### Resultant of Forces

• 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:

#### 22-2-01

The resultant of two equal vectors is equal to either of them. The angle between two vectors will be: (UP TGT Math 2016)

Solution:
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$^\circ$

### Solved Example:

#### 22-2-02

Find the resultant of the parallel force system as shown in the figure and locate the same with respect to point C:

Solution:

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

### Solved Example:

#### 22-2-03

The Law of Polygon of Forces states that: (ISRO Scientist Civil 2015)

Solution:
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.

### Solved Example:

#### 22-2-04

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.

Solution:
$A = 5\ N,\ B = 20\ N,\ \theta = 120^\circ,\ R = ?,\ \beta =?$ \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*} $\tan \beta = \dfrac{5 \sin 120^\circ}{20 + 5 \cos 120^\circ} = 0.2475$ $\beta = \tan^{-1}(0.2475)= 13 ^\circ 54'$

### Solved Example:

#### 22-2-05

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)

Solution:
\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*}

### Solved Example:

#### 22-2-06

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:

Solution:
\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*}

### Solved Example:

#### 22-2-07

If a rigid body is in equilibrium under the action of three forces, then:

### Solved Example:

#### 22-2-08

D' Alembert's principle is used for: (UPPSC AE Civil 2019)

### Solved Example:

#### 22-2-09

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.

Solution:

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,

$F^2(\cos^2\theta + \sin^2 \theta) = 46938 \mathrm{\ or,\ } F = 216.65 N$

### Solved Example:

#### 22-2-10

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:

### Solved Example:

#### 22-2-11

The resolved part of the resultant of two forces inclined at an angle 9$^\circ$ in a given direction is equal to:

### Solved Example:

#### 22-2-12

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.

Solution:
$R =\sqrt {20^{2}+30^{2}} =\sqrt {400+900} =\sqrt {1300} =36.05\ N$

### Solved Example:

#### 22-2-13

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?

Solution:
$F_x = 400 \cos 120^\circ - 200 \cos 20 = -387.94\ N$ $F_y = 400 \sin 120^\circ - 200 \sin 20= 278\ N$ $F = \sqrt{(-387.94)^2 + (278)^2}= 477.27\ N$ $\theta = \tan^{-1} \left(\dfrac{278}{-387.94}\right)= -35.62^\circ$ The force is in second quadrant as F$_x$ is negative and F$_y$ is positive. $\theta = -35.62 + 180 = 144.37^\circ$

### Solved Example:

#### 22-2-14

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$^\circ$.

Solution:
$F_x = 600 \cos 40^\circ + 800 \cos 160^\circ + 200 \cos 300^\circ= -192.13\ N$ $F_y = 600 \sin 40^\circ + 800 \sin 160^\circ + 200 \sin 300^\circ= 486.08\ N$ $F = \sqrt{(-192.13)^2 + (486.08)^2}= 522.67\ N$ $\theta = \tan^{-1} \left(\dfrac{486.08}{-192.13}\right)= -68.43^\circ$ The force is in second quadrant as F$_x$ is negative and F$_y$ is positive. $\theta = -68.43 + 180 = 111.56^\circ$

### Solved Example:

#### 22-2-15

The equilibrant of the forces 10 N at 10$^\circ$ and 15 N at 100$^\circ$ is:

Solution:
$R_x =10\cos 10^\circ + 15\cos 100^\circ=7.24\ N$ $R_y =10\sin 10^\circ + 15\sin 100^\circ=16.50\ N$ $R =\sqrt {17.24^{2} + 16.50^{2}}=18.02\ N$ $\theta =\tan ^{-1}\left( \dfrac {16.50}{7.24}\right) =66.3^\circ$ The equilibrium will be same in magnitude but opposite in direction. Equilibrant = 18.02N@180+66.3 = 246$^\circ$

A point possesses simultaneously velocities whose measures are 4, 3, 2, and 1; the angle between the first and second is 30$^\circ$, between the second and third 90$^\circ$, and between the third and fourth 120$^\circ$; find the resultant velocity and angle to the direction of the first velocity? (AAI ATC Junior Executive March 2021 Shift I)