may be replaced by a force-couple that is equivalent to the same force plus the
moment of the force about the shaft (x axis) of the sander. Then ...
:
OO O
Σ=ΣMM M


()
1.8 lb 0 0.75 in. 2.2 in.
sin 40 cos40
=
− °0− °
ijk

( )()()()( )2.5 in. 11 lb 9 in. 11 lb sin12 cos12− −×−°−°
ii jk


( )()()()( )2.5 in. 18 lb 9 in. 18 lb sin15 cos15+ −×−°−°
ii jk


()( )1.8 0.75cos 40 2.2sin 40 0.75sin 40 27.5=− °− °+ °−
 ij ki


()( ) ()( )( )99 sin12 cos12 45 162 sin15 cos15 lb in. + °− °+ + °− ° ⋅
kji kj


( )( )1.03416 27.5 45 2.5454 96.837 156.480=− − + +− − −
 ij


( )( )0.86776 20.583 41.929 lb in.+ ++ ⋅
 k


or ( )( )( )16.47 lb in. 256 lb in. 63.4 lb in.
O
=⋅−⋅+⋅Mijk W

129. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



Chapter 3, Solution 117.


Have :
x
FΣ 10 N
x x
AB−=+ 10
x x
BA⇒=−−

:
y
FΣ 0
yy
AB=+
yy
BA⇒=−

:
z
FΣ 6N
zz
AB=+ 6
zz
B A⇒=−
and
//
:
OOOAOB
Σ=×+×MMrArB
Now
()( )()
222
: 372 mm mm 72 mm
BA BA z
dd =60 +− +
or ()360 mm
BAz
d =
Then
(
)( )( )
/
135 mm 72 mm 310 mm
OA
=−+rijk


(
)( )
/
75 mm 50 mm
OB
=−rik


(
)( )( )60 N m 0.05 N m 10 N m⋅+ ⋅− ⋅ijk

() ()0.135 0.72 0.310 N m 0.075 0 0.050 N m
xyz xyz
AAA BBB
=− ⋅+ − ⋅
ijk ijk


i:
()
( )60 0.072 0.310 0.050
zy y
A AB=− − +


or 60 0.072 0.360
zy
A A=− −

(1)

j:
()
( )0.05 0.310 0.135 0.050 0.075
x zxz
A ABB=−+−−

( ) ( )0.310 0.050 10 0.135 0.075 6
x xz z
A AA A=−−−−−−


or

0 0.360 0.060
x z
A A=−


6
zx
A A=

130. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



Now 2 N
x
A= 12.00
z
A∴= N
From equation (1) ( )60 0.072 12.00 0.360
y
A=− −
or 169.1 N
y
A=−
Then 12.00 N
x
B=−
169.1 N
y
B=
6.00 N
z
B=−

( )( )( )2.00 N 169.1 N 12.00 N∴= − +Ai j k W
( )( )( )12.00 N 169.1 N 6.00 N=− + −Bijk W

131. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Chapter 3, Solution 118.


Have : Σ+=FBCR
: 3.9 lb or 3.9 lb
x xx x x
FBC B CΣ+= =− (1)
:
yy y
FC RΣ= (2)
:1.1 lb
zz
FCΣ=− (3)
Have
//
:
R
ABA CA B A
Σ++=Mr Br CMM××
() ( )( )
11
0 4.5 4 0 2.0 2 lb ft 1.5 lb ft 1.1 lb ft
12 12
00 1.1
x
xxy
xM
BCC
∴+ +⋅=+⋅−⋅
−
ijk i j k
ii j k

()
( )()2 0.166667 0.375 0.166667 0.36667 0.33333
yx x y
CB C C−++++ ijk

(
)()1.5 1.1
x
M=+ −ijk
From - coefficient 2 0.166667
yx
CM−=i (4)
- coefficient 0.375 0.166667 0.36667 1.5
xx
BC++=j (5)
- coefficient 0.33333 1.1 or 3.3 lb
yy
CC=− =−k (6)
(a) From Equations (1) and (5):
( )0.375 3.9 0.166667 1.13333
xx
CC−+ =
0.32917
1.58000 lb
0.20833
x
C==
From Equation (1): 3.9 1.58000 2.32 lb
x
B=− =

( ) 2.32 lb∴=Bi W
( )( )( )1.580 lb 3.30 lb 1.110 lb=−−Cijk W
(b) From Equation (2): 3.30 lb
yy
RC==−
( )or 3.30 lb
y
=−R W
From Equation (4): ( )0.166667 3.30 2.0 2.5500 lb ft
x
M=−−+=⋅
( )or 2.55 lb ft
x
= ⋅MW

132. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 119.

(a) Duct AB will not have a tendency to rotate about the vertical or y-axis if:
()//
0
RR
By B F B F E B E
M=Σ = + =jM jr F r F⋅⋅⋅⋅⋅ ⋅⋅⋅××××× ×××
where

()()()
/
1.125 m 0.575 m 0.7 m
FB
=−+ri
j k
()()()
/
1.35 m 0.85 m 0.7 m
EB
=−+ri j k

()( )50 N sin cos
F
αα=+

Fjk

()25 N
E
=−Fk

()
() 50 N 1.125 m 0.575 m 0.7 m 25 N 1.35 m 0.85 m 0.70
0sincos 0 01
R
B
αα
∴Σ = − + −
−
i
j ki j k
M


()()()28.75cos 35sin 21.25 56.25cos 33.75 56.25sin N mαα α α=− − + − − + ⋅

ijk

Thus,
56.25cos 33.75 0
R
By
M α=− + =

cos 0.60α=
53.130α=°

or 53.1α=° 
(b)
EF
=+RF F
where

()25 N
E
=−Fk

()( )()()50 N sin 53.130 cos53.130 40 N 30 N
F
=°+°=+F
j k j k

()() 40 N 5 N∴= +R
j k
and
() () () ()28.75 0.6 35 0.8 21.25 56.25 0.6 33.75 56.25 0.8
R
B
=Σ =− + − − − +

MM i j k


()()()24 N m 0 45 N m=− ⋅ − + ⋅i
j k

()
( )or 24.0 N m 45.0 N m=− ⋅ + ⋅Mik 


133. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Chapter 3, Solution 120.


(a) Have
F E
=Σ = +RFFF
where ( )( ) ( )( )50 N sin 60 cos60 43.301 N 25 N
F
=°+°=+
Fjkjk
( )25 N
E
=−Fk
( ) 43.301 N∴=Rj ( )or 43.3 N=Rj W
Have ( )
//
R
CFCFECE
=Σ = +MrFrFrF×××
where
( )( )
/
0.225 m 0.050 m
FC
=−rij
( )( )
/
0.450 m 0.325 m
EC
=−rij
0.225 0.050 0 N m 0.450 0.325 0 N m
0 43.301 25 0 0 25
R
C
∴= − ⋅+ − ⋅
−
ijk ijk
M

(
)( )( )6.875 N m 5.625 N m 9.7427 N m=⋅+⋅+ ⋅ ij k
( )( )( )or 6.88 N m 5.63 N m 9.74 N m
R
C
=⋅+⋅+⋅Mijk W

(b) To determine which direction duct section CD has a tendency to turn, have
R R
CD DC C
M= Mλ⋅
where
( )( )
() ()
22
0.45 m 0.1 m
0.97619 0.21693
0.45 0.1
DC
−+
==−+
−+
ij
ijλ
Then
( )( )0.97619 0.21693 6.875 5.625 9.7427 N m
R
CD
M
=−+ ++ ⋅ijijk ⋅
( )6.7113 1.22023 N m=−+ ⋅
5.4911 N m =−⋅
Since 0,
R
DC C
<Mλ⋅ duct DC tends to rotate counterclockwise relative to elbow C as viewed from D to C. W

134. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Chapter 3, Solution 121.





Have
:
A BC
Σ==FRR R λλλλ
where
()()()
42 in. 96 in. 16 in.
106 in.
BC
−−
=
ijk
λλλλ
()
21.2 lb
42 96 16
106
A
∴= −−Rijk

()( )()or 8.40 lb 19.20 lb 3.20 lb
A
=− −Ri
j kΣ
Have
/
:
A CA A
Σ+=Mr RMM××××
where
()()
()
/
1
42 in. 48 in. 42 48 ft
12
CA
=+ =+rikik

()(
)3.5 ft 4.0 ft=+ ik

()( )()8.40 lb 19.20 lb 3.20 lb=− −Ri
j k

BC
M=−M λλλλ

()
42 96 16
13.25 lb ft
106−+ +
=⋅ijk


()()()5.25 lb ft 12 lb ft 2 lb ft=− ⋅ + ⋅ + ⋅
ijk

Then ()3.5 0 4.0 lb ft 5.25 12 2 lb ft
8.40 19.20 3.20
A
⋅+− + + ⋅=
−−
ijk
ijkM
()()() 71.55 lb ft 56.80 lb ft 65.20 lb ft
A
∴= ⋅+ ⋅− ⋅Mi j k


()()()or 71.6 lb ft 56.8 lb ft 65.2 lb ft
A
=⋅+⋅−⋅Mi
j kΣ
Σ

135. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 122.


From ()
()()()
()()()
22 2
0.6 ft 4.2 ft 1.5 ft
60 lb 60 lb
0.6 ft 4.2 ft 1.5 ft
CAB
 
−+ −
 
== =
 
−+ +−
  
ijk
RR λ


( )( )( )8.00 lb 56.0 lb 20.0 lb
C
=− + −Rijk W
From
/CAC
=+Mr RM ×
where
( )( )
/
7.8 ft 1.5 ft
AC
=+rik
()()
()()()
()( )()
222
0.6 ft 4.2 ft 1.5 ft
22.5 lb ft 22.5 lb ft
0.6 ft 4.2 ft 1.5 ft
BA
 
−+
 
=⋅=⋅
 
+− +
  
ijk
M λ

(
)( )( )3 lb ft 21 lb ft 7.5 lb ft=⋅− ⋅+ ⋅ij k
() 7.8 0 1.5 lb ft 3 21 7.5 lb ft
856 20
C
∴= ⋅+−+ ⋅
−−
ijk
Mijk

(
) ( ) ( )84 3 lb ft 144 21 lb ft 436.8 7.5 lb ft  =− + ⋅ + − ⋅ + + ⋅
  
ij k
( )( )( )or 81.0 lb ft 123.0 lb ft 444 lb ft
C
=− ⋅ + ⋅ + ⋅Mijk W

136. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 123.








Have: :
AC D E
Σ+++=FFFFFR

(
)( )( )( )80 kN 40 kN 100 kN 60 kN=− − − −Rjjjj

(
)280 kN=− j
or
280 kNR= W
Have:
(
) () () ()():
x AA CC DD EE G
M Fz Fz Fz Fz RzΣ+++=

(
)()( )() ()80 kN 0 40 kN 3 m sin 60 60 kN 0+°+


(
)() ( )60 kN 3 m sin 60 280 kN
G
Z+− °=

0.185577 m
G
Z∴=−
or 0.1856 m
G
Z=− W

(
) () () ()():
zAACCDDEE G
M Fx Fx Fx Fx RxΣ+++=

(
)() ( )( ) ( )80kN 3m cos60 1.5m 40kN 1.5m 60kN1.5m−°−+ +


()
() ( )100 kN 3 m cos60 1.5 m 280 kN
Gx
+°+=
or 0.750 m
G
x= W

137. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 124.










Have:
(
) () () () () ()():
x AA BB CC DD EE FF G
M Fz Fz Fz Fz Fz Fz RzΣ+++++=

(
)() () ( )() ( )()80 kN 0 3 m sin 60 40 kN 3 m sin 60 100 kN 0
B
F+°+ °+


(
)() () ()60 kN 3 m sin 60 3 m sin 60 0
F
FR+− °+− °=

20 kN
BF
FF−= (1)
Also
(
) () () () () ()():
zAABBCCDDEEFF G
M Fx Fx Fx Fx Fx Fx RxΣ+++++=

(
)() ( )( )( )80 kN 3 m cos60 1.5 m 1.5 m 40 kN 1.5 m
B
F−°−+−+

()
() ( )( )( ) ()100 kN 3 m cos60 1.5 m 60 kN 1.5 m 1.5 m 0
FFR
+°+++−=
140 kN
BF
FF+= (2)
Solving equations (1) and (2): 80.0 kN
B
F= W
60.0 kN
F
F= W

138. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 125.









Have :
ABCD
Σ+++=FF F F F R
(
)( )( )( )116 kips 470 kips 66 kips 28 kips−− −−=jjjjR
( ) 680 kips∴=−Rj 680 kipsR= W
Have ( ) () () ()():
x AA BB CC DD E
M Fz Fz Fz Fz RzΣ+++=
()()( )()( )()( )( )( )()116 kips 24 ft 470 kips 48 ft 66 kips 18 ft 28 kips 100.5 ft 680 kips
E
z+++ =
43.156 ft
E
z∴= or 43.2 ft
E
z= W
Have (
) () () ()():
zAA BB CC DD E
M Fx Fx Fx Fx RxΣ+++=
()()( )()( )( )( )()( )()116 kips 30 ft 470 kips 96 ft 66 kips 162 ft 28 kips 96 ft 680 kips
E
x+++=
91.147
E
x∴= or 91.1 ft
E
x= W

139. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 126.








Have :
BCDE
Σ+++=FF F F F R
(
)( )( )( )470 kips 66 kips 28 kips 116 kips− −−− =jjj jR
( ) 680 kips∴=−Rj
Have () () () ()():
x BB CC DD EE B
M Fz Fz Fz Fz RzΣ+++=
()()( )()( )( )( )()( )()470 kips 48 ft 66 kips 18 ft 28 kips 100.5 ft 116 kips 680 kips 48 ftb++ +=

52.397 ftb
∴= or 52.4 ftb= W
Have () () () ()():
zBB CC DD EE B
M Fx Fx Fx Fx RxΣ+++=
()()( )( )( )()( )()( )()470 kips 96 ft 66 kips 162 ft 28 kips 96 ft 116 kips 680 kips 96 fta+++=

58.448 fta
∴= or 58.4 fta= W

140. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 127.



For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the
intersection with the center line of the trailer, the added
0.6 0.6 1.2-m
×× box should be placed adjacent to
one of the edges of the trailer with the
0.6 0.6-m
× side on the bottom. The edges to be considered are based
on the location of the resultant for the three given weights.
Have
(
)( )( ): 200 N 400 N 180 NΣ− − − =FjjjR
( ) 780 N∴=−Rj
Have ( )( )( )( )( )( )( )(): 200 N 0.3 m 400 N 1.7 m 180 N 1.7 m 780 N
z
M xΣ++=
1.34103 m x∴=
Have ( )( )( )( )( )( )( )(): 200 N 0.3 m 400 N 0.6 m 180 N 2.4 m 780 N
x
M zΣ++=

0.93846 mz
∴=
From the statement of the problem, it is known that the resultant of
R from the original loading and the
lightest load
W passes through G, the point of intersection of the two center lines. Thus, 0.
G
Σ=M
Further, since the lightest load
W is to be as small as possible, the fourth box should be placed as far from G
as possible without the box overhanging the trailer. These two requirements imply

( )( )0.3 m 1 m 1.8 m 3.7 mxz≤≤ ≤≤
continued

141. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



Let 0.3 m,x= ()( )( )( )( )( ) ( ): 200 N 0.7 m 400 N 0.7 m 180 N 0.7 m 0.7 m 0
Gz
MWΣ−−+=
380 N W∴ =
()( )( )( )( )( )( )( ): 200 N 1.5 m 400 N 1.2 m 180 N 0.6 m 380 N 1.8 m 0
Gx
MzΣ−−++−=
3.5684 m 3.7 m acceptablez∴=< ∴
Let
3.7 m,z= (
)( )( )( )( )( ) ( ): 200 N 1.5 m 400 N 1.2 m 180 N 0.6 m 1.7 m 0
Gx
MWΣ−−++=

395.29 N 380 NW
∴ =>
Since the weight W found for 0.3 mx= is less than W found for 3.7 m, 0.3 mzx= = results in the
smallest weight W.

(
)or 380 N at 0.3 m, 0, 3.57 mW= W

142. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 128.




For the largest additional weight on the trailer with the box having at least one side coinsiding with the side of
the trailer, the box must be as close as possible to point G. For
0.6 m,x
= with a small side of the box
touching the z-axis, satisfies this condition.
Let
0.6 m,x= (
)( )( )( )( )( ) ( ): 200 N 0.7 m 400 N 0.7 m 180 N 0.7 m 0.4 m 0
Gz
MWΣ−−+=
665 NW∴ =
and
(
)( )( )( )( )( )( )( ): 200 N 1.5 m 400 N 1.2 m 180 N 0.6 m 665 N 1.8 m 0
GX
MzΣ−−++−=
( ) 2.8105 m 2 m 4 m acceptablezz∴= << ∴
( )or 665 N at 0.6 m, 0, 2.81 mW= W

143. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 129.



First, reduce the given force system to a force-couple system at the origin.
Have
(
)()():2PPPΣ−+=FijjR
() 2P∴=Ri
Have ( ):
R
OO O
ΣΣ =MrFM ×
()2 2 2.5 0 0 4 1.5 5 6
210 010
R
O
Pa Pa=+=−+−
−
ijk ijk
Mijk
(
a) 2
P=Ri or Magnitude of 2P=R W
Direction of : 0 , 90 , 90
xy z
θθ θ
=°=−°=°R W
(b) Have
1

R
RO R
M
R=⋅ =
R
Mλλ

( )1.5 5 6PaPaPa=⋅− + −iijk

1.5
Pa=−
and pitch
1
1.5
0.75
2MPa
P a
RP
−
== =− or 0.75 P a=− W

144. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



(c) Have
12
R
O
=+MMM

()()
21
56
R
O
PaPa∴=−= −MMM j k
Require
2/QO
=Mr R
×

(
)()( )()()()56 2 2 2PaPayzP Py Pz−=+ =−+jkjki kj ×
From : 5 2 PaPz=i

2.5za
∴=
From : 6 2 PaPy− =−k
3
ya
∴=

∴ The axis of the wrench is parallel to the x-axis and intersects the yz-plane at 3 , 2.5yaz a== W

145. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 130.



First, reduce the given force system to a force-couple at the origin.
Have :
PPPΣ−−=FiikR

P∴ =−Rk
Have () () ( ):3 3 3
R
OO
Pa Pa Pa aΣ− − +−+=MkjijM
( ) 3
R
O
Pa∴=−−Mik
Then let vectors
(
)
1
, RM represent the components of the wrench, where their directions are the same.
(a) P=−Rk or Magnitude of P=R W
Direction of : 90 , 90 , 180
xyz
θθθ
=°=°=−°R W
(b) Have
1
R
R O
M=⋅Mλ
( )3Pa =− ⋅ − −
 
kik

3
Pa=
and pitch
1
3
3
MPa
P a
RP== = or 3
Pa=W

(c) Have
12
R
O
=+MMM

( )( )
21
33
R
O
PaPaPa∴=−=−−−− =−MMM ik k i
Require
2/QO
= ×Mr R
( )()Paxy P PxPy−=+ − = −iij k ji×
From : or PaPy ya−=− =i
:0 x=j

∴ The axis of the wrench is parallel to the z-axis and intersects the xy plane at 0,
x ya= =W

146. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 131.



First, reduce the given force system to a force-couple at the origin.
Have
()():10 N 11 NΣ− − =FjjR
() 21 N∴=−Rj
Have
():
R
OO CO
ΣΣ +Σ=MrFMM ×
()000.5Nm 00 0.375Nm 12 Nm
0100 011 0
R
O
=⋅+−⋅−⋅
−−
ijk ij k
Mj

()()0.875 N m 12 N m=⋅−⋅ ij
(
a) ()21 N=−Rj or
()21.0 N=−Rj W
(b) Have
1

R
RO R
M
R
=⋅ =
R
Mλλ

()( ) ( )0.875 N m 12 N m
 =−⋅ ⋅ − ⋅
 
jij


()
1
12 N m and 12 N m=⋅ =−⋅ Mj
and pitch
1
12 N m
0.57143 m
21 N
M
P
R
⋅
== = or
0.571 mP= W
(c) Have
12
R
O
=+MMM

()
21
0.875 N m
R
O
∴=−= ⋅MMM i
Require
2/QO
=×Mr R

()()() 0.875 N m 21 Nxz
 ∴⋅=+−
 
iik j ×


() ()0.875 21 21
x z=− +iki

147. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



From i: 0.875 21z=
0.041667 mz∴=
From
k: 0 21
x=−
0z∴=

∴ The axis of the wrench is parallel to the y-axis and intersects the xz-plane at 0, 41.7 mmxz== W

148. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 132.


(a) First, reduce the given force system to a
force-couple system.
Have
( )( )( ):50N50N50NΣ −−+ =FkjkR
( )50 N ; 50 NR=− =Rj
( )50.0 N=− jWR
Have
( )( )( ): 0.1Nm 0.1Nm 0.1Nm
R
OO
Σ⋅−⋅+⋅=MkjkM
( )( )0.1 N m 0.2 N m
R
O
=− ⋅ + ⋅Mjk
(
b) Have
1
R R
R OO
M
R=× =⋅
R
MM
λ

( )( )0.1 N m 0.2 N m =− ⋅ − ⋅ + ⋅
 
jjk
0.1 N m
=⋅
and pitch
1
0.1 N m
0.002 m
50 NM
P
R
⋅
== =
or 2.00 mm
P= W
(
c)






Have
( )( )
1
0.002 m 50 NP  == −
 
MR j

(
)0.1 N m=− ⋅ j

149. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



Note that because ( )0.2 N m ,
z
=⋅Mk the line of action of the wrench must pass through the x-axis to
compensate for
z
M as shown above:
With
(
)
1
R
O
+× =MrRM
Then ( ) ()()0.1 N m 50Nd −⋅+−×−
 
ji j

(
)( )0.1 N m 0.2 N m=− ⋅ + ⋅jk
or ( )() ( )50 N 0.2 N md =⋅
 kk
and 0.004 m
d
=
0.004 m
xd
=−=−
or 4.00 mm, 0
xz==− W

150. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 133.


First replace the given couples with an equivalent force-couple system
( ),
B
O
RM at the origin.

(
)():35lb12lbΣ=−−FR i k

(
)()() ( ): 200 lb in. 8 in. 8 in. 35 lb
R
OO
  Σ=−⋅++×−
  
MM i j k i

(
) ( )() ()140 lb in. 10 in. 4 in. 12 lb  −⋅+ + ×−
  
kij k

(
)( )( )200 48 280 120 280 140=− − +− + + −ijk

(
)( )( )248 lb in. 1600 lb in. 140 lb in.=− ⋅ − ⋅ + ⋅ijk
Now
()()
22
35 lb 12 lbR=− +−
37 lb
=
Then
()
axis
1
35 12
37
=−− ikλ
(
a)
( )( )35.0 lb 12.00=− −Rik W
(
b)
1axis
R
O
M=⋅λM
()( )()
1
35 12 248 160 140 lb in.
37
=−− ⋅− − + ⋅ik i j k

()()
1
35 248 12 140 lb in.
37
=×−× ⋅

7000
lb in.
37
=⋅
Then
1
7000
lb in.
37
37 lb
M
P
R
⋅
==
or 5.11 in.
P= W

151. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



(c) Have
1axis1
M=M λ
()
2
7000 lb in.
35 12
37 ⋅
=−−
ik
Then
12
R
O
=+MMM
or
()
()
2 2
7000
248 160 140 35 12
37
=− − + − − −Mijk ik

(
)( )( )69.037 lb in. 160 lb in. 201.36 lb in.=− ⋅ − ⋅ + ⋅ij k








Require
/zOP
=×Mr R
or
69.037 160 201.36 0
35 0 12 yz−−+ =− −
ijk
ij k
:16035 z−=−j
or 4.57 in.
z=
:201.3635
y=k
or 5.75 in.
y=

∴ The point of intersection is defined by

5.75 in.
4.57 in.
y
z
=
=
W
W

152. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 134.



First reduce the given force system to a force-couple at the origin at
B.
(
a) Have ()()
815
: 79.2 lb 51 lb
17 17

Σ−− +=


Fk ijR
( )( )( ) 24.0 lb 45.0 lb 79.2 lb∴=− − −Rijk W
and

94.2 lbR
=
Have

/
:
R
B AB A A B B
Σ×++=Mr FMMM
()
815
0 20 0 660 714 1584 660 42 8 15
17 17
00 79.2
R
B 
=− −− += −− +


−
ij k
Mkijikij
( )( )( ) 1248 lb in. 630 lb in. 660 lb in.
R
B
∴= ⋅− ⋅− ⋅Mijk
(
b) Have
1

R
RO R
M
R=⋅ =
R
Mλλ

( )()()
24.0 45.0 79.2
1248 lb in. 630 lb in. 660 lb in.
94.2
−− −
 =⋅⋅−⋅−⋅
 
ijk
ijk
537.89 lb in. = ⋅

153. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



and
11R
M=M λ
( )( )( )137.044 lb in. 256.96 lb in. 452.24 lb in.=− ⋅ − ⋅ − ⋅ijk
Then pitch
1
537.89 lb in.
5.7101 in.
94.2 lbM
p
R ⋅
== =
or 5.71 in.
p
= W
(
c) Have
12
R
B
=+MMM
( )( )
21
1248 630 660 137.044 256.96 452.24
R
B
∴=−= −− −− − −MMM i j k i j k

( )( )( )1385.04 lb in. 373.04 lb in. 207.76 lb in.=⋅−⋅−⋅ ijk
Require
2/QB
=Mr R
×
1385.04 373.04 207.76 0
24 45 79.2
x z−− =
−−−
ijk
ijk
()()( )()45 24 79.2 45z zxx=−+ −ij jk
From i:
1385.04 45 30.779 in.zz
= ∴=
From k: 207.76 45 4.6169 in.
xx
− =− ∴ =
∴ The axis of the wrench intersects the xz-plane
at

4.62 in., 30.8 in.xz
= = W

154. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 135.



(
a) First reduce the given force system to a force-couple at the origin.
Have :
BA DC DE
PP P
Σ ++=FRλλ λ












+−
−
+





−+





−= k
25
12
j
5
4
i
25
9
j
5
4
i
5
3
k
5
3
j
5
4
R
P

()
3
220
25P
∴= −−Rijk W

() ( ) ()
2223275
2201
25 25P
RP
=++=
Have

( ):
R
OO
PΣΣ =Mr M ×
()() ()
43 34 9412
24 20 20
55 55 25525
R
O
PP PP PP P
aaa
−− 
−+ −+ −+ =
 
 
jjk jij jijkM×××
()
24

5
R
O Pa
∴= −−Mik
(b) Have

1
R
R O
M=⋅Mλ
where () ()
3251
220 220
25 27 5 9 5
R
P
R P
==−− = −−
R
ijk ijkλ

155. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



Then () ()
1
1248
220
595 155
PaPa
M
−
=−− −−=ijk ik ⋅

and pitch
1
825 8
8115 5 27 5
M Pa a
p
R P
− −
== =


or 0.0988pa
=− W
(c)
() ()
11
81 8
220 220
67515 5 9 5
R
Pa Pa
M
− 
== −−=−++


Mijkijkλ
Then

() () ()
21
24 8 8
2 20 403 20 406
5675 675
R
O Pa Pa Pa
=−= −−− −++= −−−MMM ik i jk i j k
Require
2/QO
=Mr R
×
()() ()
83
403 20 406 2 20
675 25
Pa P
xz
 
−−− =+ −−
 
 
ij k ik ijk ×
()
3
20 2 20
25
P
zxz x

 =++−
 

ijk
From i:
()
3
8403 20
675 25
PaP
z 
−=


1.99012
z a∴=−
From k:
()
3
8 406 20 2.0049
675 25Pa P
x xa

−=− ∴=



∴ The axis of the wrench intersects the xz-plane at

2.00 , 1.990
x az a= =− W

156. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Chapter 3, Solution 136.


First reduce the given force-couple system to an equivalent force-couple system (
),
B
RM at point B.

()()()
22 2
480 mm 560 mm 480 mm
BD
d=− + +−
880 mm =
()
132 N
480 560 480
880
BD BD BD
Fλ==−+−Fijk

()( )12 N 6 7 6=−+− ijk

()( )()
222
240 mm 220 mm 480 mm
EB
d=+−+
580 mm =
()
145 N
240 220 480
580
EB EB EB
Fλ== −+Fijk

(
)( )5 N 12 11 24=−+ ijk
:
BDEB
Σ=+FRFF

(
)( ) ( )12 N 6 7 6 5 N 12 11 24=−+−+ −+ijk i j k

(
)( )( )12 N 29 N 48 N=− + +ijk

()()( )
22 2
340 mm 240 mm 60 mm
BF
d=++−
20 442 mm=
Then ()
20 N m
340 240 60
20 442
B
⋅
=+−Mijk
()
20 N m
17 12 3
442
⋅
=+− ijk

157. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Now determine whether R and
B
M are perpendicular

()
()
20
12 29 48 17 12 3
442
B
⋅=−++ ⋅ +−RjjkijkM

()
20
12 17 29 12 48 3
442
=−×+×−×


0=

∴R and
B
M are perpendicular so that ( ),
B
RM can be reduced to the single equivalent force
( )( )( )12.00 N 29.0 N 48.0 NR=− + + ijk W






Now require
/BBP
=×Mr R
or
() ()
20 N m
17 12 3 0.480 0.480 N m
442
12 29 48
yz
⋅
+−=− − ⋅
−
ij k
ijk
()()
20 12
: 12 0.480 0.480 48
442
z
×
=− − +j

or 1.449 m z=
()
20 3
: 0.480 29 12
442
y
−×
=− +k

or 0.922 m y=

∴ The point of intersection is defined by

0.922 my= W
1.449 m z= W

158. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 137.



First, reduce the given force system to a force-couple at the origin.
Have :
AG
Σ +=FF F R
()
()()( )
()()()
4 in. 6 in. 12 in.
10 lb 14 lb 4 lb 6 lb 2 lb
14 in.
 +−
∴= + = + − 

ij k
Rk ijk
W
and
56 lbR=
Have ( ):
R
OO CO
Σ∑ +∑=MrFMM ×

()() ()()()()
{ }12 in. 10 lb 16 in. 4 lb 6 lb 12 lb
R
O
 =++−
 
Mjkiijk ××

()
(
)( )
()
()( )()16 in. 12 in. 4 in. 12 in. 6 in.
84 lb in. 120 lb in.
20 in. 14 in.
 −−+
+⋅ + ⋅  
 ij ijk

(
)( )( )
0
221.49 lb in. 38.743 lb in. 147.429 lb in.
R
∴= ⋅+ ⋅+ ⋅Mijk

(
)( )( )18.4572 lb ft 3.2286 lb ft 12.2858 lb ft=⋅+⋅+⋅ ij k

159. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


The force-couple at O can be replaced by a single force if the direction of R is perpendicular to .
R
O
M
To be perpendicular 0
R
O
=RM⋅
Have
( )( )4 6 2 18.4572 3.2286 12.2858 0?
R
O
=+− + + =RM i j k i j k⋅⋅
73.829 19.3716 24.572 =+ −
0 ≠
∴ System cannot be reduced to a single equivalent force.
To reduce to an equivalent wrench, the moment component along the line of action of P is found.

1

R
RO R
M
R==
R
Mλ⋅ λ

(
)
()
462
18.4572 3.2286 12.2858
56
+−
=++

ijk
ij k⋅

9.1709 lb ft
= ⋅
and ( )( )
11
9.1709 lb ft 0.53452 0.80178 0.26726
R
M== ⋅ + −Mijkλ
And pitch
1
9.1709 lb ft
1.22551 ft
56 lbM
p
R
⋅
== =
or 1.226 ft
p
= W
Have
(
)( )( )
21
18.4572 3.2286 12.2858 9.1709 0.53452 0.80178 0.26726
R
O
=−= + + − + −MMM i j k i j k

(
)( )( )13.5552 lb ft 4.1244 lb ft 14.7368 lb ft=⋅−⋅+⋅ ij k
Require
2/QO
=Mr R
×

(
)( )( )13.5552 4.1244 14.7368 4 6 2yz−+ =+ +−ij kjkijk ×
( )()()26 4 4yz z y=− + + −ijk
From j: 4.1244 4 z− = or 1.0311 ftz=−
From k: 14.7368 4 or 3.6842 ft
yy
=−=−

∴ line of action of the wrench intersects the yz plane at
3.68 ft, 1.031 ft
yz
=−= W

160. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Chapter 3, Solution 138.


Define (
)()
AA A
x y
FF=+Fij

(
)()
BB B
x y
FF=+Fij
Then
(
)():0
xAB
xx
FFFΣ+=
() ()
ABx x
FF=−

(
)():
yAB
yy
FFFRΣ+=
() ()
AB
yy
FRF=−
and
() ():
ABB xy
bF F aRM

Σ×+=−++

Mk ijkjj
i:
(
)
B
y
bF aR−=
or
()
By
a
FR
b
=−
Then
()
Ay
a
FR R
b

=−−




1
a
R
b

=+



j : ()
B
x
bF M=
or
()
Bx
M
F
b
=
Then
()
Ax
M
F
b
=−
1
A
Ma
R
bb

∴=−++


Fi j W

B
Ma
R
bb
=−Fij
W

161. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 139.



First, choose a coordinate system so that the xy plane coincides with the given plane. Also, position the
coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a.
Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the
scalar components of R and M are known relative to the shown coordinate system.
A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the
given point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B.
The known components of the wrench can be expressed as
and
x yz x y z
RRR MMM M=++ = + +Rijk i jk
while the unknown forces A and B can be expressed as
and
x yz xz
AAA BB=++ =+Aijk Bik
Since the position vector of point P is given, it follows that the scalar components (x, y, z) of the position
vector
P
rare also known.
Then, for equivalence of the two systems
:
x xxx
FR ABΣ=+ (1)
:
yy y
FR AΣ= (2)
:
zz zz
FR ABΣ=+ (3)
:
x xzy
M MyAzAΣ=− (4)
:
yy xzz
M MzAxAbBΣ=−− (5)
:
zz y x
M MxAyAΣ=− (6)
continued

162. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



Based on the above six independent equations for the six unknowns ( ), , , , , ,
xyzxz
AAABBb there exists a
unique solution for A and B.
From Equation (2)
yy
AR=W

Equation (6)
()
1
x yz
A xR M
y

=−

W

Equation (1)
()
1
x xyz
BR xRM
y

=− −

W

Equation (4)
()
1
zxy
A MzR
y

=+

W

Equation (3)
()
1
zz x y
BR MzR
y

=− +

W

Equation (5)
( )
()
x yz
xzy
xM yM zM
b
MyRzR++
=
−+
W

163. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 140.



First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed
line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular
coordinate system with the axis of the wrench while one of the other axes passes through the given point.
See Figures
a and b.
Have and and are known.
RM
= =Rj M j
The unknown forces A and B can be expressed as
and
x yz xyz
AAA BBB=++ =++A i jk B i jk
The distance
a is known. It is assumed that force B intersects the xz plane at (x, 0, z). Then for equivalence
:0
x xx
FAB∑=+ (1)
:
yyy
FRAB∑=+ (2)
:0
zzz
FAB
∑ =+ (3)
:0
x y
M zB∑=− (4)
:
yzzx
M MaAxBzB∑=−−+ (5)
:0
zyy
M aA xB∑=+ (6)
Since A and B are made perpendicular,
0 or 0
xx yy zz
AB AB AB
= ++=AB⋅ (7)
There are eight unknowns: , , , , , , ,
xyzxyz
AAABBBxz
But only seven independent equations. Therefore,
there exists an infinite number of solutions. continued

164. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Next consider Equation (4): 0
y
zB=−
If 0,
y
B=Equation (7) becomes 0
xx zz
AB AB
+ =
Using Equations (1) and (3) this equation becomes
22
0
xz
AA
+=
Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that
0,
y
B≠so that from Equation (4), 0.z
=
To obtain one possible solution, arbitrarily let 0.
x
A
=
(Note: Setting , ,
yz
AA or
z
B equal to zero results in unacceptable solutions.)
The defining equations then become.
0
x
B= (1) ′

yy
RA B=+ (2)
0
zz
AB=+ (3)

zz
M aA xB=− − (5)′
0
yy
aA xB=+ (6)
0
yy zz
AB AB
+ = (7)′
Then Equation (2) can be written
yy
ARB=−
Equation (3) can be written
zz
BA
=−
Equation (6) can be written
y
y
aA
x
B
=−

Substituting into Equation (5)
′,
()
y
zz
y
RB
M aA a A
B
−
=− − − −


or
zy
M
A B
aR
=−(8)
Substituting into Equation (7)
′,
()
0
yy y y
MM
RBB B B
aR aR

− +− =



165. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



or
23
22 2
y
aR
B
aR M
=
+

Then from Equations (2), (8), and (3)

23 2
22 2 22 2
y
aR RM
AR
aR M aR M
=− =
++

23 2
22 2 22 2z
MaR aRM
A
aRaR M aR M

=− =−

++



2
22 2z
aR M
B
aR M
=
+

In summary

()
22 2
RM
MaR
aR M
=−
+Ajk
()
2
22 2
aR
aR M
aR M
=+
+
Bjk
Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at a
given point.
Lastly, if
0 and 0,R M>> it follows from the equations found for A and B that 0 and 0.
yy
AB>>
From Equation (6), 0 (assuming 0).xa<> Then, as a consequence of letting 0,
x
A=force A lies in a plane
parallel to the yz plane and to the right of the origin, while force
B lies in a plane parallel to the yz plane but to
the left of the origin, as shown in the figure below.

166. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 141.



First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and
another axis intersects the prescribed line of action ().AA′ Note that it has been assumed that the line of
action of force
B intersects the xz plane at point
( ), 0, .Pxz Denoting the known direction of line AA′ by
Ax y z
λλλ=++ijkλ
it follows that force
A can be expressed as ( )Axyz
AA λλλ== ++ ijkλA
Force
B can be expressed as x yz
BBB=++Bijk
Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows
that the distance a can be determined. In the following solution, it is assumed that a is known.
Then, for equivalence
:0
x xx
FAB λΣ=+ (1)
:
yyy
FRA BλΣ=+ (2)
:0
zzz
FAB λΣ=+ (3)
:0
x y
M zBΣ=− (4)
:
yzxz
M MaAzBxBλΣ=−+− (5)
:0
zyy
M aA xBλΣ=+ (6)
Since there are six unknowns
(), , , , ,
xyz
ABBBxz and six independent equations, it will be possible to
obtain a solution.

167. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.



Case 1: Let 0z= to satisfy Equation (4)
Now Equation (2)
yy
A RBλ=−
Equation (3)
zz
BA
λ=−
Equation (6) ()
y
y
yy
aA a
x RB
BB
λ 
=− =− − 



Substitution into Equation (5)
() ()
zyz
y
a
MaA RBA
B
λ λ
 
 =− − − − −

 
 

1

y
z
M
A B
aR
λ

∴=−



Substitution into Equation (2)
1
yy y
z
M
RBB
aR
λ
λ

=− +



2

z
y
zy
aR
B
aR Mλ
λλ
∴=
−

Then
zy
yz
MR R
A
aRaR M
M
λλ
λ λ
=− =
−
−
x
xx
zy
MR
BA
aR M
λ
λ
λλ=− =
−
z
zz
zy
MR
BA
aR M
λ
λ
λλ=− =
−
In summary

A
yz
P
aR M
λλ
=
−
A λW
()
xz z
zy
R M aR M
aR Mλλ λ
λλ=++
−
Bijk W
and
z
2
11
y
y z
aR MR
xa a R
B aRλλ
λ
 −
=− =−  

  

or
y
z
M
x
R
λ
λ
= W
Note that for this case, the lines of action of both A and B intersect the x axis.
continued

168. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Case 2: Let 0
y
B= to satisfy Equation (4)
Now Equation (2)
y
R
A
λ
=
Equation (1)
x
x
y
BR
λ
λ

=−



Equation (3)
z
z
y
BR
λ
λ

=−



Equation (6) 0 which requires 0
y
aA aλ
= =
Substitution into Equation (5)
or
xz
zx y
yy
M
MzR xR x z
Rλλ
λλλ
λλ
 

=− −− − = 

 

 

This last expression is the equation for the line of action of force
B.
In summary

A
y
R
λ

=


A
λ
()
xz
y
R
λλ
λ

=−−


Bik
Assuming that , , 0,
xyz
λλλ > the equivalent force system is as shown below.

Note that the component of A in the xz plane is parallel to B.

169. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 142.






(a) Have
/
=
B CB N
M rF
( )( )0.1 m 800 N=
80.0 N m = ⋅
or 80.0 N m
B
=⋅M

W
(b) By definition
/
sin
θ=
BAB
MrP
where ( )90 90 70θ α=°− °− ° −
90 20 10 =°− °− °
60 =°
( ) 80.0 N m 0.45 m sin 60P∴ ⋅= °

205.28 N
=P
or 205 N P=

W
(c) For P to be minimum, it must be perpendicular to the line joining
points A and B. Thus, P must be directed as shown.
Thus
min / min
==
BAB
M dP r P
or ( )
min
80.0 N m 0.45 m⋅= P
min
177.778 N∴=P
or
min
177.8 N=P
20°W

170. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 143.








Have
/CBCB
=Mr F ×
Noting the direction of the moment of each force component about C is
clockwise,

CByBx
M xF yF=+
where 144 mm 78 mm 66 mmx= −=
86 mm 108 mm 194 mmy= +=
and
() ()
()
22
78
580 N 389.65 N
78 86
Bx
F==
+

() ()
()
22
86
580 N 429.62 N
78 86
By
F==
+
( )( )( )( ) 66 mm 429.62 N 194 mm 389.65 N
C
M∴= +
103947 N mm = ⋅
103.947 N m= ⋅
or 103.9 N m
C
=⋅M

W

171. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 144.








(
a) Have
/AEADE
=Mr T ×
where
( )
/
92 in.
EA
=rj

DE DE DE
T=T λ

( )( )( )
() ( ) ( )
()
222
24 in. 132 in. 120 in.
360 lb
24 132 120 in.+−
=
++
ijk


( )( )( )48 lb 264 lb 240 lb=+ −ijk
()() 0 92 0 lb in. 22,080 lb in. 4416 lb in
48 264 240
A
∴= ⋅=− ⋅− ⋅
−
ij k
Mik
or
(
)( )1840 lb ft 368 lb ft
A
=− ⋅ − ⋅Mik W
(b) Have
/AGACG
=Mr T ×
where ( )( )
/
108 in. 92 in.
GA
=+rij
( )( )( )
()()()
()
222
24 in. 132 in. 120 in.
360 lb
24 132 120 in.
CG CG CG
T
−+ −
==
++
ijk
T λ

( )( )( )48 lb 264 lb 240 lb=− + −ijk
108 92 0 lb in.
48 264 240
A
∴ =⋅
−−
ijk
M
( )( )( )22,080 lb in. 25,920 lb in. 32,928 lb in.=− ⋅ + ⋅ + ⋅ijk
or ( )( )( )1840 lb ft 2160 lb ft 2740 lb ft
A
=− ⋅ + ⋅ + ⋅Mijk W

172. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 145.





First note

()()
22
/
2.4 1.8 m 3 m
CA
AC==−+ =r

() ( ) ()
222
/
1.2 2.4 0.3 m 2.7 m
DA
AD== +−+ =r
and
( )( )
/
2.4 m 1.8 m
CA
=− +rjk
( )( )( )
/
1.2 m 2.4 m 0.3 m
DA
=−+rijk
By definition
// //
cos
CA DA CA DA
θ=rr rr⋅
or ( )( )()()2.4 1.8 1.2 2.4 0.3 3 2.7 cos θ−+ − + =jk i jk⋅
()()()()()()0 1.2 2.4 2.4 1.8 0.3 8.1cos θ+− − + =
and
6.3
cos 0.77778
8.1
θ==

38.942
θ= °
or
38.9
θ= °W

173. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 146.








Based on
/OAOBA
=Mr T ×
where

Ox y z
MMM=++Mijk

( )( )100 lb ft 400 lb ft
x
M=+ ⋅− ⋅ijk
()()
/
6ft 4ft
AO
=+rij

BABABA
T=T λ

()()()6ft 12ft
BA
BA
a
T
d
−−
=
ijk


100 400 6 4 0
612
BA
x
BA
T
M
d
a
∴+− =
−−
ijk
ijk
() () ()4696
BA
BA
T
aa
d
 =−+ −
 
ijk
From -coefficient:j 100 6
AB BA
daT=
100
or
6
BABA
Td
a
= (1)
From -coefficient:k 400 96
AB BA
dT−=−
400
or
96
BABA
Td= (2)
Equating Equations (1) and (2) yields
()
()
100 96
6400
a=
or 4.00 ft a= W

174. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 147.








Have
()/DB DB C D CF
M= rTλ⋅ ×
where
( )( )48 in. 14 in.
0.96 0.28
50 in.
DB
−
==−
ij
ijλ


()( )
/
8 in. 16 in.
CD
=−rjk

( )( )()
()
24 in. 36 in. 8 in.
132 lb
44 in.
CF CF CF
T
−−
==
ijk
T
λ

()( )()72 lb 108 lb 24 lb=− −ijk

0.96 0.28 0
0816lbin.
72 108 24
DB
M
−
∴ =−⋅
−−

()()()( )( )()()0.96 8 24 16 108 0.28 16 72 0  = −−− − +− − −
  

1520.64 lb in.=−⋅
or 1521 lb in.
DB
M=− ⋅ W

175. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 148.






(a) Based on : 1000 N
A
FF TΣ==
or 1000 N
A
=F
20°W
( )():sin50
AA A
M MT dΣ=°
( ) ( )1000 N sin 50 2.25 m=°
1723.60 N m = ⋅
or 1724 N m
A
=⋅M

W
(b) Based on : 1000 N
B
FF TΣ==
or 1000 N
B
=F
20°W
( )():sin50
B BB
M MT dΣ=°
( ) ( )1000 N sin50 1.25 m=°
957.56 N m = ⋅
or 958 N m
B
= ⋅M

W

176. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 149.






Require the equivalent forces acting at A and C be parallel and at an angle
of
α with the vertical.
Then for equivalence,
( ):250 lbsin30 sin sin
xAB
FFF α αΣ°=+ (1)
( ):250 lbcos30 cos cos
yAB
FFF α αΣ− °=− − (2)
Dividing Equation (1) by Equation (2),
( )
()
( )
()
250 lb sin 30 sin
250 lb cos30 cos
AB
AB
FF
FF
α
α
°+
=
−°−+

Simplifying yields 30α=°
Based on
( ) ()( )(): 250 lb cos30 12 ft cos30 32 ft
CA
MFΣ°=°

93.75 lb
A
F∴=
or 93.8 lb
A
=F
60°W
Based on
( ) ()() (): 250 lb cos30 20 ft cos30 32 ft
AC
MF Σ− ° = °

156.25 lb
C
F∴=
or 156.3 lb
C
=F
60°W

177. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 150.






Have
:
ABC
Σ=FP F
where

AB AB AB
P=P λ

( )( )( )
()
2.0 in. 38 in. 24 in.
45 lb
44.989 in.
+−
=
ijk

( )( )( )or 2.00 lb 38.0 lb 24.0 lb
C
=+−Fijk W
Have
/
:
CBC AB C
Σ=Mr P M×

229.5 33 0 lbin.
11912
C
= −⋅
−
ijk
M
( )()()()(){2 lb in. 33 12 29.5 12=⋅−−− − ij

()()()()
}29.5 19 33 1 +−−
 
k
( )( )( )or 792 lb in. 708 lb in. 1187 lb in.
C
=⋅+⋅+ ⋅Mijk W

178. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 151.



For equivalence
() ( ): 90 N sin30 125 N cos40
x x
FRΣ− °+ °=
or 50.756 N
x
R= () ( ): 90 N cos30 200 N 125 N sin 40
yy
FRΣ− °− − °=
or 358.29 N
y
R=−
Then ()( )
22
50.756 358.29 361.87 NR=+−=
and
358.29
tan 7.0591 81.937
50.756y
x
R
R
θθ
−
= ==− ∴=−°

or 362 N=R
81.9°W
Also
(
) ( )( ) ( )( ) ( ): 90 N sin 35 0.6 m 200 N cos25 0.85 m 125 N sin 65 1.25 m 0
A
MM   Σ− °− ° − ° =
  
326.66 N mM
∴ =⋅
or 327 N m M= ⋅W

179. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 152.



For equivalence
:
A BC D C
Σ+++=FF F F F R
()()()()5 lb 3 lb 4 lb 7 lb
C
=− − − −Rjjki

(
)()() 7 lb 8 lb 4 lb
C
∴=− − −Rijk W
Also for equivalence
// /
:
CACABCBDCD C′′ ′Σ++=M r Fr Fr F M×××
or
0 0 1.5 in. 1 in. 0 1.5 in. 0 1.5 in. 1.5 in.
0 5 lb 0 0 3 lb 0 7 lb 0 0
C
=−+ −+
−−
ij k i j k i j k
M

( ) ( )( )7.50 lb in. 0 0 4.50 lb in. + 3.0 lb in. 0  =− ⋅− + − ⋅ − ⋅−
  
iik


( )( )10.5 lb in. 0 0 10.5 lb in. +⋅−++⋅
 
jk


( )( )( )or 12.0 lb in. 10.5 lb in. 7.5 lb in.
C
=− ⋅ + ⋅ + ⋅Mijk W

180. COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


Chapter 3, Solution 153.




Have :
ABCD
Σ +++ =FF F F F R
()( )( )( )85 lb 60 lb 90 lb 95 lb−− − − =jjjjR
( ) 330 lb∴=−Rj
Have () () () ()():
x AA BB CC DD H
M Fz Fz Fz Fz RzΣ+++=
()()( )( )( )( )( )()( )( )85 lb 9 ft 60 lb 1.5 ft 90 lb 14.25 ft 95 lb 330 lb 7.5 ft
D
z++ +=
3.5523 ft
D
z∴= or 3.55 ft
D
z= W
Have
() () () ()():
zAA BB CC DD H
M Fx Fx Fx Fx RxΣ+++=
()()( )( )( )( )( )()( )( )85 lb 3 ft 60 lb 4.5 ft 90 lb 14.25 ft 95 lb 330 lb 7.5 ft
D
x++ +=
7.0263 ft
D
x∴= or 7.03 ft
D
x= W