Resultants of Force Systems
Resolution of Forces
Learning Objectives:

Resolve forces in horizontal/vertical as well as parallel/perpendicular to a plane.
 Resolution of force means separating a force, into two mutually perpendicular components. These components are normally horizontal and vertical.
 Occasionally, a force is divided into two forces parallel and perpendicular to a plane. Resolution of force is done to simplify the process of adding or subtracting multiple forces.
 Typically when you want to add or subtract multiple forces, which are noncollinear, they are broken down into horizontal and vertical forces.
 All horizontal forces are added, similarly all vertical forces are added.
 Finally the resultant is found out from these two components.
Solved Example: 22101
The method of splitting a single force into two perpendicular components along xaxis and yaxis is called as:
A. Resolution
B. Resultant
C. Concurrent
D. Mapping
Separating a force into two forces for the purpose of addition or subtraction is called resolution.
Resultant of a force is exactly opposite process where two component forces are combined together.
concurrent forces are defined as forces that pass through a common point.
Correct Answer: A
Solved Example: 22102
What are the X and Y components of the force system shown below?
A. X = 186.00 N, Y = 464 N
B. X = 464.23 N, Y = 185 N
C. X = 466.12 N, Y = 180 N
D. None of the above are present
$\theta = \tan^{1}\left( \dfrac{2}{5}\right) = 21.8~^\circ$
Horizontal Component, $F_x = F \cos \theta = 500 \times \cos 21.8^\circ = 464.23\ N$
Vertical Component, $F_y = F \sin \theta = 500 \times \sin 21.8^\circ = 185.68~N$
Correct Answer: B
Solved Example: 22103
A force of 10 N is making an angle of 30$^\circ$ with the horizontal. Its horizontal component will be:
A. 4 N
B. 5 N
C. 7 N
D. 10 N
Horizontal Component, $F_x = F \cos \theta = 10 \times \cos 30^\circ = 5\ N$
Correct Answer: B
Solved Example: 22104
A man is pulling a trolley with 100 N weight on a horizontal road with a force of 100 N making 45$^\circ$ with the road. The horizontal and vertical components of the net force on trolley will be:
A. 52.5 N, 85.09 N
B. 55.6 N, 78.6 N
C. 60.3 N, 54.11 N
D. 70 N, 29.23 N
Horizontal Component, $F_x = F \cos \theta = 100 \times \cos 45^\circ = 70\ N$
Net Vertical Component,
\begin{align*}
F_y &= W  F\sin \theta\\ &= 100  F \sin \theta\\ &= 100  100 \times \sin 45^\circ\\ &= 29.23\ N
\end{align*}
Correct Answer: D
Resultant of Forces
Learning Objectives:

Adding two forces using the law of parallelogram.

Adding two forces using the law of triangles.

Adding more than two forces using the force polygon.

Find resultant of multiple concurrent forces acting on a point using analytical and graphical method.
Parallelogram Law:
The law of parallelogram of forces states that if two vectors acting on a particle at the same time be represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point their resultant vector is represented in magnitude and direction by the diagonal of the parallelogram drawn from the same point.
Triangle Law:
If two forces acting simultaneously on a body are represented by the sides of a triangle taken in order, their resultant is represented by the closing side of the triangle taken in the opposite order.
Polygon Law:
Polygon law of vector addition states that if a number of vector are represented completely by the side of a polygon taken in order their resultant is fully represented by the closing side of the polygon taken in opposite order.
Solved Example: 22201
The resultant of two equal vectors is equal to either of them. The angle between two vectors will be: (UP TGT Math 2016)
A. 60$^\circ$
B. 120$^\circ$
C. 180$^\circ$
D. 90$^\circ$
The resultant should make 60$^\circ$ with both forces F$_1$ and F$_2$ as shown in the following figure. Then the smaller triangles will be equilateral and resultant will be equal to either of the forces.
Angle between F$_1$ and F$_2$ = 60$^\circ$ + 60$^\circ$ = 120$^\circ$
Correct Answer: B
Solved Example: 22202
Find the resultant of the parallel force system as shown in the figure and locate the same with respect to point C:
A. 17.48 mm
B. 23.92 mm
C. 38.33 mm
D. 41.11 mm
Let R be the resultant as shown in the figure.
R = 15  60 + 10  25 = 60 N
Also,15 $\times$ 40  10 $\times$ 30 + 25 $\times$ 80 = 60 $\times$ d
d = 38.33 mm right of C
Correct Answer: C
Solved Example: 22203
The Law of Polygon of Forces states that: (ISRO Scientist Civil 2015)
A. If forces acting on a point can be represented in magnitude and direction by the sides of a polygon taken in order, then the resultant of the forces will be represented in magnitude and direction by t
B. If forces acting on a point can be represented of a polygon taken in order, their sides of a polygon taken in order, their resultant will be represented in magnitude and direction by the closing side
C. If a polygon representing the forces acting at point in a body is closed, the forces are in equilibrium.
D. If forces acting on a point can be represented in magnitude and direction by the sides of a polygon in order, the forces are in equilibrium.
Polygon law of vector addition states that if a number of vectors can be represented in magnitude and direction by the sides of a polygon taken in the same order, then their resultant is represented in magnitude and direction by the closing side of the polygon taken in the opposite order.
Correct Answer: B
Solved Example: 22204
Two forces 5 N and 20 N are acting at an angle of 120$^\circ$ between them. Find the resultant force in magnitude and direction.
A. 13.21 N, 17$^\circ$14'
B. 14.08 N, 21$^\circ$50'
C. 16.99 N, 24$^\circ$33'
D. 18.03 N, 13$^\circ$54'
\[A = 5\ N,\ B = 20\ N,\ \theta = 120^\circ,\ R = ?,\ \beta =? \] \begin{align*} R &= \sqrt{A^2+ B^2+ 2 AB \cos \theta}\\ &= \sqrt{5^2 + 20^2+ 2 \times 5 \times 20 \cos 120^\circ}\\ &= \sqrt{325}\\ &= 18.03\ N \end{align*} \[\tan \beta = \dfrac{5 \sin 120^\circ}{20 + 5 \cos 120^\circ} = 0.2475\] \[\beta = \tan^{1}(0.2475)= 13 ^\circ 54'\]
Correct Answer: D
Solved Example: 22205
If two concurrent forces A and B acting on a point are 200 N and 300 N. What is the magnitude of resultant force, if it makes an angle of 50$^\circ$ with each force? (Based on DSSSB AE Civil Sept 2021)
A. 471.08 N
B. 455.12 N
C. 400.56 N
D. Insufficient data
\begin{align*} R &= \sqrt{P^2 +Q^2 + 2PQ \cos \theta}\\ &= \sqrt{200^2 + 300^2 + 2\times 200 \times 300 \times \cos 50 ^\circ}\\ &= 455.12\ N \end{align*}
Correct Answer: B
Solved Example: 22206
Two forces with equal magnitudes F act on a body and the magnitude of the resultant force is $\dfrac{F}{3}$. The angle between the two forces is:
A. $\cos^{1}\left( \dfrac{17}{18} \right)$
B. $\cos^{1}\left( \dfrac{1}{3} \right)$
C. $\cos^{1}\left( \dfrac{2}{3} \right)$
D. $\cos^{1}\left( \dfrac{8}{9} \right)$
\begin{align*} R &= \sqrt{A^2+ B^2+ 2 AB \cos \theta}\\ \dfrac{F}{3}&= \sqrt{F^2 + F^2+ 2 \times F \times F \cos \theta}\\ \dfrac{F}{3}&= \sqrt{2F^2 + 2 \times F^2\cos \theta}\\ \dfrac{F^2}{9}&= {2F^2 + 2 F^2\cos \theta}\\ \end{align*} \begin{align*} \dfrac{1}{9}&= 2(1 + \cos \theta)\\ \dfrac{1}{18}&= {1 + \cos \theta}\\ \left(\dfrac{17}{18} \right) &= \cos \theta\\ \theta &= \cos^{1}\left( \dfrac{17}{18} \right) \end{align*}
Correct Answer: A
Solved Example: 22207
If a rigid body is in equilibrium under the action of three forces, then:
A. These forces are equal
B. The lines of action of these forces meet in a point
C. The lines of action of these forces are parallel
D. (B) and (C) both are possible
Correct Answer: D
Solved Example: 22208
D' Alembert's principle is used for: (UPPSC AE Civil 2019)
A. Reducing the problem of kinetics to equivalent statics problem
B. Determining stresses in the truss
C. Stability of floating bodies
D. Designing safe structures
Correct Answer: A
Solved Example: 22209
A ring is pulled by three forces as shown in the figure. Find the force F and the angle $\theta$ if the resultant of these forces is 100N acting in the vertical direction.
A. 197.25N, $\angle$ 27.8$^\circ$
B. 216.65N, $\angle$ 47.6$^\circ$
C. 293.32N, $\angle$ 40.4$^\circ$
D. 136.88N, $\angle$ 31.6$^\circ$
In x: $0 = 120\sin 60 F\cos\theta +250, \mathrm{\ or,\ } F\cos \theta = 146.07$,
In y: $100 = 120 \cos 60 + F\sin \theta \mathrm{\ or,\ } F \sin \theta = 160$
Dividing these two equations $\tan \theta = 0.9130 \mathrm{\ or,\ } \theta = 47.60^\circ$
Squaring and adding these two equations,
$F^2(\cos^2\theta + \sin^2 \theta) = 46938 \mathrm{\ or,\ } F = 216.65 N$
Correct Answer: B
Solved Example: 22210
The algebraic sum of the resolved parts of a number of forces in a given direction is equal to the resolved part of their resultant in the same direction. This is as per the principle of:
A. Independence of forces
B. Dependence of forces
C. Balance of force
D. Resolution of forces.
Correct Answer: D
Solved Example: 22211
The resolved part of the resultant of two forces inclined at an angle 9$^\circ$ in a given direction is equal to:
A. The algebraic sum of the resolved parts of the forces in the given direction
B. The sum of the resolved parts of the forces in the given direction
C. The difference of the forces multiplied by the cos of 9$^\circ$
D. The sum of the forces multiplied by the sin of 9$^\circ$
Correct Answer: A
Solved Example: 22212
What is the magnitude of the resultant force of the two forces which are perpendicular to each other? The two forces are 20 units and 30 units respectively.
A. 36
B. 42
C. 25
D. 40
\[R =\sqrt {20^{2}+30^{2}} =\sqrt {400+900} =\sqrt {1300} =36.05\ N\]
Correct Answer: A
Solved Example: 22213
The resultant of two forces in a plane is 400 N at 120$^\circ$. If one of the forces is 200 N at 20$^\circ$ what is the other force?
A. 347.77 N at 114.85$^\circ$
B. 435.77 N at 104.37$^\circ$
C. 357.56 N at 114.24$^\circ$
D. 477.27 N at 144.38$^\circ$
\[F_x = 400 \cos 120^\circ  200 \cos 20 = 387.94\ N\] \[F_y = 400 \sin 120^\circ  200 \sin 20= 278\ N \] \[F = \sqrt{(387.94)^2 + (278)^2}= 477.27\ N \] \[\theta = \tan^{1} \left(\dfrac{278}{387.94}\right)= 35.62^\circ\] The force is in second quadrant as F$_x$ is negative and F$_y$ is positive. \[\theta = 35.62 + 180 = 144.37^\circ\]
Correct Answer: D
Solved Example: 22214
Determine the resultant of the following forces: A = 600 N at 40$^\circ$, B = 800 N at 160$^\circ$ and C = 200 N at 300$^\circ$.
A. 532.78 N, 55.32$^\circ$
B. 435.94 N, 235.12$^\circ$
C. 522.68 N, 111.57$^\circ$
D. 627.89 N, 225.81$^\circ$
\[F_x = 600 \cos 40^\circ + 800 \cos 160^\circ + 200 \cos 300^\circ= 192.13\ N\] \[F_y = 600 \sin 40^\circ + 800 \sin 160^\circ + 200 \sin 300^\circ= 486.08\ N\] \[F = \sqrt{(192.13)^2 + (486.08)^2}= 522.67\ N\] \[\theta = \tan^{1} \left(\dfrac{486.08}{192.13}\right)= 68.43^\circ \] The force is in second quadrant as F$_x$ is negative and F$_y$ is positive. \[\theta = 68.43 + 180 = 111.56^\circ\]
Correct Answer: C
Solved Example: 22215
The equilibrant of the forces 10 N at 10$^\circ$ and 15 N at 100$^\circ$ is:
A. 18 N at 246$^\circ$
B. 18 N at 66$^\circ$
C. 25 N at 114$^\circ$
D. 25 N at 66$^\circ$
\[R_x =10\cos 10^\circ + 15\cos 100^\circ=7.24\ N\] \[R_y =10\sin 10^\circ + 15\sin 100^\circ=16.50\ N\] \[R =\sqrt {17.24^{2} + 16.50^{2}}=18.02\ N\] \[\theta =\tan ^{1}\left( \dfrac {16.50}{7.24}\right) =66.3^\circ\] The equilibrium will be same in magnitude but opposite in direction. Equilibrant = 18.02N@180+66.3 = 246$^\circ$
Correct Answer: A
Moment of a Force
Learning Objectives:

Calculate the moment of a force in various situations.
The Moment of a force is a measure of its tendency to cause a body to rotate about a specific point or axis.
Solved Example: 22301
Moment of a force is defined as: (SSC JE ME Oct 2020 Morning)
A. Product of force 'F' and perpendicular distance 'd' moved in direction of force
B. Quotient of force 'F' and perpendicular distance 'd' moved in direction of force
C. Product of force 'F' and perpendicular distance 'd' from line of action of force to pivot
D. Quotient of force 'F' and perpendicular distance 'd' from line of action of force to pivot
Correct Answer: C
Solved Example: 22302
Moment of force applied on a door is 15 N m and force applied is 3.75 N, distance of handle from pivot is: (Airforce Group X Jul 2021Shift I)
A. 11.25 m
B. 18.75 m
C. 4 m
D. 45 m
Using, $M = F \times d, \quad d = 4\ m$
Correct Answer: C
Couple
Learning Objectives:

Define which forces constitute a couple.

Define the moment of a couple.
It consists of two equal and opposite forces whose lines of action do not coincide. The forces have a turning effect or moment called a torque about an axis which is normal to the plane of the forces.
The SI unit for the torque of the couple is newton metre.
If the two forces are F and F, then the magnitude of the torque is given by the following formula: \(\tau = F \times d\) where,
\(\tau\) = torque
F = magnitude of one of the forces
d = perpendicular distance between the forces, sometimes called the arm of the couple
Solved Example: 22401
In case of forces, a couple means: (KPSC JE 2017)
A. Two unequal forces acting at two points
B. Two equal and like parallel forces acting at two points
C. Two equal and perpendicular forces acting at two points
D. Two equal and opposite forces acting at two points
Correct Answer: D
Solved Example: 22402
The distance between the forces acting in the couple is: (RRB JE ME CBT II: Aug 2019)
A. A finite variable distance
B. Infinity
C. Always a fixed distance
D. Zero
Correct Answer: A