solvedproblemsonforcesystem-240328110705-688f382f.pdf

Good one for practice

1. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

Type 1: Problems on Composition of Forces by
Parallelogram and Triangle Law
Formulas :
1. Parallelogram Law :
2. Triangle Law :
Sr. No. Examples with solution
1. Find the resultant of the given forces.
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
1
: Problems on Composition of Forces by
Parallelogram and Triangle Law
Parallelogram Law :
Triangle Law :
Examples with solution
Find the resultant of the given forces.

: Problems on Composition of Forces by
Parallelogram and Triangle Law

2. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

Solution Case 1: By Parallelogram Law
?????? = ?4
6
+ 3
6
R = 5 N
tan ∝ =
∝ =

2. Find the resultant of the given forces.

Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
2
Case 1: By Parallelogram Law
6
+ 2 × 3 × 4cos90°
R = 5 N

3 sin90°
4+3cos90°

=36.87°
Case 2: By Triangle
By Cosine rule
?????? = ?4
6
+ 3
6
−
R = 5 N
By Sine rule
??????
sin90°
sin∝
∝ =
Find the resultant of the given forces.

Triangle Law

By Cosine rule
− 2×3×4cos90°
R = 5 N
By Sine rule
°
=
3
sin∝

∝=
3
5

36.87°

3. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

Solution Case 1: By Parallelogram Law
??????
= ?50
6
+ 70
6
R =
tan ∝ =
∝

3. A block is pulled by means of
tension in the rope AB is 2500 KN. Knowing that the
resultant of the two forces at A is directed along X
determine the tension in rope AC and the magnitude of their
resultant force at A.
Solution Given Data:
Q = TAB = 2500 KN, θ = 50
To Find : R and T
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
3
Parallelogram Law

6
+ 2 × 50 × 70cos60°
R = 104.4 N

70 sin60°
50+70cos60°

∝ = 35.5°
Case 2: By Triangle
By Cosine rule
?????? = ?4
6
+ 3
6
R = 5 N
By Sine rule
??????
sin90
sin
∝ =
A block is pulled by means of two ropes as shown in fig. the
tension in the rope AB is 2500 KN. Knowing that the
resultant of the two forces at A is directed along X
determine the tension in rope AC and the magnitude of their
resultant force at A.
Given Data: Let P = TAC
= 2500 KN, θ = 50ᵒ , α = 20ᵒ
R and TAC
Triangle Law

By Cosine rule
− 2×3×4cos90°
R = 5 N
By Sine rule
90°
=
3
sin∝

sin∝=
3
5

=36.87°
two ropes as shown in fig. the
tension in the rope AB is 2500 KN. Knowing that the
resultant of the two forces at A is directed along X- axis,
determine the tension in rope AC and the magnitude of their

4. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
4

By using Parallelogram Law
Magnitude of resultant is given as

?????? = ???????
6
+ 2500
6
+ 2 ×??????× 2500cos50°
Squaring on both sides
??????
6
= ??????
6
+2500
6
+2×??????×2500cos50°………………eq
n
1
Direction of resultant is given by
tan20° =
2500 sin50°
??????+2500cos50°

Cross multiplication gives
[??????+2500cos50°]tan20° = 2500sin50°
0.3639 P + 584.88 = 1915.11
??????=
1915.11 – 584.88
0.3639

P = 3656.69 KN
Put value of P in eq
n
1 we get
??????
6
= 3656.69
6
+2500
6
+2×3656.69×2500cos50°
R = 5601.22 KN
Final Answer:
1. Tension in Rope AC = 3656.69 KN
2. Magnitude of Resultant R = 5601.22 KN
Assignment:
1. A boat is moved uniformly along a canal by two horses
pulling with forces P = 890 N and Q = 1068 N acting at
an angle α = 60ᵒ as shown in the fig. Determine magnitude of
the resultant pull on the boat and the angle β and γ as shown.

5. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM


(ANS: R = 1698 N,

???????????????????????? =
??????
??????+
2. Two forces of magnitude 300 N and 500 N is 50
500 N force being horizontal. Determine the resultant
magnitude and direction if 300 N force is a push and 500 N
force is a pull.

(ANS: R =





Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
5
(ANS: R = 1698 N, β=33ᵒ, γ = 27ᵒ)
??????????????????????????????
+??????????????????????????????

Two forces of magnitude 300 N and 500 N is 50
500 N force being horizontal. Determine the resultant
magnitude and direction if 300 N force is a push and 500 N
force is a pull.
(ANS: R = 383.61 KN, α=36.8ᵒ)



Two forces of magnitude 300 N and 500 N is 50
ᵒ
, the
500 N force being horizontal. Determine the resultant in
magnitude and direction if 300 N force is a push and 500 N

6. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
6

Type 2: Problems on Resolution of Forces by
Parallelogram and Triangle Law
1. Two forces act at an angle of 120ᵒ. The bigger force is of 40
N and the resultant is perpendicular to the smaller one. Find
the smaller force.

Solution Given Data:
1. Angle between the forces ∆AOC = θ = 120ᵒ
2. Bigger force (P) = 40 N
3. Angle between resultant (R) and smaller force (Q) ∆BOC =β=
90ᵒ
Step 1: To find the ∆AOB
From geometry of fig, we can find the ∆AOB,
α = θ- β = 120ᵒ - 90ᵒ = 30ᵒ
Step 2: To find Q
we know that,
tan ∝ =
?????? sin??????
??????+??????cos??????

tan30ᵒ =
?????? sin120°
40+??????cos120°

0.577 =
0.866 ??????
40−0.5 ??????

0.577*(40-0.5Q) = 0.866 Q

7. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
7

Q = 20 N
ANS: The smaller force Q = 20N
2. Resolve the 2000 N force into two oblique components one
acting along AB and the other acting along BC.

Solution Given Data:
1. Angle between the forces ∆AOC = 120ᵒ
2. Bigger force (F1) = 40 N
3. Angle between resultant and smaller force (F2) ∆BOC = 90ᵒ
Step 1: To find the θ
By triangle law, we have
tan θ = 2/3
θ = tan
?5

2
3

θ = 33.69ᵒ

Step 2: To Find Forces

8. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
8


By sine rule, we have
2000
sin33.69°
=
??????????????????
sin90°
=
??????????????????
sin56.31°

2000
sin33.69°
=
??????????????????
sin90°

??????????????????=
2000 sin90°
sin33.69

??????????????????=
2000 ∗ 1
0.5546

??????????????????=
2000 sin56.31°
sin33.69


FAB = 3605.56 N
FBC = 3000 N
Assignment
1. Find the magnitude of the two forces, such that
if they act at right angles, their resultant is √10
N. but if they act at 60ᵒ, their resultant is √13N.
10 = ??????
6
+??????
6
………………………eq1

13 = ??????
6
+??????
6
+????????????……………………eq2

PQ= 3……………………………….eq3
??????=
7
?
ANS: Q = 3N & P = 1 N)

9. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
9

Type 3: Problems on Resolution of Forces
into Rectangular and Oblique components of
force
Formulas for Rectangular Components
 Fx = F. Cos θ
 Fy = F. Sin θ

Formulas for Oblique Components
 ??????1=
? qgl
qgl(> )

 ??????2 =
? qgl
qgl(> )

 Or you may also apply Parallelogram law & Triangle Law
to this type of problem.
1. Resolve a force of 30N acting North-East away from the
point.
Solution Given Data:
1. F = 30 N acting N-E

10. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

Step 1: To find
Step 2: To Find
2. A force P is acting on a block as shown in fig. if horizontal
rectangular component of P is 40 N acting to the
find the y component of P.
Solution Given Data:
1. Horizontal rectangular component of P
2. tan??????=
3. FBD
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
10
To find Fx
Fx = F cos θ = 30 cos 45ᵒ
= 21.21 N
To Find Fy
Fy = F sin θ = 30 sin 45ᵒ
= 21.21 N
A force P is acting on a block as shown in fig. if horizontal
rectangular component of P is 40 N acting to the
find the y component of P.
Given Data:
Horizontal rectangular component of P = 40 N
=
7
8
= 36.87°



A force P is acting on a block as shown in fig. if horizontal
rectangular component of P is 40 N acting to the left then

= 40 N

11. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
11

Step 1: To find magnitude of P
Fx = F Cos θ
40 = P cos θ
40 = P cos (30+36.87)

??????=
84
amq::.<;°

P = 101.82 N
Step 2: To Find Py
Py = P sin θ = 101.82 sin 30ᵒ
Py = 50.91 N
3. A force 360 N is acting on a block as shown in fig. find the
components of forces along the x-y axis which are parallel
and perpendicular to the inclined.


Solution Given Data:
1. Force P = 360 N
2. Θ1 = tan
-1
¾ = 36.87ᵒ
3. Θ2 = tan
-1
2/3 = 33.69ᵒ
4

12. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
12


To Find : Px & Py
Step 1: To find magnitude of Px

Px = P cos (θ1+ θ2)
= 360 cos (70.56)
Px = 119.82 N
Step 2: To Find magnitude of Py
Py = P sin (θ1+ θ2)
= 360 sin (70.56)
Py = 339.48 N
ASSIGNMENT
1. Resolve a force of 40 KN inclined at 150ᵒ with x-axis
acting towards the point.
(ANS: Fx= 34.64 KN Fy= -20 KN)
2. Resolve the 2500 N force acting vertically on wedge
having inclination 22ᵒ as shown in fig. into two
components one acting along inclined 22ᵒ and the other

13. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

perpendicular to the inclined.
(ANS: Fx=
Hint FBD:
Fx = F sin 22
Fy = F cos 22
OR Angle from x
Fx= F cos 68
Fy= F sin 68
Problems on non
1. A force of 80 N is acting on a body. Find its components such
that one component has an angle of 45
angle of 30ᵒ
Solution Given Data:
1. Force F
2. ?????? = 45°
3. ?????? = 30°
To Find : F1 &
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
13
perpendicular to the inclined.
(ANS: Fx= 936.52 N Fy= 2318 N)
Hint FBD:
Fx = F sin 22
Fy = F cos 22
OR Angle from x-axis 90-22 =68
Fx= F cos 68
Fy= F sin 68
on non-perpendicular component
A force of 80 N is acting on a body. Find its components such
that one component has an angle of 45ᵒ & the other has an
with the force 80 N.
Given Data:
F = 80 N
°
°
F1 & F2

perpendicular component
A force of 80 N is acting on a body. Find its components such
& the other has an

14. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

Step 1: To find magnitude of
Step 2: To Find magnitude of
2. A force of 2000 N is acts at an angle of 60°
Find its components along 105° & 330° with X
Solution Given Data:
1) F = 2000 N
2) α = 105
3) β = (360
Step 1: To find magnitude of F1
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
14
To find magnitude of F1
????????????=
?????? ????????????????????????
??????????????????(?????? +??????)

??????1 =
80 sin30°
sin(45°+30°)

F1 = 41.41 N
To Find magnitude of F2
???????????? =
?????? ????????????????????????
??????????????????(?????? +??????)

??????2 =
80 sin45°
sin(45°+30°)

F2 = 58.56 N
A force of 2000 N is acts at an angle of 60° with X
Find its components along 105° & 330° with X
Given Data:
= 2000 N
105 – 60 = 45°
(360-330) + 60 = 90°
To find magnitude of F1
???????????? =
?????? ????????????????????????
??????????????????(?????? +??????)

??????1=
2000 sin90°
sin(45°+90°)

with X-axis.
Find its components along 105° & 330° with X-axis.

15. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

Step 2: To Find magnitude of F2
Ans:
1. Component along 105°
2. Component along 330° = 2000 N
3. The resultant of two forces in a plane is 800 N at 60° with x
axis. One force is 160N at 30° with x axis. Determine the
missing force & its inclination.
Solution Given Data:
1. F = 800 N
2. F1 = 160 N
3. α = 60
To find: F2 & β

Step 1: To find β

Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
15
F1 = 2828.43 N
To Find magnitude of F2
???????????? =
?????? ????????????????????????
??????????????????(?????? +??????)

??????2 =
2000 sin45°
sin(45°+90°)

F2 = 2000 N
Component along 105° = 2828.43 N
Component along 330° = 2000 N
The resultant of two forces in a plane is 800 N at 60° with x
force is 160N at 30° with x axis. Determine the
missing force & its inclination.
Given Data:
F = 800 N
F1 = 160 N
α = 60 – 30 = 30°


F2 & β
To find β
???????????? =
?????? ????????????????????????
??????????????????(?????? +??????)


The resultant of two forces in a plane is 800 N at 60° with x-
force is 160N at 30° with x axis. Determine the

16. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
16

160=
800 sin??????
sin(30+??????)

160 (sin (30+β)) = 800 sin β
80 cos β = 661.44 sin β
sin??????
cos??????
=
80
661.44

tan??????= 0.1209
β = 6.89°
Step 2: To Find magnitude of F2
????????????=
?????? ????????????????????????
??????????????????(??????+??????)

??????2 =
800 sin30
sin(30+6.89)

F2 = 666.35 N
ASSIGNMENT
1. Resolve a force of 60 N into two components F and 80
N as shown in fig. Find the value of F and θ

(ANS: F = 90.82 N and θ = 79.46°)

17. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
17

Type 4: Problems on Resultant of concurrent
force system using Method of Resolution
 If the number of forces is more than two, then its resultant can
be found out conveniently by the METHOD OF RESOLUTION.
 STEPWISE PROCEDURE OF METHOD OF RESOLUTION:
1. Resolve all forces horizontally and find the algebraic sum of all
the horizontal components (i.e., ΣFx)
2. Resolve all forces vertically and find the algebraic sum of all
the vertical components (i.e., ΣFy).
3. The resultant R of the given forces will be given by the
equation:
??????= ?(??????????????????
??????
)+(??????????????????
??????
)
4. The resultant force will be inclined at an angle θ, with the
horizontal, such that
tan??????=,
ΣFy
ΣFx
,
5. Position of the resultant

18. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

1. The forces 20 N, 30 N, 40 N, 50 N and 60 N are acting at
one of the angular points of a regular hexagon, towards the
other five angular points, taken in order. Find the
and direction of the resultant force.
Solutio
n
Included angle of any regular polygon = 180
Step 1: Resolving all the forces horizontally (ΣFx)
?????????????????? = 20cos
+
= 36 N

Step 2: Resolving all the forces vertically (ΣFy)
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
18
The forces 20 N, 30 N, 40 N, 50 N and 60 N are acting at
one of the angular points of a regular hexagon, towards the
other five angular points, taken in order. Find the
and direction of the resultant force.

Included angle of any regular polygon = 180
= 180−
360
6
= 120°
?????? =
120
4
= 30°
Resolving all the forces horizontally (ΣFx)
cos0°+ 30 cos30°+ 40cos60°+ 50
+ 60 cos120°
= 36 N
Resolving all the forces vertically (ΣFy)

The forces 20 N, 30 N, 40 N, 50 N and 60 N are acting at
one of the angular points of a regular hexagon, towards the
other five angular points, taken in order. Find the magnitude

Included angle of any regular polygon = 180-
7:4
??.?? ?????

Resolving all the forces horizontally (ΣFx)
50 cos90°
Resolving all the forces vertically (ΣFy)

19. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

??????????????????= 20
= 151.6 N
Step 3: Magnitude of the resultant force,
Step 4: Direction of the resultant force,
Step 5: Position of the resultant force,
2. The following forces act at a point :
i. 20 N inclined at 30
ii. 25 N towards North,
iii. 30 N towards North West, and
iv. 35 N inclined at 40
Find the magnitude and direction of the resultant force.
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
19
20sin0°+ 30 sin30°+ 40sin60°+
+ 60 sin120°
= 151.6 N
Magnitude of the resultant force,
?????? = ?(??????????????????
??????
)+(??????????????????
??????
?????? = ?(36)
6
+(151.6)
6
?????? = 155.8 ??????
Direction of the resultant force,
???????????????????????? = ,
??????????????????
??????????????????
,
tan?????? =
151.6
36

?????? = 76.6°
Position of the resultant force,

The following forces act at a point :
20 N inclined at 30ᵒ towards North of East,
25 N towards North,
towards North West, and
35 N inclined at 40ᵒ towards South of West.
Find the magnitude and direction of the resultant force.
+ 50 sin90°
)
6

,
towards North of East,
towards South of West.
Find the magnitude and direction of the resultant force.

20. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
20

Solutio
n

Step 1: Resolving all the forces horizontally (ΣFx)
??????????????????= 20cos30°+ 25 cos90°+ 30cos135°+ 35 cos220°

= - 30.7 N
Step 2: Resolving all the forces vertically (ΣFy)
??????????????????= 20sin30°+ 25 sin90°+ 30sin135°+ 35 sin220°

= 33.7 N
Step 3: Magnitude of the resultant force,
??????= ?(??????????????????
??????
)+(??????????????????
??????
)
??????= ?(−30.7)
6
+(33.7)
6

??????=45.6 ??????
Step 4: Direction of the resultant force,
????????????????????????= ,
??????????????????
??????????????????
,
tan??????= ,
33.7
−30.7
,
??????=47.7°
Step 5: Position of the resultant force,

21. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

Since ΣFx is negative and ΣFy is positive, Resultant lies
between second
Therefore acute angle of the resultant = 180
3. The Striker of carom board laying on the board is being
pulled by four players as shown in fig. the players are sitting
exactly at the centre of the four sides. Determine the
resultant of forces in magnitude and direction.
Solutio
n
General calculation
?????????????????????????????? =
??????????????????
??????????????????
??????????????????????????????=
??????????????????
??????????????????
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
21
Since ΣFx is negative and ΣFy is positive, Resultant lies
between second quadrants.

Therefore acute angle of the resultant = 180
Striker of carom board laying on the board is being
pulled by four players as shown in fig. the players are sitting
exactly at the centre of the four sides. Determine the
resultant of forces in magnitude and direction.

General calculation : Respective angles of each player
??????????????????
??????????????????
= ????????????.??????°
??????????????????
??????????????????
=????????????.??????°
Since ΣFx is negative and ΣFy is positive, Resultant lies

Therefore acute angle of the resultant = 180ᵒ - 47.7ᵒ = 132.3ᵒ
Striker of carom board laying on the board is being
pulled by four players as shown in fig. the players are sitting
exactly at the centre of the four sides. Determine the
resultant of forces in magnitude and direction.

Respective angles of each player

22. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

??????????????????????????????=
??????????????????
??????????????????
?????????????????????????????? =
??????????????????
??????????????????

Step 1: Resolving all the forces
horizontally (ΣFx)
?????????????????? = 20
= 20.25
Step 2: Resolving all the forces vertically (ΣFy)
?????????????????? = 20sin
= 20.89
Step 3: Magnitude of the resultant
Step 4: Direction of the resultant force,
Step 5: Position of the resultant force,
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
22
??????????????????
??????????????????
=????????????.????????????°
??????????????????
??????????????????
= ????????????.??????°

Resolving all the forces
horizontally (ΣFx)
20cos16.7°+ 25 cos79.7°− 10
+ 15 cos68.2°
20.25 N
Resolving all the forces vertically (ΣFy)
sin16.7°+ 25 sin79.7°+ 10sin26.56
20.89 N
Magnitude of the resultant force,
?????? = ?(??????????????????
??????
)+(??????????????????
??????
?????? = ?(20.25)
6
+(20.89
?????? = 29.09 ??????
Direction of the resultant force,
???????????????????????? = ,
??????????????????
??????????????????
,
tan?????? = ,
20.89
20.25
,
?????? = 45.89°
Position of the resultant force,
10cos26.56°
Resolving all the forces vertically (ΣFy)
56°− 15 sin68.2°

)
89)
6

,
,

23. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

Since ΣFx and ΣFy is positive, Resultant lies
quadrants.
4. For the system shown, determine
i) The required value of α if resultant of three forces
is to be vertical and
ii) The corresponding magnitude of resultant.
Step I) ƩFx = 0
100cos?????? + 150
150(cos??????.cos
30 cos?????? − 75
Step ii) R = Ʃ
?????? = −100
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
23
Since ΣFx and ΣFy is positive, Resultant lies
the system shown, determine
The required value of α if resultant of three forces
is to be vertical and
The corresponding magnitude of resultant.

Fx = 0
150cos(?????? +30)−200cos?????? = 0
cos30−sin??????.sin30)−100cos?????? = 0
75sin?????? = 0
tan?????? =
sin??????
cos??????
=
30
75
= 21.8°
ƩFy
100 sin21.8−150sin(21.8+30)−
R = -229.29 N
Since ΣFx and ΣFy is positive, Resultant lies in first

The required value of α if resultant of three forces
The corresponding magnitude of resultant.



200sin21.8

24. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
24


(ANS: α = 21.8ᵒ & R = 229.29 N)
5.

25. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

Type 5: Problems on
1. Find the moment of force 500 N about
respectively as shown in fig
Solution Step 1: Free Body Diagram
Step 2: Finding moments
1. Moment about point O
2. Moment about point A
??????@??????=
3. Moment about
??????@
4. Moment about point C
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
25
: Problems on Moment of force
Find the moment of force 500 N about point O, A,B & C
respectively as shown in fig
Free Body Diagram

Finding moments
Moment about point O
??????@?????? = 500cos36.87 ×3 =1200
Moment about point A
= −500sin36.87×2+500cos36.87
Moment about point B
@?????? = −500sin36.87×4+500cos
Moment about point C
Moment of force
point O, A,B & C


1200 ??????.??????
87×3=600 ??????.??????
36.87×3=0

26. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
26

??????@??????= −500sin36.87×4+500cos36.87×1.5=−600??????.??????
M@C = 600 N.m
5. Moment about point D
??????@??????= −500sin36.87×4= −1200 ??????.??????
M@D = 1200 N.m
6. Moment about point E
??????@??????= 0
2. Find the moment of force 2000 N about point ‘O’ as shown in
fig

Solution Step 1: Free Body Diagram

Step 2: Finding moments
Moment about point O
??????@??????=−2000sin30 ×5=−5000 ??????.??????

27. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
27

3. Find the moment of force 50 N about point ‘O’ as shown in fig


Solution Step 1: Free Body Diagram

Step 2: Finding moments
Moment about point O
??????@??????=−2000sin30 ×5=−5000 ??????.??????

4. Find the moment of forces as shown in fig. on lever about
point “O”.

28. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
28


Solution Step 1: Free Body Diagram

Step 2: Finding moments
Moment about point O

29. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

Type
1. Find the Couple of system shown in fig
Solution Step 1: Finding
Step 2: Finding moments
I. Moment about point A
II. Moment about point B
III. Moment about point C
IV. Moment about point
From moments I, II, III & IV
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
29
Type 6: Problems on Couple
Couple of system shown in fig
Finding Moment of Couple
Finding moments about point
Moment about point A
Moment about point B
Moment about point C
Moment about point D
moments I, II, III & IV
Couple

30. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
30

Note: The above example shows that moment
of couple is constant. Hence the couple is
treated as free vector which can be
represented anywhere on a rigid body.
2. Find the Couple of system shown in fig

Solution Step 1: Finding Moment of Couple


3. Find the Couple of system shown in fig

Solution Step 1: Finding Moment of Couple

4. Find the Couple of system shown in fig

31. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
31


Solution Step 1: Finding Moment of Couple

32. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

Type 7: Problems on
force system using Method of Resolution
 STEPWISE PROCEDURE OF METHOD OF RESOLUTION:
1. Find resultant R
parallel forces considering proper sign convention.
2. Find ΣMo. Take the algebraic sum moment of forces
point (say O)
(+????????????
3. Apply Varignon’s theorem,
Where d = perpendicular distance between line of action of
R and reference point O
4. Position of the resultant
Resultant may
O at a distance d, depending on the sign of ΣF & ΣMo.
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
32
: Problems on Resultant of parallel
force system using Method of Resolution
STEPWISE PROCEDURE OF METHOD OF RESOLUTION:
resultant R = Σ F. Take the algebraic sum of all the
parallel forces considering proper sign convention.
(+???????????? ↑) ( −???????????? ↓)
Find ΣMo. Take the algebraic sum moment of forces
(say O) considering proper sign convention.
????????????↻ −??????????????????????????????????????????????????????) ( −???????????? ↺ −??????????????????????????????????????????????????????????????????????????????
Apply Varignon’s theorem, ????????????
??????= ?????? ×??????
Where d = perpendicular distance between line of action of
R and reference point O.
Position of the resultant with respect to point O.
Resultant may lie to the right or left of the reference point
O at a distance d, depending on the sign of ΣF & ΣMo.
Resultant of parallel
force system using Method of Resolution
STEPWISE PROCEDURE OF METHOD OF RESOLUTION:
= Σ F. Take the algebraic sum of all the
parallel forces considering proper sign convention.
Find ΣMo. Take the algebraic sum moment of forces about a
ing proper sign convention.
??????????????????????????????????????????????????????????????????????????????)
Where d = perpendicular distance between line of action of
with respect to point O.
or left of the reference point
O at a distance d, depending on the sign of ΣF & ΣMo.

33. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

1. Find the resultant of following force system and also find the
equivalent force and couple at point A of the same force
system shown in fig.
Solution Case I : To find resultant of
Step 1: To find magnitude of Resultant R
R = ΣFy = - 70 + 100 + 50
= 50 N
Step 2: To find ΣMo
Taking moment about point “O”
????????????
?=100
= 394 N
Step 3: Applying Varignon’s theorem
Step 4: Position
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
33
Find the resultant of following force system and also find the
equivalent force and couple at point A of the same force
system shown in fig.
Case I : To find resultant of force system
To find magnitude of Resultant R
70 + 100 + 50 – 86-34 + 90
= 50 N (↑)
To find ΣMo
Taking moment about point “O”
100×1.5+50×3.5−86×6.5−34
= 394 N-m (↺)
Applying Varignon’s theorem,
????????????
?= ?????? ×??????
?????? =
????????????
?
??????
=
394
50
= 7.88 ??????
Position of the resultant force,
Find the resultant of following force system and also find the
equivalent force and couple at point A of the same force

×8+90×10

34. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

Case II : To find
Step 1: To find magnitude of Resultant R
R = ΣF = - 70 + 100 + 50
= 50 N
Step 2: To find ΣM
Taking moment about point “A”
????????????
?=70
= 219
Step 3: Position of the equivalent force and couple,
2. Find the resultant of
w.r.t. point B.
Solution Step 1: To find magnitude of Resultant R
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
34
: To find equivalent force and couple at point A
To find magnitude of Resultant R
70 + 100 + 50 – 86 - 34 + 90
= 50 N (↑)
To find ΣMA
Taking moment about point “A”
70×3.5 −100×2−86×3−34× 4
219 N-m (↺)
Position of the equivalent force and couple,
Find the resultant of given active forces as shown in fig.
w.r.t. point B.

To find magnitude of Resultant R
equivalent force and couple at point A
4.5+90×6.5
Position of the equivalent force and couple,

given active forces as shown in fig.

35. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
35

R = ΣF = 100 + 200 -150
= 150 N (→)
Step 2: To find ΣMB
Taking moment about point “B”
????????????
?= 150+100×5−150×3.5+200×1.5
= 425 N-m = 425 N-m (↻)
Step 3: Applying Varignon’s theorem,
????????????
?=??????×ℎ
ℎ =
????????????
?
??????
=
425
150
= 2.83 ??????
Step 4: Position of the resultant force w.r.t point B,

3. Find the resultant of given active forces as shown in fig. and
show its position w.r.t. point A.

36. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
36

Solution Step 1: To find magnitude of Resultant R
Here we will consider AC as x-axis
R = ΣF = - 50 + 60 – 70 + 80 – 90
= - 70 N
= 70 N
Step 2: To find ΣMA
Taking moment about point “A”
????????????
?= −60×3+70×6−80×7.5 +90×9
= 450 N-m = 450 N-m (↻)
Step 3: Applying Varignon’s theorem,
????????????
?=??????×??????
??????=
????????????
?
??????
=
450
70
= 6.429 ??????
Step 4: Position of the resultant force w.r.t point A,

4. Determine the resultant of the parallel forces as shown in fig
and locate it w.r.t. O, radius is 1m.

37. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
37


Solution Step 1: To draw a FBD of given fig

For 80 N force and 150 N force
sin??????=
??????????????????
????????????????????????

sin30° =
??????
5
1

x1 = 0.5 m
For 100 N force
sin60° =
??????
6
1

x2 = 0.866 m
For 50 N force
sin45°=
??????
7
1

38. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
38

x3 = 0.707 m
Step 2: To find magnitude of Resultant R
R = ΣF = - 100 – 80 + 150 - 50
= 80 N (←)
Step 3: To find ΣMo
Taking moment about point “O”
????????????
?= 100×0.866+80×0.5+150×0.5−50×0.707
= 166.25 N-m (↺)
Step 3: Applying Varignon’s theorem,
????????????
?=??????×??????
??????=
????????????
?
??????
=
166.25
80
= 2.078 ??????
Step 4: Position of the resultant force,

5. A part roof truss is acted by wind and other forces as shown
in figure. All the forces form a parallel force system and are
perpendicular to portion AB of the truss. Find the resultant
of the force and its location w.r.t. hinge A.

39. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
39


(Ans : R = 1000 N and d = 0)
6. Replace the force system acting on a bar as shown in fig. by a
single force.

(Ans : R = 80 N and d = 2.375 m)

40. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
40

Type 6: Problems on Resultant of general
force system using Method of Resolution
 STEPWISE PROCEDURE OF METHOD OF RESOLUTION:
1. Resolve all forces horizontally and find the algebraic sum of all
the horizontal components (i.e., ΣFx)
2. Resolve all forces vertically and find the algebraic sum of all
the vertical components (i.e., ΣFy).
3. Find the magnitude of Resultant R
The resultant R of the given forces will be given by the
equation:
??????= ?(??????????????????
??????
)+(??????????????????
??????
)
4. Find direction θ. The resultant force will be inclined at an
angle θ, with the horizontal, such that
tan??????=,
ΣFy
ΣFx
,
5. Find ΣMo. Take the algebraic sum moment of forces about a
point (say O) considering proper sign convention.
(+????????????↻ −??????????????????????????????????????????????????????) ( −???????????? ↺ −??????????????????????????????????????????????????????????????????????????????)
6. Apply Varignon’s theorem,
I. ????????????
??????=?????? ×??????,Where d = perpendicular distance between
line of action of R and reference point O. OR

41. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
41

II. ????????????
??????=??????????????????×??????, Where y = distance between point O and
the intersection of line of action of resultant R with y-axis.
OR
III. ????????????
??????=??????????????????×??????, Where y = distance between point O and
the intersection of line of action of resultant R with x-axis.
7. Position of the resultant with respect to point O.
Depending upon the sign of ΣFx, ΣFy and ΣMo any one possible
position of R among the following eight may arise.
1

2

42. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
42

3

4

5

6

43. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
43

7

8

44. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
44

1. Replace the system of forces and couple shown in fig. by a
single force couple system at A.

Solution Step 1: Resolving all the forces horizontally (ΣFx)
??????????????????= −100cos36.87°−75
= 155 N (←)
Step 2: Resolving all the forces vertically (ΣFy)
??????????????????= −200+ 30− 100 sin36.87°
= 210 N (↓)
Step 3: Magnitude of the resultant force,
??????= ?(??????????????????
??????
)+(??????????????????
??????
)
??????= ?(155)
6
+(210)
6

??????= 261 ??????
Step 4: Direction of the resultant force,
????????????????????????= ,
??????????????????
??????????????????
,
tan??????=
210
155

??????= 53.57°
Step 5: Find ΣMA
????????????
?= 50×2+80− 100sin36.87°

45. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
45

2. Replace the system of forces shown in fig. by a single force.

Solution General Calculations:
tan??????1 =
??????????????????
??????????????????
=
2
1.5

1) ??????1 = tan
?5
6
5.9
= 53.13°
2) ??????2 = tan
?5
6
7
= 33.69°
3) ??????3 = tan
?5
5
7
= 18.44°
4) ??????4 = tan
?5
6
5.9
= 53.13°
5) ??????5 = tan
?5
7
6.9
= 50.13°
Step 1: Resolving all the forces horizontally (ΣFx)
??????????????????= 65 cos53.13°+ 140cos33.69−90cos18.44
−100cos53.13+50cos50.13
= 42.12 N (→)
Step 2: Resolving all the forces vertically (ΣFy)
??????????????????= −65 sin53.13° + 140sin33.69−90sin18.44
+100sin53.13−50sin50.13
= 38.78 N (↑)

46. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
46

Step 3: Magnitude of the resultant force,
??????= ?(??????????????????
??????
)+(??????????????????
??????
)
??????= ?(42.12)
6
+(38.78)
6

??????= 57.25 ??????
Step 4: Direction of the resultant force,
????????????????????????= ,
??????????????????
??????????????????
,
tan??????= ,
38.78
42.12
,
??????= 42.64°
Step 5: To find ƩMo,
Ʃ????????????= 65cos53.13 ×3
−140sin33.39×1.5
−90cos18.44×3
+90sin18.44 ×4.5
−100cos53.13×1
−100sin53.13×6
+50cos50.19×3+50sin50.19×6

47. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

Step 6: Apply Varignon’s theorem
Step 7: Position of Resultant
3. Find the resultant of the force system shown in fig. and
replace it by a single force and couple system at A.
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
47
Ʃ????????????=341.03 ??????.?????? ↺
Apply Varignon’s theorem
Ʃ???????????? = Ʃfy x X
?????? =
341.03
38.78
= 8.79??????
Position of Resultant
Find the resultant of the force system shown in fig. and
replace it by a single force and couple system at A.
↺

Find the resultant of the force system shown in fig. and
replace it by a single force and couple system at A.

48. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
48

Solution Step 1: Resolving all the forces horizontally (ΣFx)
??????????????????= 100cos40+85cos50+70sin40
= 176.24 N (→)
Step 2: Resolving all the forces vertically (ΣFy)
??????????????????= 100sin40−85sin50−90−70cos40
= 144.46 N (↓)
Step 3: Magnitude of the resultant force,
??????= ?(??????????????????
??????
)+(??????????????????
??????
)
??????= ?(176.24)
6
+(144.46)
6

??????= 227.88 ??????
Step 4: Direction of the resultant force,
????????????????????????= ,
??????????????????
??????????????????
,
tan??????=
144.46
176.24

??????= 39.34°
Step 5: Find ΣMO
????????????
?= −100cos40×4
+100sin40×4
+85cos50×5 −85sin50×2 +90×3+175−150
ΣMO = 388.65 N.m (↻)
Step 6: Apply Varignon’s theorem
R x d
??????=
388.65
227.88
=1.70 ??????

49. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

Step 7: Position of Resultant R W.r.t. O
Step 8: Find ΣM
????????????
?= −175
−
−
−
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
49
ƩFx . y
?????? =
388.65
176.24
= 2.20??????
ƩFy . x
?????? =
388.65
144.46
= 2.69 ??????
Position of Resultant R W.r.t. O
Find ΣMA
+150
−70sin40 ×4
−70cos40×4
−85sin50 ×2− 85cos50 ×9− 90
ΣMA=1671.43 N.m (↺

90×7
↺)

50. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

4. Find the resultant of the force
2.5 m.
Solution Free Body Dia.
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
50
Find the resultant of the force system shown in fig Radius =
Free Body Dia.

system shown in fig Radius =

51. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
51


Step 1: Resolving all the forces horizontally (ΣFx)
??????????????????= 84cos40−55cos35+79cos60+50
= 108.79 N (→)
Step 2: Resolving all the forces vertically (ΣFy)
??????????????????= −84sin40−55sin35−79sin60+123−60
= 90.96 N (↓)
Step 3: Magnitude of the resultant force,
??????= ?(??????????????????
??????
)+(??????????????????
??????
)
??????= ?(108.79)
6
+(90.96)
6

??????= 141.80 ??????
Step 4: Direction of the resultant force,
????????????????????????= ,
??????????????????
??????????????????
,
tan??????=
90.96
108.79

??????=39.89°

52. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
52

Step 5: Find ΣMO
????????????
?=84∗2.5−123∗2.5+55∗2.5−79∗2.5+50
∗2.5cos40−60∗2.5sin40
ΣMO = 158.16 N.m (↺)
Step 6: Apply Varignon’s theorem
R x d
??????=
158.16
141.80
= 1.12 ??????
Step 7: Position of Resultant R W.r.t. O

Assignment 3
Q1. A triangular plate ABC is subjected to four coplanar
forces as shown in fig. Find the resultant completely and
locate its position with respect to A.

53. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM

Ans: R = 12.71 KN, θ=86.48°,
Hint:
Fig.1. FBD
Q2. Determine the resultant of the following force system
Ans: R = 199.64 N & θ=4.10 °
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
53
R = 12.71 KN, θ=86.48°, ƩMA= 77.66 KN.m and x= 6.12 m

Fig.1. FBD Fig.2. Position
Determine the resultant of the following force system
R = 199.64 N & θ=4.10 °

= 77.66 KN.m and x= 6.12 m

Fig.2. Position

Determine the resultant of the following force system

54. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
54

Q3. The resultant of the three pulls applied through the
three chain attached to bracket is θ as shown in fig.
Determine the magnitude and θ of the resultant

Ans: R = 623.24 N & θ =75.4 °
Q.4. State and Explain a) Parallelogram Law B) Varignon’s
Theorem
Q.5. Classify the force System