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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
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM
Solution Case 1: By Parallelogram Law
๐‘… = 4 + 3
R = 5 N
tan โˆ =
โˆ =
2. Find the resultant of the given forces.
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
2
Case 1: By Parallelogram Law
+ 2 ร— 3 ร— 4 cos 90ยฐ
R = 5 N
3 sin 90ยฐ
4 + 3 cos 90ยฐ
= 36.87ยฐ
Case 2: By Triangle
By Cosine rule
๐‘… = 4 + 3 โˆ’
R = 5 N
By Sine rule
๐‘…
sin 90ยฐ
sin โˆ
โˆ =
Find the resultant of the given forces.
Triangle Law
By Cosine rule
โˆ’ 2 ร— 3 ร— 4 cos 90ยฐ
R = 5 N
By Sine rule
ยฐ
=
3
sin โˆ
โˆ =
3
5
36.87ยฐ
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM
Solution Case 1: By Parallelogram Law
๐‘…
= 50 + 70
R =
tan โˆ =
โˆ
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, ฮธ = 50
To Find : R and T
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
3
Parallelogram Law
+ 2 ร— 50 ร— 70 cos 60ยฐ
R = 104.4 N
70 sin 60ยฐ
50 + 70 cos 60ยฐ
โˆ = 35.5ยฐ
Case 2: By Triangle
By Cosine rule
๐‘… = 4 + 3
R = 5 N
By Sine rule
๐‘…
sin 90
sin
โˆ =
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, ฮธ = 50แต’ , ฮฑ = 20แต’
R and TAC
Triangle Law
By Cosine rule
โˆ’ 2 ร— 3 ร— 4 cos 90ยฐ
R = 5 N
By Sine rule
90ยฐ
=
3
sin โˆ
sin โˆ =
3
5
= 36.87ยฐ
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
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
4
By using Parallelogram Law
Magnitude of resultant is given as
๐‘… = ๐‘ƒ + 2500 + 2 ร— ๐‘ƒ ร— 2500 cos 50ยฐ
Squaring on both sides
๐‘… = ๐‘ƒ + 2500 + 2 ร— ๐‘ƒ ร— 2500 cos 50ยฐโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆeqn
1
Direction of resultant is given by
tan 20ยฐ =
2500 sin 50ยฐ
๐‘ƒ + 2500 cos 50ยฐ
Cross multiplication gives
[๐‘ƒ + 2500 cos 50ยฐ] tan 20ยฐ = 2500 sin 50ยฐ
0.3639 P + 584.88 = 1915.11
๐‘ƒ =
1915.11 โ€“ 584.88
0.3639
P = 3656.69 KN
Put value of P in eqn
1 we get
๐‘… = 3656.69 + 2500 + 2 ร— 3656.69 ร— 2500 cos 50ยฐ
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 ฮฑ = 60แต’ as shown in the fig. Determine magnitude of
the resultant pull on the boat and the angle ฮฒ and ฮณ as shown.
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM
(ANS: R = 1698 N,
๐ญ๐š๐ง ๐œท =
๐‘ธ
๐‘ท +
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, ฮฒ=33แต’, ฮณ = 27แต’)
๐‘ธ ๐ฌ๐ข๐ง ๐œถ
+ ๐‘ธ ๐œ๐จ๐ฌ๐œถ
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, ฮฑ=36.8แต’)
Two forces of magnitude 300 N and 500 N is 50แต’
, the
500 N force being horizontal. Determine the resultant in
magnitude and direction if 300 N force is a push and 500 N
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 120แต’. 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 โˆ†AOC = ฮธ = 120แต’
2. Bigger force (P) = 40 N
3. Angle between resultant (R) and smaller force (Q) โˆ†BOC =ฮฒ=
90แต’
Step 1: To find the โˆ†AOB
From geometry of fig, we can find the โˆ†AOB,
ฮฑ = ฮธ- ฮฒ = 120แต’ - 90แต’ = 30แต’
Step 2: To find Q
we know that,
tan โˆ =
๐‘„ sin ๐œƒ
๐‘ƒ + ๐‘„ cos ๐œƒ
tan 30แต’ =
๐‘„ sin 120ยฐ
40 + ๐‘„ cos 120ยฐ
0.577 =
0.866 ๐‘„
40 โˆ’ 0.5 ๐‘„
0.577*(40-0.5Q) = 0.866 Q
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 โˆ†AOC = 120แต’
2. Bigger force (F1) = 40 N
3. Angle between resultant and smaller force (F2) โˆ†BOC = 90แต’
Step 1: To find the ฮธ
By triangle law, we have
tan ฮธ = 2/3
ฮธ = tan
2
3
ฮธ = 33.69แต’
Step 2: To Find Forces
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
8
By sine rule, we have
2000
sin 33.69ยฐ
=
๐น๐ด๐ต
sin 90ยฐ
=
๐น๐ต๐ถ
sin 56.31ยฐ
2000
sin 33.69ยฐ
=
๐น๐ด๐ต
sin 90ยฐ
๐น๐ด๐ต =
2000 sin 90ยฐ
sin 33.69
๐น๐ด๐ต =
2000 โˆ— 1
0.5546
๐น๐ต๐ถ =
2000 sin 56.31ยฐ
sin 33.69
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 โˆš10
N. but if they act at 60แต’, their resultant is โˆš13N.
10 = ๐‘ƒ + ๐‘„ โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆeq1
13 = ๐‘ƒ + ๐‘„ + ๐‘ƒ๐‘„โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆeq2
PQ= 3โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ.eq3
๐‘„ = ANS: Q = 3N & P = 1 N)
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
๏‚ท Fx = F. Cos ฮธ
๏‚ท Fy = F. Sin ฮธ
๏‚ท
Formulas for Oblique Components
๏‚ท ๐น1 =
( )
๏‚ท ๐น2 =
( )
๏‚ท 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
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 ๐œƒ =
3. FBD
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
10
To find Fx
Fx = F cos ฮธ = 30 cos 45แต’
= 21.21 N
To Find Fy
Fy = F sin ฮธ = 30 sin 45แต’
= 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
= = 36.87ยฐ
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
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
11
Step 1: To find magnitude of P
Fx = F Cos ฮธ
40 = P cos ฮธ
40 = P cos (30+36.87)
๐‘ƒ =
. ยฐ
P = 101.82 N
Step 2: To Find Py
Py = P sin ฮธ = 101.82 sin 30แต’
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. ฮ˜1 = tan-1
ยพ = 36.87แต’
3. ฮ˜2 = tan-1
2/3 = 33.69แต’
4
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 (ฮธ1+ ฮธ2)
= 360 cos (70.56)
Px = 119.82 N
Step 2: To Find magnitude of Py
Py = P sin (ฮธ1+ ฮธ2)
= 360 sin (70.56)
Py = 339.48 N
ASSIGNMENT
1. Resolve a force of 40 KN inclined at 150แต’ 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 22แต’ as shown in fig. into two
components one acting along inclined 22แต’ and the other
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 30แต’
Solution Given Data:
1. Force F
2. ๐›ผ = 45ยฐ
3. ๐›ฝ = 30ยฐ
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 45แต’ & the other has an
with the force 80 N.
Given Data:
F = 80 N
ยฐ
ยฐ
F1 & F2
perpendicular component
A force of 80 N is acting on a body. Find its components such
& the other has an
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 60ยฐ
Find its components along 105ยฐ & 330ยฐ with X
Solution Given Data:
1) F = 2000 N
2) ฮฑ = 105
3) ฮฒ = (360
Step 1: To find magnitude of F1
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
14
To find magnitude of F1
๐‘ญ๐Ÿ =
๐‘ญ ๐ฌ๐ข๐ง ๐œท
๐ฌ๐ข๐ง(๐œถ + ๐œท)
๐น1 =
80 sin 30ยฐ
sin(45ยฐ + 30ยฐ)
F1 = 41.41 N
To Find magnitude of F2
๐‘ญ๐Ÿ =
๐‘ญ ๐ฌ๐ข๐ง ๐œถ
๐ฌ๐ข๐ง(๐œถ + ๐œท)
๐น2 =
80 sin 45ยฐ
sin(45ยฐ + 30ยฐ)
F2 = 58.56 N
A force of 2000 N is acts at an angle of 60ยฐ with X
Find its components along 105ยฐ & 330ยฐ with X
Given Data:
= 2000 N
105 โ€“ 60 = 45ยฐ
(360-330) + 60 = 90ยฐ
To find magnitude of F1
๐‘ญ๐Ÿ =
๐‘ญ ๐ฌ๐ข๐ง ๐œท
๐ฌ๐ข๐ง(๐œถ + ๐œท)
๐น1 =
2000 sin 90ยฐ
sin(45ยฐ + 90ยฐ)
with X-axis.
Find its components along 105ยฐ & 330ยฐ with X-axis.
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM
Step 2: To Find magnitude of F2
Ans:
1. Component along 105ยฐ
2. Component along 330ยฐ = 2000 N
3. The resultant of two forces in a plane is 800 N at 60ยฐ with x
axis. One force is 160N at 30ยฐ with x axis. Determine the
missing force & its inclination.
Solution Given Data:
1. F = 800 N
2. F1 = 160 N
3. ฮฑ = 60
To find: F2 & ฮฒ
Step 1: To find ฮฒ
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
15
F1 = 2828.43 N
To Find magnitude of F2
๐‘ญ๐Ÿ =
๐‘ญ ๐ฌ๐ข๐ง ๐œถ
๐ฌ๐ข๐ง(๐œถ + ๐œท)
๐น2 =
2000 sin 45ยฐ
sin(45ยฐ + 90ยฐ)
F2 = 2000 N
Component along 105ยฐ = 2828.43 N
Component along 330ยฐ = 2000 N
The resultant of two forces in a plane is 800 N at 60ยฐ with x
force is 160N at 30ยฐ with x axis. Determine the
missing force & its inclination.
Given Data:
F = 800 N
F1 = 160 N
ฮฑ = 60 โ€“ 30 = 30ยฐ
F2 & ฮฒ
To find ฮฒ
๐‘ญ๐Ÿ =
๐‘ญ ๐ฌ๐ข๐ง ๐œท
๐ฌ๐ข๐ง(๐œถ + ๐œท)
The resultant of two forces in a plane is 800 N at 60ยฐ with x-
force is 160N at 30ยฐ with x axis. Determine the
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
16
160 =
800 sin ๐›ฝ
sin(30 + ๐›ฝ)
160 (sin (30+ฮฒ)) = 800 sin ฮฒ
80 cos ฮฒ = 661.44 sin ฮฒ
sin ๐›ฝ
cos ๐›ฝ
=
80
661.44
tan๐›ฝ = 0.1209
ฮฒ = 6.89ยฐ
Step 2: To Find magnitude of F2
๐‘ญ๐Ÿ =
๐‘ญ ๐ฌ๐ข๐ง ๐œถ
๐ฌ๐ข๐ง(๐œถ + ๐œท)
๐น2 =
800 sin 30
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 ฮธ
(ANS: F = 90.82 N and ฮธ = 79.46ยฐ)
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
17
Type 4: Problems on Resultant of concurrent
force system using Method of Resolution
๏ถ If the number of forces is more than two, then its resultant can
be found out conveniently by the METHOD OF RESOLUTION.
๏ถ STEPWISE PROCEDURE OF METHOD OF RESOLUTION:
1. Resolve all forces horizontally and find the algebraic sum of all
the horizontal components (i.e., ฮฃFx)
2. Resolve all forces vertically and find the algebraic sum of all
the vertical components (i.e., ฮฃFy).
3. The resultant R of the given forces will be given by the
equation:
๐‘น = (๐šบ๐…๐ฑ๐Ÿ
) + (๐šบ๐…๐ฒ๐Ÿ
)
4. The resultant force will be inclined at an angle ฮธ, with the
horizontal, such that
tan๐œƒ =
ฮฃFy
ฮฃFx
5. Position of the resultant
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 (ฮฃFx)
๐›ด๐น๐‘ฅ = 20 cos
+
= 36 N
Step 2: Resolving all the forces vertically (ฮฃFy)
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 โˆ’
360
6
= 120ยฐ
๐œƒ =
120
4
= 30ยฐ
Resolving all the forces horizontally (ฮฃFx)
cos 0ยฐ + 30 cos 30ยฐ + 40 cos 60ยฐ + 50
+ 60 cos 120ยฐ
= 36 N
Resolving all the forces vertically (ฮฃFy)
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-
.
Resolving all the forces horizontally (ฮฃFx)
50 cos 90ยฐ
Resolving all the forces vertically (ฮฃFy)
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM
๐›ด๐น๐‘ฆ = 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
20 sin 0ยฐ + 30 sin 30ยฐ + 40 sin 60ยฐ +
+ 60 sin 120ยฐ
= 151.6 N
Magnitude of the resultant force,
๐‘น = (๐šบ๐…๐ฑ๐Ÿ
) + (๐šบ๐…๐ฒ๐Ÿ
๐‘… = (36) + (151.6)
๐‘… = 155.8 ๐‘
Direction of the resultant force,
๐ญ๐š๐ง ๐œฝ =
๐šบ๐…๐ฒ
๐šบ๐…๐ฑ
tan ๐œƒ =
151.6
36
๐œƒ = 76.6ยฐ
Position of the resultant force,
The following forces act at a point :
20 N inclined at 30แต’ towards North of East,
25 N towards North,
towards North West, and
35 N inclined at 40แต’ towards South of West.
Find the magnitude and direction of the resultant force.
+ 50 sin 90ยฐ
)
towards North of East,
towards South of West.
Find the magnitude and direction of the resultant force.
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
20
Solutio
n
Step 1: Resolving all the forces horizontally (ฮฃFx)
๐›ด๐น๐‘ฅ = 20 cos 30ยฐ + 25 cos 90ยฐ + 30 cos 135ยฐ + 35 cos 220ยฐ
= - 30.7 N
Step 2: Resolving all the forces vertically (ฮฃFy)
๐›ด๐น๐‘ฆ = 20 sin 30ยฐ + 25 sin 90ยฐ + 30 sin 135ยฐ + 35 sin 220ยฐ
= 33.7 N
Step 3: Magnitude of the resultant force,
๐‘น = (๐šบ๐…๐ฑ๐Ÿ
) + (๐šบ๐…๐ฒ๐Ÿ
)
๐‘… = (โˆ’30.7) + (33.7)
๐‘… = 45.6 ๐‘
Step 4: Direction of the resultant force,
๐ญ๐š๐ง ๐œฝ =
๐šบ๐…๐ฒ
๐šบ๐…๐ฑ
tan ๐œƒ =
33.7
โˆ’30.7
๐œƒ = 47.7ยฐ
Step 5: Position of the resultant force,
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM
Since ฮฃFx is negative and ฮฃFy 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
๐ญ๐š๐ง ๐œฝ๐Ÿ =
๐Ÿ๐Ÿ“๐ŸŽ
๐Ÿ“๐ŸŽ๐ŸŽ
๐ญ๐š๐ง ๐œฝ๐Ÿ =
๐Ÿ“๐Ÿ“๐ŸŽ
๐Ÿ๐ŸŽ๐ŸŽ
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
21
Since ฮฃFx is negative and ฮฃFy 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
๐Ÿ๐Ÿ“๐ŸŽ
๐Ÿ“๐ŸŽ๐ŸŽ
= ๐Ÿ๐Ÿ”. ๐Ÿ•ยฐ
๐Ÿ“๐Ÿ“๐ŸŽ
๐Ÿ๐ŸŽ๐ŸŽ
= ๐Ÿ•๐Ÿ—. ๐Ÿ•ยฐ
Since ฮฃFx is negative and ฮฃFy is positive, Resultant lies
Therefore acute angle of the resultant = 180แต’ - 47.7แต’ = 132.3แต’
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
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM
๐ญ๐š๐ง ๐œฝ๐Ÿ‘ =
๐Ÿ๐Ÿ“๐ŸŽ
๐Ÿ‘๐ŸŽ๐ŸŽ
๐ญ๐š๐ง ๐œฝ๐Ÿ’ =
๐Ÿ๐Ÿ“๐ŸŽ
๐Ÿ๐ŸŽ๐ŸŽ
Step 1: Resolving all the forces
horizontally (ฮฃFx)
๐›ด๐น๐‘ฅ = 20
= 20.25
Step 2: Resolving all the forces vertically (ฮฃFy)
๐›ด๐น๐‘ฆ = 20 sin
= 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
๐Ÿ๐Ÿ“๐ŸŽ
๐Ÿ‘๐ŸŽ๐ŸŽ
= ๐Ÿ๐Ÿ”. ๐Ÿ“๐Ÿ”ยฐ
๐Ÿ๐Ÿ“๐ŸŽ
๐Ÿ๐ŸŽ๐ŸŽ
= ๐Ÿ”๐Ÿ–. ๐Ÿยฐ
Resolving all the forces
horizontally (ฮฃFx)
20 cos 16.7ยฐ + 25 cos 79.7ยฐ โˆ’ 10
+ 15 cos 68.2ยฐ
20.25 N
Resolving all the forces vertically (ฮฃFy)
sin 16.7ยฐ + 25 sin 79.7ยฐ + 10 sin 26.56
20.89 N
Magnitude of the resultant force,
๐‘น = (๐šบ๐…๐ฑ๐Ÿ
) + (๐šบ๐…๐ฒ๐Ÿ
๐‘… = (20.25) + (20.89
๐‘… = 29.09 ๐‘
Direction of the resultant force,
๐ญ๐š๐ง ๐œฝ =
๐šบ๐…๐ฒ
๐šบ๐…๐ฑ
tan ๐œƒ =
20.89
20.25
๐œƒ = 45.89ยฐ
Position of the resultant force,
10 cos 26.56ยฐ
Resolving all the forces vertically (ฮฃFy)
56ยฐ โˆ’ 15 sin 68.2ยฐ
)
89)
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM
Since ฮฃFx and ฮฃFy is positive, Resultant lies
quadrants.
4. For the system shown, determine
i) The required value of ฮฑ if resultant of three forces
is to be vertical and
ii) The corresponding magnitude of resultant.
Step I) ฦฉFx = 0
100 cos ๐›ผ + 150
150(cos ๐›ผ. cos
30 cos ๐›ผ โˆ’ 75
Step ii) R = ฦฉ
๐‘… = โˆ’100
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
23
Since ฮฃFx and ฮฃFy is positive, Resultant lies
the system shown, determine
The required value of ฮฑ if resultant of three forces
is to be vertical and
The corresponding magnitude of resultant.
Fx = 0
150 cos(๐›ผ + 30) โˆ’ 200cos ๐›ผ = 0
cos 30 โˆ’ sin ๐›ผ. sin 30) โˆ’ 100 cos ๐›ผ = 0
75 sin ๐›ผ = 0
tan ๐›ผ =
sin ๐›ผ
cos ๐›ผ
=
30
75
= 21.8ยฐ
ฦฉFy
100 sin 21.8 โˆ’ 150 sin(21.8 + 30) โˆ’
R = -229.29 N
Since ฮฃFx and ฮฃFy is positive, Resultant lies in first
The required value of ฮฑ if resultant of three forces
The corresponding magnitude of resultant.
200 sin 21.8
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
24
(ANS: ฮฑ = 21.8แต’ & R = 229.29 N)
5.
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
๐‘€@๐ด =
3. Moment about
๐‘€@
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
๐‘€@๐‘‚ = 500 cos 36.87 ร— 3 = 1200
Moment about point A
= โˆ’500 sin 36.87 ร— 2 + 500 cos 36.87
Moment about point B
@๐ต = โˆ’500 sin 36.87 ร— 4 + 500 cos
Moment about point C
Moment of force
point O, A,B & C
1200 ๐‘. ๐‘š
87 ร— 3 = 600 ๐‘. ๐‘š
36.87 ร— 3 = 0
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
26
๐‘€@๐ถ = โˆ’500sin 36.87 ร— 4 + 500cos 36.87 ร— 1.5 = โˆ’600๐‘. ๐‘š
M@C = 600 N.m
5. Moment about point D
๐‘€@๐ท = โˆ’500sin 36.87 ร— 4 = โˆ’1200 ๐‘. ๐‘š
M@D = 1200 N.m
6. Moment about point E
๐‘€@๐ธ = 0
2. Find the moment of force 2000 N about point โ€˜Oโ€™ as shown in
fig
Solution Step 1: Free Body Diagram
Step 2: Finding moments
Moment about point O
๐‘€@๐‘‚ = โˆ’2000 sin 30 ร— 5 = โˆ’5000 ๐‘. ๐‘š
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
27
3. Find the moment of force 50 N about point โ€˜Oโ€™ as shown in fig
Solution Step 1: Free Body Diagram
Step 2: Finding moments
Moment about point O
๐‘€@๐‘‚ = โˆ’2000 sin 30 ร— 5 = โˆ’5000 ๐‘. ๐‘š
4. Find the moment of forces as shown in fig. on lever about
point โ€œOโ€.
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
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
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
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
31
Solution Step 1: Finding Moment of Couple
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM
Type 7: Problems on
force system using Method of Resolution
๏ถ STEPWISE PROCEDURE OF METHOD OF RESOLUTION:
1. Find resultant R
parallel forces considering proper sign convention.
2. Find ฮฃMo. Take the algebraic sum moment of forces
point (say O)
(+๐’—๐’†
3. Apply Varignonโ€™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 ฮฃF & ฮฃMo.
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 = ฮฃ F. Take the algebraic sum of all the
parallel forces considering proper sign convention.
(+๐’—๐’† โ†‘) ( โˆ’๐’—๐’† โ†“)
Find ฮฃMo. Take the algebraic sum moment of forces
(say O) considering proper sign convention.
๐’—๐’† โ†ป โˆ’๐‘ช๐’๐’๐’„๐’Œ๐’˜๐’Š๐’”๐’†) ( โˆ’๐’—๐’† โ†บ โˆ’๐’‚๐’๐’•๐’Š๐’„๐’๐’๐’„๐’Œ๐’˜๐’Š๐’”๐’†
Apply Varignonโ€™s theorem, ๐œฎ๐‘ด๐’ = ๐‘น ร— ๐’…
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 ฮฃF & ฮฃMo.
Resultant of parallel
force system using Method of Resolution
STEPWISE PROCEDURE OF METHOD OF RESOLUTION:
= ฮฃ F. Take the algebraic sum of all the
parallel forces considering proper sign convention.
Find ฮฃMo. Take the algebraic sum moment of forces about a
ing proper sign convention.
๐’‚๐’๐’•๐’Š๐’„๐’๐’๐’„๐’Œ๐’˜๐’Š๐’”๐’†)
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 ฮฃF & ฮฃMo.
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 = ฮฃFy = - 70 + 100 + 50
= 50 N
Step 2: To find ฮฃMo
Taking moment about point โ€œOโ€
๐›ด๐‘€ = 100
= 394 N
Step 3: Applying Varignonโ€™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 โ€“ 86-34 + 90
= 50 N (โ†‘)
To find ฮฃMo
Taking moment about point โ€œOโ€
100 ร— 1.5 + 50 ร— 3.5 โˆ’ 86 ร— 6.5 โˆ’ 34
= 394 N-m (โ†บ)
Applying Varignonโ€™s theorem,
๐›ด๐‘€ = ๐‘… ร— ๐‘‘
๐‘‘ =
๐›ด๐‘€
๐‘…
=
394
50
= 7.88 ๐‘š
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
ร— 8 + 90 ร— 10
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM
Case II : To find
Step 1: To find magnitude of Resultant R
R = ฮฃF = - 70 + 100 + 50
= 50 N
Step 2: To find ฮฃM
Taking moment about point โ€œAโ€
๐›ด๐‘€ = 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 โ€“ 86 - 34 + 90
= 50 N (โ†‘)
To find ฮฃMA
Taking moment about point โ€œAโ€
70 ร— 3.5 โˆ’ 100 ร— 2 โˆ’ 86 ร— 3 โˆ’ 34 ร— 4
219 N-m (โ†บ)
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 ร— 6.5
Position of the equivalent force and couple,
given active forces as shown in fig.
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
35
R = ฮฃF = 100 + 200 -150
= 150 N (โ†’)
Step 2: To find ฮฃMB
Taking moment about point โ€œBโ€
๐›ด๐‘€ = 150 + 100 ร— 5 โˆ’ 150 ร— 3.5 + 200 ร— 1.5
= 425 N-m = 425 N-m (โ†ป)
Step 3: Applying Varignonโ€™s theorem,
๐›ด๐‘€ = ๐‘… ร— โ„Ž
โ„Ž =
๐›ด๐‘€
๐‘…
=
425
150
= 2.83 ๐‘š
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.
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 = ฮฃF = - 50 + 60 โ€“ 70 + 80 โ€“ 90
= - 70 N
= 70 N
Step 2: To find ฮฃMA
Taking moment about point โ€œAโ€
๐›ด๐‘€ = โˆ’60 ร— 3 + 70 ร— 6 โˆ’ 80 ร— 7.5 + 90 ร— 9
= 450 N-m = 450 N-m (โ†ป)
Step 3: Applying Varignonโ€™s theorem,
๐›ด๐‘€ = ๐‘… ร— ๐‘‘
๐‘‘ =
๐›ด๐‘€
๐‘…
=
450
70
= 6.429 ๐‘š
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.
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 ๐œƒ =
๐‘‚๐‘๐‘
๐ป๐‘ฆ๐‘๐‘œ
sin 30ยฐ =
๐‘ฅ
1
x1 = 0.5 m
For 100 N force
sin 60ยฐ =
๐‘ฅ
1
x2 = 0.866 m
For 50 N force
sin 45ยฐ =
๐‘ฅ
1
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
38
x3 = 0.707 m
Step 2: To find magnitude of Resultant R
R = ฮฃF = - 100 โ€“ 80 + 150 - 50
= 80 N (โ†)
Step 3: To find ฮฃMo
Taking moment about point โ€œOโ€
๐›ด๐‘€ = 100 ร— 0.866 + 80 ร— 0.5 + 150 ร— 0.5 โˆ’ 50 ร— 0.707
= 166.25 N-m (โ†บ)
Step 3: Applying Varignonโ€™s theorem,
๐›ด๐‘€ = ๐‘… ร— ๐‘‘
๐‘‘ =
๐›ด๐‘€
๐‘…
=
166.25
80
= 2.078 ๐‘š
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.
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)
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
40
Type 6: Problems on Resultant of general
force system using Method of Resolution
๏ถ STEPWISE PROCEDURE OF METHOD OF RESOLUTION:
1. Resolve all forces horizontally and find the algebraic sum of all
the horizontal components (i.e., ฮฃFx)
2. Resolve all forces vertically and find the algebraic sum of all
the vertical components (i.e., ฮฃFy).
3. Find the magnitude of Resultant R
The resultant R of the given forces will be given by the
equation:
๐‘น = (๐šบ๐…๐ฑ๐Ÿ
) + (๐šบ๐…๐ฒ๐Ÿ
)
4. Find direction ฮธ. The resultant force will be inclined at an
angle ฮธ, with the horizontal, such that
tan๐œƒ =
ฮฃFy
ฮฃFx
5. Find ฮฃMo. Take the algebraic sum moment of forces about a
point (say O) considering proper sign convention.
(+๐’—๐’† โ†ป โˆ’๐‘ช๐’๐’๐’„๐’Œ๐’˜๐’Š๐’”๐’†) ( โˆ’๐’—๐’† โ†บ โˆ’๐’‚๐’๐’•๐’Š๐’„๐’๐’๐’„๐’Œ๐’˜๐’Š๐’”๐’†)
6. Apply Varignonโ€™s theorem,
I. ๐œฎ๐‘ด๐’ = ๐‘น ร— ๐’…,Where d = perpendicular distance between
line of action of R and reference point O. OR
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
41
II. ๐œฎ๐‘ด๐’ = ๐œฎ๐‘ญ๐’™ ร— ๐’š, Where y = distance between point O and
the intersection of line of action of resultant R with y-axis.
OR
III. ๐œฎ๐‘ด๐’ = ๐œฎ๐‘ญ๐’š ร— ๐’™, 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 ฮฃFx, ฮฃFy and ฮฃMo any one possible
position of R among the following eight may arise.
1
2
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
42
3
4
5
6
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
43
7
8
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 (ฮฃFx)
๐›ด๐น๐‘ฅ = โˆ’100 cos 36.87ยฐ โˆ’ 75
= 155 N (โ†)
Step 2: Resolving all the forces vertically (ฮฃFy)
๐›ด๐น๐‘ฆ = โˆ’200 + 30 โˆ’ 100 sin 36.87ยฐ
= 210 N (โ†“)
Step 3: Magnitude of the resultant force,
๐‘น = (๐šบ๐…๐ฑ๐Ÿ
) + (๐šบ๐…๐ฒ๐Ÿ
)
๐‘… = (155) + (210)
๐‘… = 261 ๐‘
Step 4: Direction of the resultant force,
๐ญ๐š๐ง ๐œฝ =
๐šบ๐…๐ฒ
๐šบ๐…๐ฑ
tan ๐œƒ =
210
155
๐œƒ = 53.57ยฐ
Step 5: Find ฮฃMA
๐›ด๐‘€ = 50 ร— 2 + 80 โˆ’ 100 sin 36.87ยฐ
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 ๐œƒ1 =
๐‘‚๐‘๐‘
๐ด๐‘‘๐‘—
=
2
1.5
1) ๐œƒ1 = tan
.
= 53.13ยฐ
2) ๐œƒ2 = tan = 33.69ยฐ
3) ๐œƒ3 = tan = 18.44ยฐ
4) ๐œƒ4 = tan
.
= 53.13ยฐ
5) ๐œƒ5 = tan
.
= 50.13ยฐ
Step 1: Resolving all the forces horizontally (ฮฃFx)
๐›ด๐น๐‘ฅ = 65 cos 53.13ยฐ + 140 cos 33.69 โˆ’ 90 cos 18.44
โˆ’ 100 cos 53.13 + 50 cos 50.13
= 42.12 N (โ†’)
Step 2: Resolving all the forces vertically (ฮฃFy)
๐›ด๐น๐‘ฆ = โˆ’65 sin 53.13ยฐ + 140 sin 33.69 โˆ’ 90 sin 18.44
+ 100 sin 53.13 โˆ’ 50 sin 50.13
= 38.78 N (โ†‘)
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
46
Step 3: Magnitude of the resultant force,
๐‘น = (๐šบ๐…๐ฑ๐Ÿ
) + (๐šบ๐…๐ฒ๐Ÿ
)
๐‘… = (42.12) + (38.78)
๐‘… = 57.25 ๐‘
Step 4: Direction of the resultant force,
๐ญ๐š๐ง ๐œฝ =
๐šบ๐…๐ฒ
๐šบ๐…๐ฑ
tan ๐œƒ =
38.78
42.12
๐œƒ = 42.64ยฐ
Step 5: To find ฦฉMo,
ฦฉ๐‘€๐‘œ = 65 cos 53.13 ร— 3
โˆ’ 140 sin 33.39 ร— 1.5
โˆ’ 90 cos 18.44 ร— 3
+ 90 sin 18.44 ร— 4.5
โˆ’ 100 cos 53.13 ร— 1
โˆ’ 100 sin 53.13 ร— 6
+ 50 cos 50.19 ร— 3 + 50 sin 50.19 ร— 6
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM
Step 6: Apply Varignonโ€™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
ฦฉ๐‘€๐‘œ = 341.03 ๐‘. ๐‘š โ†บ
Apply Varignonโ€™s theorem
ฦฉ๐‘€๐‘œ = ฦฉfy x X
๐‘ฅ =
341.03
38.78
= 8.79๐‘š
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.
โ†บ
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
48
Solution Step 1: Resolving all the forces horizontally (ฮฃFx)
๐›ด๐น๐‘ฅ = 100 cos 40 + 85 cos 50 + 70 sin 40
= 176.24 N (โ†’)
Step 2: Resolving all the forces vertically (ฮฃFy)
๐›ด๐น๐‘ฆ = 100 sin 40 โˆ’ 85 sin 50 โˆ’ 90 โˆ’ 70cos 40
= 144.46 N (โ†“)
Step 3: Magnitude of the resultant force,
๐‘น = (๐šบ๐…๐ฑ๐Ÿ
) + (๐šบ๐…๐ฒ๐Ÿ
)
๐‘… = (176.24) + (144.46)
๐‘… = 227.88 ๐‘
Step 4: Direction of the resultant force,
๐ญ๐š๐ง ๐œฝ =
๐šบ๐…๐ฒ
๐šบ๐…๐ฑ
tan ๐œƒ =
144.46
176.24
๐œƒ = 39.34ยฐ
Step 5: Find ฮฃMO
๐›ด๐‘€ = โˆ’100 cos 40 ร— 4
+ 100 sin 40 ร— 4
+ 85 cos 50 ร— 5 โˆ’ 85 sin 50 ร— 2 + 90 ร— 3 + 175 โˆ’ 150
ฮฃMO = 388.65 N. m (โ†ป)
Step 6: Apply Varignonโ€™s theorem
R x d
๐‘‘ =
388.65
227.88
= 1.70 ๐‘š
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM
Step 7: Position of Resultant R W.r.t. O
Step 8: Find ฮฃM
๐›ด๐‘€ = โˆ’175
โˆ’
โˆ’
โˆ’
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
49
ฦฉFx . y
๐‘ฆ =
388.65
176.24
= 2.20๐‘š
ฦฉFy . x
๐‘ฅ =
388.65
144.46
= 2.69 ๐‘š
Position of Resultant R W.r.t. O
Find ฮฃMA
+ 150
โˆ’ 70 sin 40 ร— 4
โˆ’ 70 cos 40 ร— 4
โˆ’ 85 sin 50 ร— 2 โˆ’ 85 cos 50 ร— 9 โˆ’ 90
ฮฃMA = 1671.43 N. m (โ†บ
90 ร— 7
โ†บ)
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 =
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
51
Step 1: Resolving all the forces horizontally (ฮฃFx)
๐›ด๐น๐‘ฅ = 84 cos 40 โˆ’ 55 cos 35 + 79 cos 60 + 50
= 108.79 N (โ†’)
Step 2: Resolving all the forces vertically (ฮฃFy)
๐›ด๐น๐‘ฆ = โˆ’84 sin 40 โˆ’ 55 sin 35 โˆ’ 79 sin 60 + 123 โˆ’ 60
= 90.96 N (โ†“)
Step 3: Magnitude of the resultant force,
๐‘น = (๐šบ๐…๐ฑ๐Ÿ
) + (๐šบ๐…๐ฒ๐Ÿ
)
๐‘… = (108.79) + (90.96)
๐‘… = 141.80 ๐‘
Step 4: Direction of the resultant force,
๐ญ๐š๐ง ๐œฝ =
๐šบ๐…๐ฒ
๐šบ๐…๐ฑ
tan ๐œƒ =
90.96
108.79
๐œƒ = 39.89ยฐ
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
52
Step 5: Find ฮฃMO
๐›ด๐‘€ = 84 โˆ— 2.5 โˆ’ 123 โˆ— 2.5 + 55 โˆ— 2.5 โˆ’ 79 โˆ— 2.5 + 50
โˆ— 2.5 cos 40 โˆ’ 60 โˆ— 2.5 sin 40
ฮฃMO = 158.16 N. m (โ†บ)
Step 6: Apply Varignonโ€™s theorem
R x d
๐‘‘ =
158.16
141.80
= 1.12 ๐‘š
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.
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM
Ans: R = 12.71 KN, ฮธ=86.48ยฐ,
Hint:
Fig.1. FBD
Q2. Determine the resultant of the following force system
Ans: R = 199.64 N & ฮธ=4.10 ยฐ
Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University
53
R = 12.71 KN, ฮธ=86.48ยฐ, ฦฉMA= 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 & ฮธ=4.10 ยฐ
= 77.66 KN.m and x= 6.12 m
Fig.2. Position
Determine the resultant of the following force system
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 ฮธ as shown in fig.
Determine the magnitude and ฮธ of the resultant
Ans: R = 623.24 N & ฮธ =75.4 ยฐ
Q.4. State and Explain a) Parallelogram Law B) Varignonโ€™s
Theorem
Q.5. Classify the force System

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Unit 1. force system, solved problems on force system.pdf

  • 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 ๐‘… = 4 + 3 R = 5 N tan โˆ = โˆ = 2. Find the resultant of the given forces. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University 2 Case 1: By Parallelogram Law + 2 ร— 3 ร— 4 cos 90ยฐ R = 5 N 3 sin 90ยฐ 4 + 3 cos 90ยฐ = 36.87ยฐ Case 2: By Triangle By Cosine rule ๐‘… = 4 + 3 โˆ’ R = 5 N By Sine rule ๐‘… sin 90ยฐ sin โˆ โˆ = Find the resultant of the given forces. Triangle Law By Cosine rule โˆ’ 2 ร— 3 ร— 4 cos 90ยฐ R = 5 N By Sine rule ยฐ = 3 sin โˆ โˆ = 3 5 36.87ยฐ
  • 3. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM Solution Case 1: By Parallelogram Law ๐‘… = 50 + 70 R = tan โˆ = โˆ 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, ฮธ = 50 To Find : R and T Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University 3 Parallelogram Law + 2 ร— 50 ร— 70 cos 60ยฐ R = 104.4 N 70 sin 60ยฐ 50 + 70 cos 60ยฐ โˆ = 35.5ยฐ Case 2: By Triangle By Cosine rule ๐‘… = 4 + 3 R = 5 N By Sine rule ๐‘… sin 90 sin โˆ = 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, ฮธ = 50แต’ , ฮฑ = 20แต’ R and TAC Triangle Law By Cosine rule โˆ’ 2 ร— 3 ร— 4 cos 90ยฐ R = 5 N By Sine rule 90ยฐ = 3 sin โˆ sin โˆ = 3 5 = 36.87ยฐ 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 ๐‘… = ๐‘ƒ + 2500 + 2 ร— ๐‘ƒ ร— 2500 cos 50ยฐ Squaring on both sides ๐‘… = ๐‘ƒ + 2500 + 2 ร— ๐‘ƒ ร— 2500 cos 50ยฐโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆeqn 1 Direction of resultant is given by tan 20ยฐ = 2500 sin 50ยฐ ๐‘ƒ + 2500 cos 50ยฐ Cross multiplication gives [๐‘ƒ + 2500 cos 50ยฐ] tan 20ยฐ = 2500 sin 50ยฐ 0.3639 P + 584.88 = 1915.11 ๐‘ƒ = 1915.11 โ€“ 584.88 0.3639 P = 3656.69 KN Put value of P in eqn 1 we get ๐‘… = 3656.69 + 2500 + 2 ร— 3656.69 ร— 2500 cos 50ยฐ 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 ฮฑ = 60แต’ as shown in the fig. Determine magnitude of the resultant pull on the boat and the angle ฮฒ and ฮณ as shown.
  • 5. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM (ANS: R = 1698 N, ๐ญ๐š๐ง ๐œท = ๐‘ธ ๐‘ท + 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, ฮฒ=33แต’, ฮณ = 27แต’) ๐‘ธ ๐ฌ๐ข๐ง ๐œถ + ๐‘ธ ๐œ๐จ๐ฌ๐œถ 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, ฮฑ=36.8แต’) Two forces of magnitude 300 N and 500 N is 50แต’ , 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 120แต’. 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 โˆ†AOC = ฮธ = 120แต’ 2. Bigger force (P) = 40 N 3. Angle between resultant (R) and smaller force (Q) โˆ†BOC =ฮฒ= 90แต’ Step 1: To find the โˆ†AOB From geometry of fig, we can find the โˆ†AOB, ฮฑ = ฮธ- ฮฒ = 120แต’ - 90แต’ = 30แต’ Step 2: To find Q we know that, tan โˆ = ๐‘„ sin ๐œƒ ๐‘ƒ + ๐‘„ cos ๐œƒ tan 30แต’ = ๐‘„ sin 120ยฐ 40 + ๐‘„ cos 120ยฐ 0.577 = 0.866 ๐‘„ 40 โˆ’ 0.5 ๐‘„ 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 โˆ†AOC = 120แต’ 2. Bigger force (F1) = 40 N 3. Angle between resultant and smaller force (F2) โˆ†BOC = 90แต’ Step 1: To find the ฮธ By triangle law, we have tan ฮธ = 2/3 ฮธ = tan 2 3 ฮธ = 33.69แต’ Step 2: To Find Forces
  • 8. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University 8 By sine rule, we have 2000 sin 33.69ยฐ = ๐น๐ด๐ต sin 90ยฐ = ๐น๐ต๐ถ sin 56.31ยฐ 2000 sin 33.69ยฐ = ๐น๐ด๐ต sin 90ยฐ ๐น๐ด๐ต = 2000 sin 90ยฐ sin 33.69 ๐น๐ด๐ต = 2000 โˆ— 1 0.5546 ๐น๐ต๐ถ = 2000 sin 56.31ยฐ sin 33.69 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 โˆš10 N. but if they act at 60แต’, their resultant is โˆš13N. 10 = ๐‘ƒ + ๐‘„ โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆeq1 13 = ๐‘ƒ + ๐‘„ + ๐‘ƒ๐‘„โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆeq2 PQ= 3โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ.eq3 ๐‘„ = 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 ๏‚ท Fx = F. Cos ฮธ ๏‚ท Fy = F. Sin ฮธ ๏‚ท Formulas for Oblique Components ๏‚ท ๐น1 = ( ) ๏‚ท ๐น2 = ( ) ๏‚ท 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 ๐œƒ = 3. FBD Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University 10 To find Fx Fx = F cos ฮธ = 30 cos 45แต’ = 21.21 N To Find Fy Fy = F sin ฮธ = 30 sin 45แต’ = 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 = = 36.87ยฐ 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 ฮธ 40 = P cos ฮธ 40 = P cos (30+36.87) ๐‘ƒ = . ยฐ P = 101.82 N Step 2: To Find Py Py = P sin ฮธ = 101.82 sin 30แต’ 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. ฮ˜1 = tan-1 ยพ = 36.87แต’ 3. ฮ˜2 = tan-1 2/3 = 33.69แต’ 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 (ฮธ1+ ฮธ2) = 360 cos (70.56) Px = 119.82 N Step 2: To Find magnitude of Py Py = P sin (ฮธ1+ ฮธ2) = 360 sin (70.56) Py = 339.48 N ASSIGNMENT 1. Resolve a force of 40 KN inclined at 150แต’ 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 22แต’ as shown in fig. into two components one acting along inclined 22แต’ 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 30แต’ Solution Given Data: 1. Force F 2. ๐›ผ = 45ยฐ 3. ๐›ฝ = 30ยฐ 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 45แต’ & the other has an with the force 80 N. Given Data: F = 80 N ยฐ ยฐ 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 60ยฐ Find its components along 105ยฐ & 330ยฐ with X Solution Given Data: 1) F = 2000 N 2) ฮฑ = 105 3) ฮฒ = (360 Step 1: To find magnitude of F1 Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University 14 To find magnitude of F1 ๐‘ญ๐Ÿ = ๐‘ญ ๐ฌ๐ข๐ง ๐œท ๐ฌ๐ข๐ง(๐œถ + ๐œท) ๐น1 = 80 sin 30ยฐ sin(45ยฐ + 30ยฐ) F1 = 41.41 N To Find magnitude of F2 ๐‘ญ๐Ÿ = ๐‘ญ ๐ฌ๐ข๐ง ๐œถ ๐ฌ๐ข๐ง(๐œถ + ๐œท) ๐น2 = 80 sin 45ยฐ sin(45ยฐ + 30ยฐ) F2 = 58.56 N A force of 2000 N is acts at an angle of 60ยฐ with X Find its components along 105ยฐ & 330ยฐ with X Given Data: = 2000 N 105 โ€“ 60 = 45ยฐ (360-330) + 60 = 90ยฐ To find magnitude of F1 ๐‘ญ๐Ÿ = ๐‘ญ ๐ฌ๐ข๐ง ๐œท ๐ฌ๐ข๐ง(๐œถ + ๐œท) ๐น1 = 2000 sin 90ยฐ sin(45ยฐ + 90ยฐ) with X-axis. Find its components along 105ยฐ & 330ยฐ with X-axis.
  • 15. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM Step 2: To Find magnitude of F2 Ans: 1. Component along 105ยฐ 2. Component along 330ยฐ = 2000 N 3. The resultant of two forces in a plane is 800 N at 60ยฐ with x axis. One force is 160N at 30ยฐ with x axis. Determine the missing force & its inclination. Solution Given Data: 1. F = 800 N 2. F1 = 160 N 3. ฮฑ = 60 To find: F2 & ฮฒ Step 1: To find ฮฒ Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University 15 F1 = 2828.43 N To Find magnitude of F2 ๐‘ญ๐Ÿ = ๐‘ญ ๐ฌ๐ข๐ง ๐œถ ๐ฌ๐ข๐ง(๐œถ + ๐œท) ๐น2 = 2000 sin 45ยฐ sin(45ยฐ + 90ยฐ) F2 = 2000 N Component along 105ยฐ = 2828.43 N Component along 330ยฐ = 2000 N The resultant of two forces in a plane is 800 N at 60ยฐ with x force is 160N at 30ยฐ with x axis. Determine the missing force & its inclination. Given Data: F = 800 N F1 = 160 N ฮฑ = 60 โ€“ 30 = 30ยฐ F2 & ฮฒ To find ฮฒ ๐‘ญ๐Ÿ = ๐‘ญ ๐ฌ๐ข๐ง ๐œท ๐ฌ๐ข๐ง(๐œถ + ๐œท) The resultant of two forces in a plane is 800 N at 60ยฐ with x- force is 160N at 30ยฐ with x axis. Determine the
  • 16. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University 16 160 = 800 sin ๐›ฝ sin(30 + ๐›ฝ) 160 (sin (30+ฮฒ)) = 800 sin ฮฒ 80 cos ฮฒ = 661.44 sin ฮฒ sin ๐›ฝ cos ๐›ฝ = 80 661.44 tan๐›ฝ = 0.1209 ฮฒ = 6.89ยฐ Step 2: To Find magnitude of F2 ๐‘ญ๐Ÿ = ๐‘ญ ๐ฌ๐ข๐ง ๐œถ ๐ฌ๐ข๐ง(๐œถ + ๐œท) ๐น2 = 800 sin 30 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 ฮธ (ANS: F = 90.82 N and ฮธ = 79.46ยฐ)
  • 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 ๏ถ If the number of forces is more than two, then its resultant can be found out conveniently by the METHOD OF RESOLUTION. ๏ถ STEPWISE PROCEDURE OF METHOD OF RESOLUTION: 1. Resolve all forces horizontally and find the algebraic sum of all the horizontal components (i.e., ฮฃFx) 2. Resolve all forces vertically and find the algebraic sum of all the vertical components (i.e., ฮฃFy). 3. The resultant R of the given forces will be given by the equation: ๐‘น = (๐šบ๐…๐ฑ๐Ÿ ) + (๐šบ๐…๐ฒ๐Ÿ ) 4. The resultant force will be inclined at an angle ฮธ, with the horizontal, such that tan๐œƒ = ฮฃFy ฮฃFx 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 (ฮฃFx) ๐›ด๐น๐‘ฅ = 20 cos + = 36 N Step 2: Resolving all the forces vertically (ฮฃFy) 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 โˆ’ 360 6 = 120ยฐ ๐œƒ = 120 4 = 30ยฐ Resolving all the forces horizontally (ฮฃFx) cos 0ยฐ + 30 cos 30ยฐ + 40 cos 60ยฐ + 50 + 60 cos 120ยฐ = 36 N Resolving all the forces vertically (ฮฃFy) 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- . Resolving all the forces horizontally (ฮฃFx) 50 cos 90ยฐ Resolving all the forces vertically (ฮฃFy)
  • 19. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM ๐›ด๐น๐‘ฆ = 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 20 sin 0ยฐ + 30 sin 30ยฐ + 40 sin 60ยฐ + + 60 sin 120ยฐ = 151.6 N Magnitude of the resultant force, ๐‘น = (๐šบ๐…๐ฑ๐Ÿ ) + (๐šบ๐…๐ฒ๐Ÿ ๐‘… = (36) + (151.6) ๐‘… = 155.8 ๐‘ Direction of the resultant force, ๐ญ๐š๐ง ๐œฝ = ๐šบ๐…๐ฒ ๐šบ๐…๐ฑ tan ๐œƒ = 151.6 36 ๐œƒ = 76.6ยฐ Position of the resultant force, The following forces act at a point : 20 N inclined at 30แต’ towards North of East, 25 N towards North, towards North West, and 35 N inclined at 40แต’ towards South of West. Find the magnitude and direction of the resultant force. + 50 sin 90ยฐ ) 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 (ฮฃFx) ๐›ด๐น๐‘ฅ = 20 cos 30ยฐ + 25 cos 90ยฐ + 30 cos 135ยฐ + 35 cos 220ยฐ = - 30.7 N Step 2: Resolving all the forces vertically (ฮฃFy) ๐›ด๐น๐‘ฆ = 20 sin 30ยฐ + 25 sin 90ยฐ + 30 sin 135ยฐ + 35 sin 220ยฐ = 33.7 N Step 3: Magnitude of the resultant force, ๐‘น = (๐šบ๐…๐ฑ๐Ÿ ) + (๐šบ๐…๐ฒ๐Ÿ ) ๐‘… = (โˆ’30.7) + (33.7) ๐‘… = 45.6 ๐‘ Step 4: Direction of the resultant force, ๐ญ๐š๐ง ๐œฝ = ๐šบ๐…๐ฒ ๐šบ๐…๐ฑ tan ๐œƒ = 33.7 โˆ’30.7 ๐œƒ = 47.7ยฐ Step 5: Position of the resultant force,
  • 21. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM Since ฮฃFx is negative and ฮฃFy 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 ๐ญ๐š๐ง ๐œฝ๐Ÿ = ๐Ÿ๐Ÿ“๐ŸŽ ๐Ÿ“๐ŸŽ๐ŸŽ ๐ญ๐š๐ง ๐œฝ๐Ÿ = ๐Ÿ“๐Ÿ“๐ŸŽ ๐Ÿ๐ŸŽ๐ŸŽ Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University 21 Since ฮฃFx is negative and ฮฃFy 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 ๐Ÿ๐Ÿ“๐ŸŽ ๐Ÿ“๐ŸŽ๐ŸŽ = ๐Ÿ๐Ÿ”. ๐Ÿ•ยฐ ๐Ÿ“๐Ÿ“๐ŸŽ ๐Ÿ๐ŸŽ๐ŸŽ = ๐Ÿ•๐Ÿ—. ๐Ÿ•ยฐ Since ฮฃFx is negative and ฮฃFy is positive, Resultant lies Therefore acute angle of the resultant = 180แต’ - 47.7แต’ = 132.3แต’ 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 ๐ญ๐š๐ง ๐œฝ๐Ÿ‘ = ๐Ÿ๐Ÿ“๐ŸŽ ๐Ÿ‘๐ŸŽ๐ŸŽ ๐ญ๐š๐ง ๐œฝ๐Ÿ’ = ๐Ÿ๐Ÿ“๐ŸŽ ๐Ÿ๐ŸŽ๐ŸŽ Step 1: Resolving all the forces horizontally (ฮฃFx) ๐›ด๐น๐‘ฅ = 20 = 20.25 Step 2: Resolving all the forces vertically (ฮฃFy) ๐›ด๐น๐‘ฆ = 20 sin = 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 ๐Ÿ๐Ÿ“๐ŸŽ ๐Ÿ‘๐ŸŽ๐ŸŽ = ๐Ÿ๐Ÿ”. ๐Ÿ“๐Ÿ”ยฐ ๐Ÿ๐Ÿ“๐ŸŽ ๐Ÿ๐ŸŽ๐ŸŽ = ๐Ÿ”๐Ÿ–. ๐Ÿยฐ Resolving all the forces horizontally (ฮฃFx) 20 cos 16.7ยฐ + 25 cos 79.7ยฐ โˆ’ 10 + 15 cos 68.2ยฐ 20.25 N Resolving all the forces vertically (ฮฃFy) sin 16.7ยฐ + 25 sin 79.7ยฐ + 10 sin 26.56 20.89 N Magnitude of the resultant force, ๐‘น = (๐šบ๐…๐ฑ๐Ÿ ) + (๐šบ๐…๐ฒ๐Ÿ ๐‘… = (20.25) + (20.89 ๐‘… = 29.09 ๐‘ Direction of the resultant force, ๐ญ๐š๐ง ๐œฝ = ๐šบ๐…๐ฒ ๐šบ๐…๐ฑ tan ๐œƒ = 20.89 20.25 ๐œƒ = 45.89ยฐ Position of the resultant force, 10 cos 26.56ยฐ Resolving all the forces vertically (ฮฃFy) 56ยฐ โˆ’ 15 sin 68.2ยฐ ) 89)
  • 23. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM Since ฮฃFx and ฮฃFy is positive, Resultant lies quadrants. 4. For the system shown, determine i) The required value of ฮฑ if resultant of three forces is to be vertical and ii) The corresponding magnitude of resultant. Step I) ฦฉFx = 0 100 cos ๐›ผ + 150 150(cos ๐›ผ. cos 30 cos ๐›ผ โˆ’ 75 Step ii) R = ฦฉ ๐‘… = โˆ’100 Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University 23 Since ฮฃFx and ฮฃFy is positive, Resultant lies the system shown, determine The required value of ฮฑ if resultant of three forces is to be vertical and The corresponding magnitude of resultant. Fx = 0 150 cos(๐›ผ + 30) โˆ’ 200cos ๐›ผ = 0 cos 30 โˆ’ sin ๐›ผ. sin 30) โˆ’ 100 cos ๐›ผ = 0 75 sin ๐›ผ = 0 tan ๐›ผ = sin ๐›ผ cos ๐›ผ = 30 75 = 21.8ยฐ ฦฉFy 100 sin 21.8 โˆ’ 150 sin(21.8 + 30) โˆ’ R = -229.29 N Since ฮฃFx and ฮฃFy is positive, Resultant lies in first The required value of ฮฑ if resultant of three forces The corresponding magnitude of resultant. 200 sin 21.8
  • 24. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University 24 (ANS: ฮฑ = 21.8แต’ & 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 ๐‘€@๐ด = 3. Moment about ๐‘€@ 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 ๐‘€@๐‘‚ = 500 cos 36.87 ร— 3 = 1200 Moment about point A = โˆ’500 sin 36.87 ร— 2 + 500 cos 36.87 Moment about point B @๐ต = โˆ’500 sin 36.87 ร— 4 + 500 cos Moment about point C Moment of force point O, A,B & C 1200 ๐‘. ๐‘š 87 ร— 3 = 600 ๐‘. ๐‘š 36.87 ร— 3 = 0
  • 26. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University 26 ๐‘€@๐ถ = โˆ’500sin 36.87 ร— 4 + 500cos 36.87 ร— 1.5 = โˆ’600๐‘. ๐‘š M@C = 600 N.m 5. Moment about point D ๐‘€@๐ท = โˆ’500sin 36.87 ร— 4 = โˆ’1200 ๐‘. ๐‘š M@D = 1200 N.m 6. Moment about point E ๐‘€@๐ธ = 0 2. Find the moment of force 2000 N about point โ€˜Oโ€™ as shown in fig Solution Step 1: Free Body Diagram Step 2: Finding moments Moment about point O ๐‘€@๐‘‚ = โˆ’2000 sin 30 ร— 5 = โˆ’5000 ๐‘. ๐‘š
  • 27. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University 27 3. Find the moment of force 50 N about point โ€˜Oโ€™ as shown in fig Solution Step 1: Free Body Diagram Step 2: Finding moments Moment about point O ๐‘€@๐‘‚ = โˆ’2000 sin 30 ร— 5 = โˆ’5000 ๐‘. ๐‘š 4. Find the moment of forces as shown in fig. on lever about point โ€œOโ€.
  • 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 ๏ถ STEPWISE PROCEDURE OF METHOD OF RESOLUTION: 1. Find resultant R parallel forces considering proper sign convention. 2. Find ฮฃMo. Take the algebraic sum moment of forces point (say O) (+๐’—๐’† 3. Apply Varignonโ€™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 ฮฃF & ฮฃMo. 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 = ฮฃ F. Take the algebraic sum of all the parallel forces considering proper sign convention. (+๐’—๐’† โ†‘) ( โˆ’๐’—๐’† โ†“) Find ฮฃMo. Take the algebraic sum moment of forces (say O) considering proper sign convention. ๐’—๐’† โ†ป โˆ’๐‘ช๐’๐’๐’„๐’Œ๐’˜๐’Š๐’”๐’†) ( โˆ’๐’—๐’† โ†บ โˆ’๐’‚๐’๐’•๐’Š๐’„๐’๐’๐’„๐’Œ๐’˜๐’Š๐’”๐’† Apply Varignonโ€™s theorem, ๐œฎ๐‘ด๐’ = ๐‘น ร— ๐’… 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 ฮฃF & ฮฃMo. Resultant of parallel force system using Method of Resolution STEPWISE PROCEDURE OF METHOD OF RESOLUTION: = ฮฃ F. Take the algebraic sum of all the parallel forces considering proper sign convention. Find ฮฃMo. Take the algebraic sum moment of forces about a ing proper sign convention. ๐’‚๐’๐’•๐’Š๐’„๐’๐’๐’„๐’Œ๐’˜๐’Š๐’”๐’†) 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 ฮฃF & ฮฃMo.
  • 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 = ฮฃFy = - 70 + 100 + 50 = 50 N Step 2: To find ฮฃMo Taking moment about point โ€œOโ€ ๐›ด๐‘€ = 100 = 394 N Step 3: Applying Varignonโ€™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 โ€“ 86-34 + 90 = 50 N (โ†‘) To find ฮฃMo Taking moment about point โ€œOโ€ 100 ร— 1.5 + 50 ร— 3.5 โˆ’ 86 ร— 6.5 โˆ’ 34 = 394 N-m (โ†บ) Applying Varignonโ€™s theorem, ๐›ด๐‘€ = ๐‘… ร— ๐‘‘ ๐‘‘ = ๐›ด๐‘€ ๐‘… = 394 50 = 7.88 ๐‘š 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 ร— 8 + 90 ร— 10
  • 34. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM Case II : To find Step 1: To find magnitude of Resultant R R = ฮฃF = - 70 + 100 + 50 = 50 N Step 2: To find ฮฃM Taking moment about point โ€œAโ€ ๐›ด๐‘€ = 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 โ€“ 86 - 34 + 90 = 50 N (โ†‘) To find ฮฃMA Taking moment about point โ€œAโ€ 70 ร— 3.5 โˆ’ 100 ร— 2 โˆ’ 86 ร— 3 โˆ’ 34 ร— 4 219 N-m (โ†บ) 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 ร— 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 = ฮฃF = 100 + 200 -150 = 150 N (โ†’) Step 2: To find ฮฃMB Taking moment about point โ€œBโ€ ๐›ด๐‘€ = 150 + 100 ร— 5 โˆ’ 150 ร— 3.5 + 200 ร— 1.5 = 425 N-m = 425 N-m (โ†ป) Step 3: Applying Varignonโ€™s theorem, ๐›ด๐‘€ = ๐‘… ร— โ„Ž โ„Ž = ๐›ด๐‘€ ๐‘… = 425 150 = 2.83 ๐‘š 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 = ฮฃF = - 50 + 60 โ€“ 70 + 80 โ€“ 90 = - 70 N = 70 N Step 2: To find ฮฃMA Taking moment about point โ€œAโ€ ๐›ด๐‘€ = โˆ’60 ร— 3 + 70 ร— 6 โˆ’ 80 ร— 7.5 + 90 ร— 9 = 450 N-m = 450 N-m (โ†ป) Step 3: Applying Varignonโ€™s theorem, ๐›ด๐‘€ = ๐‘… ร— ๐‘‘ ๐‘‘ = ๐›ด๐‘€ ๐‘… = 450 70 = 6.429 ๐‘š 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 ๐œƒ = ๐‘‚๐‘๐‘ ๐ป๐‘ฆ๐‘๐‘œ sin 30ยฐ = ๐‘ฅ 1 x1 = 0.5 m For 100 N force sin 60ยฐ = ๐‘ฅ 1 x2 = 0.866 m For 50 N force sin 45ยฐ = ๐‘ฅ 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 = ฮฃF = - 100 โ€“ 80 + 150 - 50 = 80 N (โ†) Step 3: To find ฮฃMo Taking moment about point โ€œOโ€ ๐›ด๐‘€ = 100 ร— 0.866 + 80 ร— 0.5 + 150 ร— 0.5 โˆ’ 50 ร— 0.707 = 166.25 N-m (โ†บ) Step 3: Applying Varignonโ€™s theorem, ๐›ด๐‘€ = ๐‘… ร— ๐‘‘ ๐‘‘ = ๐›ด๐‘€ ๐‘… = 166.25 80 = 2.078 ๐‘š 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 ๏ถ STEPWISE PROCEDURE OF METHOD OF RESOLUTION: 1. Resolve all forces horizontally and find the algebraic sum of all the horizontal components (i.e., ฮฃFx) 2. Resolve all forces vertically and find the algebraic sum of all the vertical components (i.e., ฮฃFy). 3. Find the magnitude of Resultant R The resultant R of the given forces will be given by the equation: ๐‘น = (๐šบ๐…๐ฑ๐Ÿ ) + (๐šบ๐…๐ฒ๐Ÿ ) 4. Find direction ฮธ. The resultant force will be inclined at an angle ฮธ, with the horizontal, such that tan๐œƒ = ฮฃFy ฮฃFx 5. Find ฮฃMo. Take the algebraic sum moment of forces about a point (say O) considering proper sign convention. (+๐’—๐’† โ†ป โˆ’๐‘ช๐’๐’๐’„๐’Œ๐’˜๐’Š๐’”๐’†) ( โˆ’๐’—๐’† โ†บ โˆ’๐’‚๐’๐’•๐’Š๐’„๐’๐’๐’„๐’Œ๐’˜๐’Š๐’”๐’†) 6. Apply Varignonโ€™s theorem, I. ๐œฎ๐‘ด๐’ = ๐‘น ร— ๐’…,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. ๐œฎ๐‘ด๐’ = ๐œฎ๐‘ญ๐’™ ร— ๐’š, Where y = distance between point O and the intersection of line of action of resultant R with y-axis. OR III. ๐œฎ๐‘ด๐’ = ๐œฎ๐‘ญ๐’š ร— ๐’™, 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 ฮฃFx, ฮฃFy and ฮฃMo 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 (ฮฃFx) ๐›ด๐น๐‘ฅ = โˆ’100 cos 36.87ยฐ โˆ’ 75 = 155 N (โ†) Step 2: Resolving all the forces vertically (ฮฃFy) ๐›ด๐น๐‘ฆ = โˆ’200 + 30 โˆ’ 100 sin 36.87ยฐ = 210 N (โ†“) Step 3: Magnitude of the resultant force, ๐‘น = (๐šบ๐…๐ฑ๐Ÿ ) + (๐šบ๐…๐ฒ๐Ÿ ) ๐‘… = (155) + (210) ๐‘… = 261 ๐‘ Step 4: Direction of the resultant force, ๐ญ๐š๐ง ๐œฝ = ๐šบ๐…๐ฒ ๐šบ๐…๐ฑ tan ๐œƒ = 210 155 ๐œƒ = 53.57ยฐ Step 5: Find ฮฃMA ๐›ด๐‘€ = 50 ร— 2 + 80 โˆ’ 100 sin 36.87ยฐ
  • 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 ๐œƒ1 = ๐‘‚๐‘๐‘ ๐ด๐‘‘๐‘— = 2 1.5 1) ๐œƒ1 = tan . = 53.13ยฐ 2) ๐œƒ2 = tan = 33.69ยฐ 3) ๐œƒ3 = tan = 18.44ยฐ 4) ๐œƒ4 = tan . = 53.13ยฐ 5) ๐œƒ5 = tan . = 50.13ยฐ Step 1: Resolving all the forces horizontally (ฮฃFx) ๐›ด๐น๐‘ฅ = 65 cos 53.13ยฐ + 140 cos 33.69 โˆ’ 90 cos 18.44 โˆ’ 100 cos 53.13 + 50 cos 50.13 = 42.12 N (โ†’) Step 2: Resolving all the forces vertically (ฮฃFy) ๐›ด๐น๐‘ฆ = โˆ’65 sin 53.13ยฐ + 140 sin 33.69 โˆ’ 90 sin 18.44 + 100 sin 53.13 โˆ’ 50 sin 50.13 = 38.78 N (โ†‘)
  • 46. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University 46 Step 3: Magnitude of the resultant force, ๐‘น = (๐šบ๐…๐ฑ๐Ÿ ) + (๐šบ๐…๐ฒ๐Ÿ ) ๐‘… = (42.12) + (38.78) ๐‘… = 57.25 ๐‘ Step 4: Direction of the resultant force, ๐ญ๐š๐ง ๐œฝ = ๐šบ๐…๐ฒ ๐šบ๐…๐ฑ tan ๐œƒ = 38.78 42.12 ๐œƒ = 42.64ยฐ Step 5: To find ฦฉMo, ฦฉ๐‘€๐‘œ = 65 cos 53.13 ร— 3 โˆ’ 140 sin 33.39 ร— 1.5 โˆ’ 90 cos 18.44 ร— 3 + 90 sin 18.44 ร— 4.5 โˆ’ 100 cos 53.13 ร— 1 โˆ’ 100 sin 53.13 ร— 6 + 50 cos 50.19 ร— 3 + 50 sin 50.19 ร— 6
  • 47. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM Step 6: Apply Varignonโ€™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 ฦฉ๐‘€๐‘œ = 341.03 ๐‘. ๐‘š โ†บ Apply Varignonโ€™s theorem ฦฉ๐‘€๐‘œ = ฦฉfy x X ๐‘ฅ = 341.03 38.78 = 8.79๐‘š 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. โ†บ 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 (ฮฃFx) ๐›ด๐น๐‘ฅ = 100 cos 40 + 85 cos 50 + 70 sin 40 = 176.24 N (โ†’) Step 2: Resolving all the forces vertically (ฮฃFy) ๐›ด๐น๐‘ฆ = 100 sin 40 โˆ’ 85 sin 50 โˆ’ 90 โˆ’ 70cos 40 = 144.46 N (โ†“) Step 3: Magnitude of the resultant force, ๐‘น = (๐šบ๐…๐ฑ๐Ÿ ) + (๐šบ๐…๐ฒ๐Ÿ ) ๐‘… = (176.24) + (144.46) ๐‘… = 227.88 ๐‘ Step 4: Direction of the resultant force, ๐ญ๐š๐ง ๐œฝ = ๐šบ๐…๐ฒ ๐šบ๐…๐ฑ tan ๐œƒ = 144.46 176.24 ๐œƒ = 39.34ยฐ Step 5: Find ฮฃMO ๐›ด๐‘€ = โˆ’100 cos 40 ร— 4 + 100 sin 40 ร— 4 + 85 cos 50 ร— 5 โˆ’ 85 sin 50 ร— 2 + 90 ร— 3 + 175 โˆ’ 150 ฮฃMO = 388.65 N. m (โ†ป) Step 6: Apply Varignonโ€™s theorem R x d ๐‘‘ = 388.65 227.88 = 1.70 ๐‘š
  • 49. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM Step 7: Position of Resultant R W.r.t. O Step 8: Find ฮฃM ๐›ด๐‘€ = โˆ’175 โˆ’ โˆ’ โˆ’ Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University 49 ฦฉFx . y ๐‘ฆ = 388.65 176.24 = 2.20๐‘š ฦฉFy . x ๐‘ฅ = 388.65 144.46 = 2.69 ๐‘š Position of Resultant R W.r.t. O Find ฮฃMA + 150 โˆ’ 70 sin 40 ร— 4 โˆ’ 70 cos 40 ร— 4 โˆ’ 85 sin 50 ร— 2 โˆ’ 85 cos 50 ร— 9 โˆ’ 90 ฮฃMA = 1671.43 N. m (โ†บ 90 ร— 7 โ†บ)
  • 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 (ฮฃFx) ๐›ด๐น๐‘ฅ = 84 cos 40 โˆ’ 55 cos 35 + 79 cos 60 + 50 = 108.79 N (โ†’) Step 2: Resolving all the forces vertically (ฮฃFy) ๐›ด๐น๐‘ฆ = โˆ’84 sin 40 โˆ’ 55 sin 35 โˆ’ 79 sin 60 + 123 โˆ’ 60 = 90.96 N (โ†“) Step 3: Magnitude of the resultant force, ๐‘น = (๐šบ๐…๐ฑ๐Ÿ ) + (๐šบ๐…๐ฒ๐Ÿ ) ๐‘… = (108.79) + (90.96) ๐‘… = 141.80 ๐‘ Step 4: Direction of the resultant force, ๐ญ๐š๐ง ๐œฝ = ๐šบ๐…๐ฒ ๐šบ๐…๐ฑ tan ๐œƒ = 90.96 108.79 ๐œƒ = 39.89ยฐ
  • 52. Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University 52 Step 5: Find ฮฃMO ๐›ด๐‘€ = 84 โˆ— 2.5 โˆ’ 123 โˆ— 2.5 + 55 โˆ— 2.5 โˆ’ 79 โˆ— 2.5 + 50 โˆ— 2.5 cos 40 โˆ’ 60 โˆ— 2.5 sin 40 ฮฃMO = 158.16 N. m (โ†บ) Step 6: Apply Varignonโ€™s theorem R x d ๐‘‘ = 158.16 141.80 = 1.12 ๐‘š 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, ฮธ=86.48ยฐ, Hint: Fig.1. FBD Q2. Determine the resultant of the following force system Ans: R = 199.64 N & ฮธ=4.10 ยฐ Prepared By: Prof. V.V. Nalawade , Assistant Professor, MGM University 53 R = 12.71 KN, ฮธ=86.48ยฐ, ฦฉMA= 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 & ฮธ=4.10 ยฐ = 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 ฮธ as shown in fig. Determine the magnitude and ฮธ of the resultant Ans: R = 623.24 N & ฮธ =75.4 ยฐ Q.4. State and Explain a) Parallelogram Law B) Varignonโ€™s Theorem Q.5. Classify the force System