## Statics – Equilibrium Of Body

## The Conditions Of Equilibrium

A body is said to be in equilibrium when

- The linear acceleration of the body is zero and
- The angular acceleration of any axis is zero.

If the linear and angular accelerations are zero, the body is not necessarily at rest. A body having uniform velocity or uniform angular velocity also has zero linear acceleration and zero angular acceleration, respectively.

**Equilibrium Of Body Definition:** A body is in equilibrium if it is either at rest or in motion with uniform linear or angular velocity, or with a combination of both uniform linear and angular velocities.

**Read and Learn More: Class 11 Physics Notes**

- From Newton’s second law of motion, we know that the acceleration of a body is related to the applied external force. Whether any external force is acting on the body cannot be determined from the observation of the equilibrium of the body.
- There may or may not be any external force acting on the body when the body is in equilibrium.
- No object can remain in equilibrium under the action of a single force but a number of forces together may keep the object in equilibrium. Again, only external forces can change the state of equilibrium of a body.
- Internal forces like interatomic forces cannot shift a body from its equilibrium position as internal forces occur in pairs and balance each other. Hence, internal forces play no part in determining the equilibrium state of a body.

From the above discussions, it is evident that there are two conditions under which a body can be kept in equilibrium when forces are applied at different points on it. We consider only the action of coplanar forces in our discussion.

**Equilibrium Of Body 1st Condition**: For equilibrium, the vector sum (i.e., resultant) of all coplanar forces acting on a body must be zero.

Mathematically, we can express this condition as \(\vec{R}=\vec{F}_1+\vec{F}_2+\vec{F}_3+\cdots=\sum_i \vec{F}_i=\overrightarrow{0}\)…(1)

Resolving the forces into mutually perpendicular components along suitably chosen x and y-axes, we get,

⇒ \(R_x =F_{1 x}+F_{2 x}+F_{3 x}+\cdots=\sum_i F_{i x}=0\)….(2)

and \(R_y=F_{1 y}+F_{2 y}+F_{3 y}+\cdots=\sum_i F_{i y}=0\)….(3)

Hence, The Condition Can Be Expressed In Another Form: For coplanar forces acting on a body, the sum of components along a given direction and along the direction perpendicular to it should separately be zero.

When this condition is fulfilled, the body will not have a linear acceleration.

**Equilibrium Of Body 1st Condition Vector Form:** Equation (1) is the vector form of the 1st condition.

**Alternative Description Of The 1st Condition:** Applying the law of polygon of vectors it can be stated that if the magnitude and direction of forces (F_{1}, F_{2}, F_{3}, etc.), acting on a body, can represent the sides of a closed polygon taken in order, then the body remains in equilibrium under the action of the forces.

**Alternative Description Of The 2nd Condition:** The algebraic sum of moments of the coplanar forces acting on a body, taken about any point, should be zero.

- The mathematical expression for the above condition may be written by taking moments about any algebraic point O. The sum of the moments of the forces is, τ = F
_{1}r_{1}+ F_{2}r_{2}+ F_{3}r_{3}+…. = 0 - If the algebraic sum of the moments of the coplanar forces is equal to zero about a point on that plane, the sum will also be zero about all other points on that plane. For example, the sum will be zero about the point O’ also.
- When this condition is fulfilled, the angular acceleration of the body will be zero. A body will remain in equilibrium when both conditions are fulfilled together.
- Shows an incomplete polygon formed by the vectors. Clearly, in this case, the first condition has not been fulfilled.

**Alternative Description Vector form:** Vector form of the 2nd condition,

∴ \(\vec{\tau}=\overrightarrow{r_1} \times \vec{F}_1+\vec{r}_2 \times \vec{F}_2+\vec{r}_3 \times \vec{F}_3+\cdots\)

= \(\sum_i \overrightarrow{r_i} \times \vec{F}_i=\overrightarrow{0}\)….(4)

**Equilibrium Under Two Forces:** When two equal and oppositely directed forces act along the same straight line, the resultant force becomes zero. The condition for equilibrium is that both forces acting on the body should lie on the same straight line, and be equal in magnitude but directed opposite to each other.

**Equilibrium Under Two Forces Example:** A pendulum is suspended from a rigid support using a string. When the string is vertical, the pendulum remains in equilibrium. In this case, the weight (W) of the pendulum and the tension (T) in the string are equal and opposite.

But if the bob is displaced from the equilibrium position, the weight W and the tension T are neither equal nor opposite. Hence, the system cannot be in equilibrium.

**Equilibrium Under Three Parallel Forces: **Conditions for the equilibrium of a body under the action of three parallel forces are

- They must be coplanar,
- The resultant of any two forces must be equal and opposite to the third force and
- The resultant of two forces and the third force should have the same line of action.

A body in equilibrium under the action of three coplanar forces \(\vec{F}_1\), \(\vec{F}_2\), and \(\vec{F}_3\) is shown. The line of action of the resultant \(\vec{R}\), of \(\vec{F}_1\) and \(\vec{F}_2\), is the same as that of \(\vec{F}_3\).

∴ \(\vec{R}\) = –\(\vec{F}_3\)

The conditions for equilibrium under three non-parallel forces can be expressed in any of the following four ways:

1. **Law Of Resultant:** The three forces should be coplanar and the resultant of any two of the forces must be equal and opposite to the third one and the lines of action of all the three forces should pass through the same point

2. **Law Of Triangle Of Forces:** For three non-parallel, coplanar forces acting on a body, if the lines of action of the forces pass through the same point and the magnitudes and directions of these forces can be represented by the three sides of a triangle taken in order, then the body remains in equilibrium.

Suppose the forces P, Q, R act on a body at a single point and the magnitude and direction of the forces \(\vec{P}\) and \(\vec{Q}\) are represented by the two sides AB and BC of a triangle ABC,

i.e., \(\vec{A B}\) = \(\vec{P}\) and \(\vec{B C}\) = \(\vec{Q}\)

From the triangle law of vector addition \(\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C}\)

or, \(\vec{P}+\vec{Q}=\overrightarrow{A C} \text { or, } \vec{P}+\vec{Q}+\vec{R}=\overrightarrow{A C}+\vec{R}\)

The first condition of equilibrium requires that the resultant, \(\vec{P}\) + \(\vec{Q}\) + \(\vec{R}\) = \(\vec{0}\)

∴ \(\overrightarrow{A C}+\vec{R}=\overrightarrow{0} \text { or, } \vec{R}=-\overrightarrow{A C}=\overrightarrow{C A}\)

Hence, the magnitude and direction of the third force \(\vec{R}\), is represented by the third side \(\overrightarrow{C A}\), of triangle ABC.

3. **Law Of Orthogonal Components:** A body under the action of three non-parallel, coplanar forces remains in equilibrium if

- The lines of action of the forces pass through the same point and
- The algebraic sum of the components of the forces taken along two mutually perpendicular axes is zero separately.

Three coplanar forces \(\vec{P}\), \(\vec{Q}\), \(\vec{R}\) act at the point O. x and y-axes are chosen along two mutually perpendicular lines OX and OY. Suppose P, Q, and R, the three forces, make angles α, β and γ respectively with the x-axis.

Now, the components of the three forces along the x and y-axes are,

⇒ \(P_x=P \cos \alpha, Q_x=Q \cos \beta, R_x=R \cos \gamma\)

⇒ \(P_y=P \sin \alpha, Q_y=Q \sin \beta, R_y=R \sin \gamma\)

Hence, the sum of the components along the x-axis, \(S_x=P_x+Q_x+R_x=P \cos \alpha+Q \cos \beta+R \cos \gamma\)

and some of the components along y-axis \(S_y=P_y+Q_y+R_y=P \sin \alpha+\mathrm{Q} \sin \beta+\mathrm{R} \sin \gamma\)

According to the condition for equilibrium of an object under the action of three forces, the resultant S of S_{x} and S_{y} should be zero,

i.e„ S = \(\sqrt{s_x^2+s_y^2}=0\)

This is true only when S_{x} = 0 and S_{y} = 0.

∴ S_{x} = 0 and S_{y} = 0 is the required condition.

4. **Larni’s Theorem:** Anybody, under the action of three coplanar forces acting at a point, remains in equilibrium when each force is directly proportional to the sine of the angle between the other two forces.

OX, OY, OZ are the lines of action of three forces \(\vec{P}\), \(\vec{Q}\), \(\vec{R}\) acting at a point O of the body.

Let \(\overrightarrow{O A}=\vec{P}\) and \(\overrightarrow{O B}=\vec{Q}\). So, the diagonal \(\overrightarrow{O D}\) of the parallelogram OADB represents the resultant of forces \(\vec{P}\) and \(\vec{Q}\). In other words, \(\vec{P}\) + \(\vec{Q}\) = \(\overrightarrow{O D}\).

∴ \(\vec{P}\) + \(\vec{Q}\) = \(\overrightarrow{O D}\) + \(\vec{R}\)

When the body is in equilibrium, \(\vec{P}+\vec{Q}+\vec{R}=\overrightarrow{0} \text { or, } \overrightarrow{O D}+\vec{R}=\overrightarrow{0} \text { or, } \vec{R}=-\overrightarrow{O D}=\overrightarrow{D O}\)

Clearly, \(\overrightarrow{O A}, \overrightarrow{A D}, \overrightarrow{D O}\) form the arms of the triangle OAD, which, taken in order, represent the three forces \(\vec{P}, \vec{Q}, \vec{R}\)

From the properties of triangle, \(\frac{P}{\sin \angle O D A}=\frac{Q}{\sin \angle A O D}=\frac{R}{\sin \angle O A D}\)

Here, sin∠ODA = sin∠BOD =sin( 180° – ∠YOZ) = sin ∠YOZ = sin(Q, R), where (Q, R) is the angle between Q and R.

Similarly, sin ∠AOD = sin(R, P) and sin ∠OAD = sin (P, Q)

∴ \(\frac{P}{\sin (Q, R)}=\frac{Q}{\sin (R, P)}=\frac{R}{\sin (P, Q)}\)

This is the mathematical representation of Larni’s theorem for the equilibrium of a body under the action of three coplanar forces.

**Equilibrant:** The single force that balances the resultant of all other forces, is called the equilibrant. F_{3} is the equilibrant of F_{1} and F_{2}. It can also be said that F_{1} is the equilibrant of F_{2} and F_{3}, or F_{2} is the equilibrant of F_{1} and F_{3}.

## Statics – Equilibrium Numerical Examples

**Example 1. Three forces F, F and √2F acting at a point are in equilibrium. Find the angles between the forces.**

**Solution:**

Show the three forces. Let the angles between the forces be α, β and γ, as shown.

∴ α + β + γ = 360°

or, β + γ = 360°- α

∴ sin(α) = -sinα

By Lami’s theorem \(\frac{\sqrt{2} F}{\sin \alpha}=\frac{F}{\sin \beta}=\frac{F}{\sin \gamma}\)

∴ β = γ

Also, \(\frac{F}{\sin \beta}=\frac{\sqrt{2 F}}{\sin \alpha}\) or, \(\sin \alpha=\sqrt{2} \sin \beta\).

Now, \(\sin \alpha=-\sin (\beta+\gamma)=-\sin 2 \beta=-2 \sin \beta \cos \beta\)

Hence, \(-2 \sin \beta \cos \beta=\sqrt{2} \sin \beta\)

or, \(\cos \beta=-\frac{1}{\sqrt{2}}=\cos 135^{\circ}\)

∴ \(\beta=135^{\circ}=\gamma\)

and \(\alpha=360^{\circ}-\left(135^{\circ}+135^{\circ}\right)=90^{\circ}\).

**Example 2. One end of a light string is fixed to the ceiling and the other end is tied to the walk A body of mass 500 g is suspended from that string in such a way that the part of the string towards the wall remains horizontal and the portion towards the ceiling, makes an angle of 30° with the ceiling. Find the tension on the two parts of the string.**

**Solution:**

Let the positive y-axis be vertically upwards and the x-axis be horizontal. Let T_{1} = tension in the horizontal part of the string, T_{2} = tension in the other part of the string, W = weight of the suspended body = 500 x 981 dyn.

Sum of the components of the forces along the y-axis = T_{2 }sin30°- W

Sum of the components of the forces along the x-axis = T_{1 }– T_{2}cos30°

As the system is in equilibrium T_{2} sin30°- W= 0….(1)

T_{1 }– 2 cos30° = 0…(2)

From (1), \(\frac{1}{2} T_2=W\)

or, \(T_2=2 W\)

= \(2 \times 500 \times 981 \mathrm{dyn}=2 \times \frac{500 \times 981}{10^5} \mathrm{~N}=9.81 \mathrm{~N}\)

From (2), \(T_1=T_2 \cos 30^{\circ}=9.81 \times \cos 30^{\circ}=8.5 \mathrm{~N} \text {. }\)

**Example 3. A weight W is suspended by using two strings. One of the strings makes an angle of 30° with the vertical. What should be the direction of the other string; so that the tension in it becomes minimum? Find the tension in each string at this position.**

**Solution:**

Let the angle between the two strings be α.

Given that the angle between the weight W and the first string = 180° -30° = 150°

Hence, the angle between the weight and the second string

=360° – 150°- α

= 210°- α

For equilibrium, using Lami’s theorem, \(\frac{T_2}{\sin 150^{\circ}}=\frac{W}{\sin \alpha} \text { or, } \frac{T_2}{0.5}=\frac{W}{\sin \alpha} \text { or, } T_2=\frac{W}{2 \sin \alpha}\)

For T_{2} to be minimum, sina should be maximum i.e., sinα = 1 or α = 90°. Hence, the tension in the second string will be minimal when it is at right angles to the first string.

At this setting, \(T_2=\frac{W}{2}\)

Also, \(\frac{W}{\sin 90^{\circ}}=\frac{T_1}{\sin \left(210-90^{\circ}\right)}=\frac{T_1}{\sin 120^{\circ}}\)

or, W = \(\frac{T_1}{\sin 60^{\circ}}\)

∴ \(T_1=W \sin 60^{\circ}=W \times \frac{\sqrt{3}}{2}\)

∴ \(T_1=\frac{\sqrt{3} W}{2} \text { and } T_2=\frac{W}{2} .\)

**Example 4. From the markings, 20 cm, 40 cm, 60 cm, 80 cm and 100 cm on a light metre ruler, masses of 1 g, 2 g, 3 g, 4 g, and 5 g respectively are suspended. At which mark should the ruler be pivoted so that it remains horizontal?**

**Solution:**

The gravitational forces on 1g, 2g,3g,4g, and 5g masses are downward parallel forces. Suppose their resultant acts at the x cm mark on the scale. Hence,

x = \(\frac{\sum_{i=1}^5 F_i x_i}{\sum_{i=1}^5 F_i}\)

= \(\frac{1 \times 20+2 \times 40+3 \times 60+4 \times 80+5 \times 100}{1+2+3+4+5} \)

= \(\frac{1100}{15}=73.3\)

When the downward resultant and the upward normal force of the support, act at the same point, the metre ruler remains in equilibrium. Hence, the ruler should be pivoted at the 73.3 cm mark.

**Example 5. Two parallel forces P and Q (P> Q) are acting at two points, A and B, of a body. If the forces Interchange their positions, by how much will the point of action of the resultant of the two forces shift along the line AB?**

**Solution:**

Suppose C is the point of action of the resultant in the first case.

∴ P x AC = Q x BC…(1)

If the new position of the point of action of the resultant is then Q x AC’ = P x BC’ …(2)

From (1) we get, P x AC = Q(AB – AC) or,(P + Q) x AC = Q x AB

∴ Ac = \(\frac{Q}{P+Q} \times A B\)…(3)

Similarly from (2) we get, \(A C^{\prime}=\frac{P}{P+Q} \times A B\)…(4)

∴ Subtracting equation (3) from equation (4) we get, \(A C^{\prime}-A C=\frac{A B}{(P+Q)}(P-Q) \text { or, } C C^{\prime}=\frac{P-Q}{P+Q} \times A B \text {. }\)

**Example 6. The result of two parallel forces P and Q is P.If the force P shifts by a distance x, parallel to Itself, prove that R will shift by \(\frac{P x}{P+Q}\).**

**Solution:**

Suppose the parallel forces P and Q act at points A and B and the resultant R acts at C.

∴ P x AC = Q x BC

or, \(\frac{P}{Q}=\frac{B C}{A C} \text { or, } \frac{P+Q}{Q}=\frac{B C+A C}{A C}=\frac{A B}{A C}\)

∴ AC = \(A B \times \frac{Q}{P+Q}\)

Now if the force P acts at point D such that AD = x, then the resultant also shifts and acts at E instead of C.

Hence, Px DE = Qx BE

or, \(\frac{P}{Q}=\frac{B E}{D E} or, \frac{P+Q}{Q}=\frac{B E+D E}{D E}=\frac{B D}{D E}=\frac{A B-x}{A E-x}\)

∴ \((A E-x)(P+Q)=(A B-x) Q\)

AE = \(\frac{Q \cdot A B+P \cdot x}{P+Q}\)

∴ Displacement of the line of action of the resultant, CE = \(A E-A C=\frac{Q \cdot A B+P \cdot x}{P+Q}-\frac{Q \cdot A B}{P+Q}=\frac{P x}{P+Q}\)

**Example 7. The mass of a uniform rod of length 10 m is 10 kg. The rod is placed over knife edges A and B. One end of the rod is resting on the knife edge A and the other end is 2 m outside the knife edge B. A 30-kg weight is now suspended 2 m away from end A. Find the magnitude of the normal forces on the knife edges.**

**Solution:**

Let R_{1} and R_{2} be the normal forces at knife edges A and B respectively. The weight of the rod, 10 kg x 9.8 m · s^{-2} = 98 N, is acting at O, the mid-point of the rod.

From the conditions of equilibrium, R_{1} + R_{2} = (30 + 10) x 9.8 = 392 N

Taking moments about the point A, R_{2 }x AB = 30 x 9.8 x AD+ 10 x 9.8 x AO

or, R_{2} x 8 = 294 x 2 + 98 x 5

∴ R_{2} = 134.75 N

∴ R_{1} = 392-134.75 = 257.25 N

**Example 8. A ladder weighing W rests with one end against a rough vertical wall and the other end on a rough floor. It is inclined at an angle of 45° with the horizontal. The coefficient of friction of the ladder with the floor and the wall are μ and μ’ respectively. Show that the minimum horizontal force that can move the base of the ladder towards the wall is \(\frac{W\left(1+2 \mu-\mu \mu^{\prime}\right)}{2\left(1-\mu^{\prime}\right)}\)**

**Solution:**

A horizontal force F is applied at point B of the ladder AB, such that the ladder is at the limiting stage of equilibrium. The figure shows the forces acting at this stage.

At equilibrium, the horizontal components give, F= R’ + μR…(1)

and the vertical components give, W+μ’R’ = R…(2)

∴ ∠ABO = 45° , we have OA = OB = 2L (say). At equilibrium, the moments of the forces about B give \(W L+\mu^{\prime} R^{\prime} \cdot 2 L=R^{\prime} \cdot 2 L\)

∴ \(W=2 R^{\prime}\left(1-\mu^{\prime}\right)\)

or, \(R^{\prime}=\frac{W}{2\left(1-\mu^{\prime}\right)}\)….(3)

From (2) we get, R = \(W\left[1+\frac{\mu^{\prime}}{2\left(1-\mu^{\prime}\right)}\right]\)

and from (1) we get, F = \(W\left[\frac{1}{2\left(1-\mu^{\prime}\right)}+\mu+\frac{\mu^{\prime} \mu}{2\left(1-\mu^{\prime}\right)}\right]\)

= \(W \frac{\left(1+2 \mu-\mu \mu^{\prime}\right)}{2\left(1-\mu^{\prime}\right)}\)

**Example 9. A uniform ladder of length L and mass M leans against a frictionless vertical wall. The ladder makes an angle θ with the ground and the coefficient of the static friction between the ground and the foot of the ladder is μ. What maximum height a man of mass m can climb up the ladder so that the ladder will not skid?**

**Solution:**

The value of limiting friction between the ground and the ladder is μn.

Let the man climb a maximum height x up the ladder. The point P in the figure denotes the position of the man.

Here frictional force between the ladder and ground, f = μn

where n = contact or normal force on the ladder by ground = (Mg + mg) and the contact force on the ladder by the wall is n’.

At equilibrium, net force on the ladder is zero, i.e., n’ – f = 0 or, n’ = μ(M+ m)g

Also, the moment about the point B is zero.

i.e., \(n^{\prime} L \sin \theta-M g \frac{L}{2} \cos \theta-m g \cdot x \cos \theta=0\)

or, \( m x \cos \theta=\mu(M+m) L \sin \theta-\frac{M L}{2} \cos \theta\)

∴ x = \(\frac{\mu(M+m) L \sin \theta-\frac{M L}{2} \cos \theta}{m \cos \theta}\)