Engineering Mechanics- Sttatics - Concurrent Forces

Three

1. ميحرلا نمحرلا للها مسب

2. Concurrent Forces
in Three Dimensions

3. 3D relations as generalizations of 2D
relations.

4. y
z
x
O
j
i
k
x
y
O
j
i
),,(
AAA
zyxA( , )
A A
Ax y

5. A = A
x i + A
y j + A
z k
222
zyx
AAAA 
A = A
x i + A
y j
2 2
x y
A A A 
cos
x
x
A
A

cos
y
y
A
A

cos
z
z
A
A

cos
x
x
A
A

cos
y
y
A
A


6. ( ) ( ) ( )
AB B A B A B Ax x y y z z     r i j k
( ) ( )
AB B A B Ax x y y   r i j
yx
A
AA
A A A
  
A
u i j
yx z
A
AA A
A A A A
   
A
u i j k

7. cos cos
x yA   u i j
cos cos cos
x y zA     u i j k
2 2
cos cos 1
x y  
2 2 2
cos cos cos 1
x y z    
2 2
( ) ( )
( ) ( )
AB B A B A
AB
AB
A B A B
x x y y
x x y y
  
 
  
r i j
u
r
2 2 2
( ) ( ) ( )
( ) ( ) ( )
AB B A B A B A
AB
AB
A B A B B A
x x y y z z k
x x y y z z
    
 
    
r i j
u
r

8. ( ) ( )
x yF F  R i j
( ) ( ) ( )
x y zF F F    R i j k
0
0
x
y
F
F




0
0
0
x
y
z
F
F
F







9. double projection case
x
y
z


F
O
A
B
C

10. x
y
z
0
90
Fz
F
xyF
O
A
C
B

cos
sin
z
xy
F F
F F





11. x
y
z

zF
x yF
yF
x
F
O
C
D
E
cos
sin cos
x xyF F
F

 


sin
sin sin
y xyF F
F

 



12. Example 3.4Express the force vector F shown in Figure
(a) as a Cartesian vector. Also define its
direction.

13. 6.8660sin100 
o
zF
4.3545cos5045cos 
oo
xyxFF
5060cos100 
o
xyF
4.3545sin5045sin 
oo
xyyFF

14.  N 6.86 5.43 5.43 kjiF 
N 100)6.86(5.4)3(5.4)3(
222
F
354.0 cos
o
3.69
354.0 cos
o
111
866.0 cos
o
30

15. Example 3.2
Among the wires attached to the tower
shown, three wires AB, AC and AD are
attached to point A which lies on z axis. The
magnitudes of the tensions in these cables
are 50 N, 300 N and 400 N, respectively. If
cable AD is parallel to the y axis, determine
the magnitude and direction angles of the
resultant of these tensions at point A.

16. 2
4
1
2
60
o
45
o
A
B
C
D
x
y
z
A(0,0,12)
C(2,-4,0)

17. 4
1
2
60
o
45
o
A
B
C
D
x
y
z
as cable AD is parallel to the y axis yet with
opposite sense then
T
AD = -400 j

18. A(0,0,12)
C(2,-4,0)
AC = 2 i - 4 j - 12 k
2 4 12
4 16 144
AC
 

 
i j k
u
300
(2 4 12 )
12.81
AC
  T i j k
T
AC
= (46.85 i -93.7 j - 281.1 k) N

19. 60
o45
o
A
B
x
y
z
5
0
s
i
n
6
0
o
T
AB(z = -50 sin (60
o
) = - 43.3 N
T
AB(xy =50 cos (60
o
) = 25 N

20. 45
o
Bx
y
z
25sin45
o
T
AB(x =25 cos (45o) =17.68 N
T
AB(y =25 sin (45o) =17.68 N
T
AB =(17.68 i + 17.68 j -43.3 k) N

21. R =( 17.68 +46.85) i + (17.68-93.7-400) j + (-43.3 - 281.1) k N
R = (64.53 i – 476 j - 324.4 k) N
R = 579.6 N
64.53
cos
579.6

1
cos (0.11) 83.6
o


 
476
cos
579.6



1
cos ( 0.82) 145.2
o


  
324.4
cos
579.6



1
cos ( 0.56) 124
o


  

22. Example
3.7
The 100 kg crate shown is supported as
shown. Determine the tension in cords AC
and AD and the stretch of the spring.
)0 ,0 ,0(A
( 1, 2, 2)D

23. N 981 )81.9)(100( kkW 

B BF NF i
 cos 120 cos 135 cos 60
o o o
C C
F  F i j k
 N 5.0 707.0 5.0 kji 
CF

24.  N 667.0 667.0 333.0 kji 
DF





 

3
2 2 1

kji
uF
DADDD
FF
)0 ,0 ,0(A
( 1, 2, 2)D

25. 0
x
F 0333.0 5.0 
DCB
FFF
0
y
F 0 667.0 707.0 
DC FF
0
z
F 0981 667.0 5.0 
DC
FF
CD
FF 06.1
N 813
CF
N 628 
DF
694 N
B
F

26. LkF
B

694
0.462 m
1500
L  
The force F
B
is expressed in newtons
(N) while the spring constant is
expressed in kilo newton per meter
(kN/m).