Informative line

### Equilibrium

Learn concept of equilibrium in physics and components of forces. Practice to find tension in the string shown in figure, if the body is in equilibrium.

# Free Body Diagram to Calculate acm and Torque

• If two or more forces act on a body at different points, then two free body diagrams are used to determine the acceleration of center of mass and the torque.

### Free body diagram for center of mass

• Consider a system on which three forces are acting.
• To study acm, the position of forces does not make any difference. All the forces can be drawn at a single point on the body.

### Free body diagram for torque

• Torque on a body is dependent on the position of force or on the point of application.
• To study the rotation or torque, all forces on the body should be drawn at their original point of application.

#### A rod of mass m and length $$\ell$$ is hanging with the help of strings as shown in figure. The correct free body diagram, to calculate acm is-

A

B

C

D None of these

×

For acm

The position of force is not important, its easy to calculate acm if all forces are drawn at the same point.

### A rod of mass m and length $$\ell$$ is hanging with the help of strings as shown in figure. The correct free body diagram, to calculate acm is-

A
B
C
D

None of these

Option B is Correct

# Equilibrium

• A body is said to be in equilibrium,

when, $$\vec F_{net} = 0$$     {Translational equilibrium}

And    $$\vec \tau_{net} = 0$$     {Rotational equilibrium}

Important point

If    $$\vec F_{net} = 0$$, then $$\overrightarrow\tau_{net}$$ calculated about any point, comes out to be same.

#### Which one of the following bodies is not in equilibrium?

A

B

C

D

×

Net force on body A

$$\vec F_{net }= 10\, N - 10 \, N = 0$$

Net force on body B

$$\vec F_{net }= 20\, N - 10 \, N - 10 \, N$$

$$\vec F_{net }= 0$$

Net force on body C

$$\vec F_{net }= 10\, N + 30 \, N - 40 \, N$$

$$\vec F_{net }= 0$$

Net force on body D

as the body is at rest,

$$\vec F_{net }= 0$$

All bodies are in translational equilibrium.

Rotational equilibrium of body A

Torque  about point A on body A due to F1 and F2

$$\tau_A = \dfrac{\ell}{2} × 10 +\dfrac{\ell}{2} × 10$$  [Taking anticlockwise as positive, $$ACW\rightarrow+Ve$$]

$$\tau_A = 10 \,\ell$$

Since $$\tau_A \neq 0$$

$$\tau_{net} \neq 0$$, Body is not in rotational equilibrium.

Rotational equilibrium of body B

Torque about point C due to F1, F2 and F3

$$(\tau_c) = \ell × 10 - 10\, \ell$$

= 0

Since $$\tau_c = 0, \vec \tau_{net} = 0$$

Body is in equilibrium.

Rotational equilibrium of body C

$$\tau_A = 0× 10 + \ell × 30 (Clockwise) + \dfrac{3\ell}{4} × 40 (Anticlockwise)$$

$$= - 30 \,\ell + 30 \, \ell$$

$$= 0$$

Since, $$\vec \tau_A = 0$$$$\vec \tau_{net} = 0$$

$$\therefore$$ Body is in rotational equilibrium.

Rotational equilibrium of body D

Torque about point C = 0

because neither N nor mg causing rotation.

Since $$\tau_{net} = 0 ;$$ body is in rotational equilibrium.

### Which one of the following bodies is not in equilibrium?

A
B
C
D

Option A is Correct

# Calculating Tension in the String

## Equilibrium

• A body is said to be in equilibrium

when  $$\vec F_{net} =0$$    {Translational equilibrium}

and  $$\vec \tau_{net} =0$$      {Rotational equilibrium}

Important Point

If, $$\vec F_{net} = 0$$, Then $$\vec \tau_{net}$$ calculated about any point comes out to be same.

#### A rod AB of mass M and length $$\ell$$ hangs with the help of two strings as shown in figure. If a mass $$m =\dfrac{M}{2}$$, is also hanging from end A, as shown, the system still is in equilibrium. find T1 and T2.

A $$T_1 = \dfrac{3\,Mg}{2} ;\,T_2 = 0$$

B $$T_1 = 0 ; \,T_2 = 2\, mg$$

C $$T_1 = \dfrac{3\,Mg}{2} ;\, T_2 =\dfrac{3\,Mg}{2}$$

D $$T_1 = 2\,Mg ;\, \,T_2 =\dfrac{3\,Mg}{2}$$

×

Free body diagram of rod of mass M and length $$\ell$$

As Rod is in equilibrium

$$\therefore$$ For the Rod in translational equilibrium

$$\vec F_{net} = 0$$

Free body diagram can be altered such that all the forces are drawn at the same point.

For translational equilibrium

$$\vec F_{net} = 0$$

$$T= mg$$

$$T= \dfrac{M}{2}g$$       $$\left(as \, \, m = \dfrac{M}{2}\right)$$ .......(i)

As Rod is at rest

$$T_1 + T_2 = Mg +T$$

Substitute the value of T from (i)

$$T_1 + T_2 = \dfrac{3\,Mg}{2}$$     .....(ii)

As Rod is in rotational equilibrium

$$\tau_{net}=0$$ ; $$\overrightarrow\tau$$ about any point = 0

$$\vec \tau$$ about point P

$$\tau_P = \dfrac{-T\,\ell}{3} + Mg \left(\dfrac{\ell}{2} -\dfrac{\ell}{3} \right) - T_2\left(\dfrac{\ell}{2} -\dfrac{\ell}{3}+ \dfrac{\ell}{4}\right)$$

As $$\tau_P = 0$$

$$0 = -\dfrac{Mg}{2}.\dfrac{\ell}{3} + Mg \dfrac{\ell}{6} - T_2\dfrac{5\,\ell}{12}$$

$$\dfrac{Mg}{6} -\dfrac{Mg}{6}+\dfrac{5\, T_2}{12} = 0$$

$$T_2 = 0$$  .........(iii)

Substituting the value of  $$T_2$$ from (iii) in (ii)

$$T_1 + 0 = \dfrac{3\,Mg}{2}$$

$$T_1 = \dfrac{3\,Mg}{2}$$

$$T_2 = 0$$

### A rod AB of mass M and length $$\ell$$ hangs with the help of two strings as shown in figure. If a mass $$m =\dfrac{M}{2}$$, is also hanging from end A, as shown, the system still is in equilibrium. find T1 and T2.

A

$$T_1 = \dfrac{3\,Mg}{2} ;\,T_2 = 0$$

.

B

$$T_1 = 0 ; \,T_2 = 2\, mg$$

C

$$T_1 = \dfrac{3\,Mg}{2} ;\, T_2 =\dfrac{3\,Mg}{2}$$

D

$$T_1 = 2\,Mg ;\, \,T_2 =\dfrac{3\,Mg}{2}$$

Option A is Correct

#### A rod AB of mass M and length $$\ell ,$$ is kept horizontal on a support, which is at a distance $$\dfrac{\ell} {3}$$ from end A, as shown in figure. What should be the value of mass of a particle sticked at end A so that the system remains in equilibrium?

A $$m= \dfrac{M}{2}$$

B $$m= \dfrac{M}{3}$$

C $$m= \dfrac{M}{4}$$

D $$m= \dfrac{M}{5}$$

×

Free Body Diagram

If rod is in equilibrium, then Free body diagram can be altered such that all forces can be drawn at the same point.

Free Body Diagram at Center of mass

$$\therefore$$ If  $$F_{net} =0$$

$$Mg+mg =N$$

For rotational equilibrium

$$\vec\tau_{net} =0,$$ taking point P,

where  $$\vec \tau_P$$  is to be calculated.

$$\tau_P = mg\dfrac{\ell}{3} -Mg \dfrac{\ell}{6}$$

$$0 = mg\dfrac{\ell}{3} - Mg\dfrac{\ell}{6}$$

$$\Rightarrow mg\dfrac{\ell}{3}=Mg\dfrac{\ell}{6}$$

$$\Rightarrow m = \dfrac{M}{2}$$

### A rod AB of mass M and length $$\ell ,$$ is kept horizontal on a support, which is at a distance $$\dfrac{\ell} {3}$$ from end A, as shown in figure. What should be the value of mass of a particle sticked at end A so that the system remains in equilibrium?

A

$$m= \dfrac{M}{2}$$

.

B

$$m= \dfrac{M}{3}$$

C

$$m= \dfrac{M}{4}$$

D

$$m= \dfrac{M}{5}$$

Option A is Correct

# Calculation of Tension in Strings when a Uniform Rod is in Equilibrium

Equilibrium

A body is said to be in equilibrium,

when $$\vec F_{net} = 0$$ {Translational equilibrium}

and $$\vec\tau_{net} = 0$$ {Rotational equilibrium}

Important point

• If $$\vec F _{net} = 0,$$ then $$\vec \tau_{net}$$ calculated about any point comes out to be same.

#### A rod of mass m and length $$\ell$$ is in equilibrium as shown in figure. Find the value of Tension forces T1 and T2.

A $$\dfrac{3}{4} \,mg ,\, \dfrac{mg}{4}$$

B $$\dfrac{mg}{4} ,\, \dfrac{mg}{2}$$

C $$\dfrac{mg}{2} ,\, \dfrac{mg}{2}$$

D $$3 \, mg , \, 2 \, mg$$

×

For the rod to be in translational equilibrium

$$\vec F_{net} = 0$$

Free Body Diagram

For $$\vec F_{net}$$, free body diagram can be altered such that all the forces are drawn at the same point.

$$T_1 + T_2 - mg = 0$$   .......(i)

For the rod to be in rotational equilibrium;

$$\overrightarrow\tau_{net} = 0$$

$$\vec \tau_B = (\vec \tau_B )_{T_1}+ (\vec \tau_B )_{mg} + (\vec \tau_B )_{T_2}$$

$$= (r_\bot)_{T_1} . T_1 + (r_\bot)_{mg} . mg +(r_\bot)_{T_2} . T_2$$

$$=\dfrac{2\,\ell}{3} (T_1)(clockwise) +\dfrac{\,\ell}{2} (mg) (anticlockwise)+ 0. T_2$$

Taking anticlockwise as positive

$$\tau_B = -\dfrac{2\,\ell}{3} \,T_1 + \dfrac{mg\,\ell}{2}$$

If body is in equilibrium

$$\vec \tau_B = 0$$

$$0 = \dfrac{mg\,\ell}{2} -\dfrac{2\,\ell\, T_1 }{3} \,$$   ......(ii)

from (ii)

$$T_1 = \dfrac{3}{4} \, mg$$

Substituting the value of T1 in (i)

$$T_2 = \dfrac{mg}{4}$$

### A rod of mass m and length $$\ell$$ is in equilibrium as shown in figure. Find the value of Tension forces T1 and T2.

A

$$\dfrac{3}{4} \,mg ,\, \dfrac{mg}{4}$$

.

B

$$\dfrac{mg}{4} ,\, \dfrac{mg}{2}$$

C

$$\dfrac{mg}{2} ,\, \dfrac{mg}{2}$$

D

$$3 \, mg , \, 2 \, mg$$

Option A is Correct

# Force on a Hinged Rod

## Equilibrium

• A body is said to be in equilibrium.

when, $$\vec F_{net} =0$$  {Translational equilibrium}

and $$\vec \tau_{net} =0$$  {Rotational equilibrium}

Important Point

• If $$\vec F_{net} = 0$$, then $$\vec \tau_{net}$$ calculated about any point comes out to be same.

#### A rod of mass m and length $$\ell$$, hinged at one end, is kept at equilibrium using a string of length $$\ell$$, tied at the other end as shown in figure. Find the tension in the string and normal force due to the hinge.

A $$T= \dfrac{mg}{2\sqrt3},\; N_x =\dfrac{mg}{4\sqrt3},\;N_y = \dfrac{3\,mg}{4}$$

B $$T= \dfrac{mg}{2},\, N_x =mg, \,N_y = mg$$

C $$T= mg,\, N_x =\dfrac{mg}{4}, \,N_y = 3mg$$

D $$T= 3\,mg,\, N_x = mg, \,N_y = mg$$

×

The direction of normal force due to hinge can not be determined directly because of the complicated structure at hinge.

Therefore, we assume two components of normal force as $$N_x, \,N_y$$.

Free body diagram

Here $$\theta = \dfrac{\pi}{3}$$ as structure forms an equilateral triangle.

As the rod is at rest,

$$\therefore$$ Rod is in translational as well as in rotational equilibrium.

For the rod in translational equilibrium

$$\vec F_{net} = 0;$$

Free body diagram can be altered such that all the forces are drawn at the same point.

Resolving forces in horizontal and vertical directions

for $$\vec F_{net} = 0$$

$$\vec F_{x} = 0$$$$\vec F_{y} = 0$$

$$F_x =$$ forces in $$(x)$$ horizontal direction

$$F_y=$$ forces in $$(y)$$ vertical direction

$$F_x = T \, sin \dfrac{\pi}{6} - N_x$$

as $$F_x = 0;$$

$$T \, sin \dfrac{\pi}{6} - N_x = 0$$     ........(i)

$$F_y = T \, cos \left(\dfrac{\pi}{6}\right) + N_y- mg$$

as $$F_y = 0$$

$$T \, cos \left(\dfrac{\pi}{6}\right) + N_y- mg =0$$    ........(ii)

For the rod in rotational equilibrium.

$$\vec \tau_{net } = 0$$

In equilibrium, $$\vec \tau$$ about any point is zero.

It is convenient to take torque about a point from which maximum number of unknown forces pass.

$$\therefore$$  Here, hinge is that point.

$$\tau_H = - mg \times \dfrac{\ell}{2}\, cos \left(\dfrac{\pi}{3}\right) + T. \ell \, cos \dfrac{\pi}{6}$$

$$\tau_H = -\dfrac{mg \ell}{4} +\dfrac{\sqrt3 \ell\, T}{2}$$

As $$\tau_H = 0$$

$$-\dfrac{mg}{4} + \dfrac{\sqrt3 \,T}{2} = 0$$       ........(iii)

Solving equation (i),(ii) and (iii),

$$T = \dfrac{mg}{2\sqrt3}$$

$$N_x = \dfrac{mg}{4\sqrt3}$$

$$N_y = \dfrac{3\,mg}{4}$$

### A rod of mass m and length $$\ell$$, hinged at one end, is kept at equilibrium using a string of length $$\ell$$, tied at the other end as shown in figure. Find the tension in the string and normal force due to the hinge.

A

$$T= \dfrac{mg}{2\sqrt3},\; N_x =\dfrac{mg}{4\sqrt3},\;N_y = \dfrac{3\,mg}{4}$$

.

B

$$T= \dfrac{mg}{2},\, N_x =mg, \,N_y = mg$$

C

$$T= mg,\, N_x =\dfrac{mg}{4}, \,N_y = 3mg$$

D

$$T= 3\,mg,\, N_x = mg, \,N_y = mg$$

Option A is Correct

# Calculation of Friction Force on the Base of Inclined Ladder (Assuming Wall is Frictionless)

## Equilibrium

A body is said to be in equilibrium,

when $$\vec F_{net} = 0$$ {Translational equilibrium}

and $$\vec\tau_{net} = 0$$ {Rotational equilibrium}

Important point

• If $$\vec F _{net} = 0,$$ then $$\vec \tau_{net}$$ calculated about any point comes out to be same.
• If the inclined ladder is in equilibrium, then friction is static, and the ladder adjusts itself according to the tendency of relative motion between the surfaces in contact.

For example

• To calculate friction force provided by ground on a ladder of mass m and length $$\ell$$, inclined against a smooth wall at an angle $$\theta$$ with the horizontal, if the ladder is in equilibrium as shown in figure.

• Free body diagram of ladder of mass m and length $$\ell$$ inclined against a smooth wall.

• As rod is in translational as well as rotational equilibrium.
• For $$\vec F_{net} =0;$$ Free body diagram can be altered such that all the forces are drawn at the same point.
• Free body diagram of all forces acting on the ladder at a point is shown in figure.
• For $$\vec F_{net} =0;\,\,\, \vec F_x=0; \,\,\,\vec F_y=0$$
• $$\vec F_x=$$ forces in $$x$$ direction

$$\vec F_y=$$ forces in $$y$$ direction

$$f_x = f- N_2$$

$$f_x =0; \,\,f-N_2=0$$

$$f= N_2$$ .........(i)

$$f_y= N_1 -mg$$

$$f_y=0; \,\, N_1-mg =0$$

$$N_1 = mg$$ ........(ii)

• For body to be in rotational equilibrium

$$\vec \tau_{net} =0;$$

• If $$\vec \tau _{B}$$ is calculated, then more number of unknown variables can be eliminated.

$$(r_{\bot})_{N_2} = \ell\,sin\,\theta$$

$$(r_{\bot})_{mg} = \dfrac{\ell}{2}\,cos\,\theta$$

$$\Rightarrow\tau_B = -(r_\bot)_{N_2}× N_2 +(r_\bot)_{mg} × mg$$

$$= - N_2 \,\ell \,sin\,\theta + mg \dfrac{\ell}{2}\,cos\theta$$        (taking anticlockwise positive)

As $$\tau_B =0$$

$$0= \dfrac{mg}{2} cos\,\theta - N_2\,sin \,\theta$$

$$N_2 = \dfrac{mg}{2} \,cot \,\theta$$      ........(iii)

From (i) and (iii)

$$\Rightarrow f= \dfrac{mg}{2} \,cot \,\theta$$

#### A ladder of mass m = 5 kg and length $$\ell = 2\,m$$, is kept inclined at an angle $$\theta=\dfrac{\pi}{4}$$ against a smooth wall, as shown in figure. Find the friction force on the ladder by the ground. (Given: $$g= 10 \,m/s^2$$)

A 20 N

B 28 N

C 25 N

D 30 N

×

where N1 and N= Normal contact forces

f = friction force

As rod is in translational as well as rotational equilibrium.

For $$\vec F _{net} =0 ;$$ free body diagram can be altered such that all the forces are drawn at the same point.

Free body diagram of all forces acting at a point on ladder

For $$\vec F_{net} =0$$                                  { $$F_x =$$ forces in $$x$$ direction}

$$\vec F_x =0 , \,\,\vec F_y =0$$                            { $$F_y =$$ forces in $$y$$ direction }

$$F_x = f-N_2$$

$$F_x= 0; \,\,f-N_2=0$$

$$f=N_2$$      ........(i)

$$F_y = N_1 -mg$$

$$F_y=0; \,\,N_1-mg=0$$

$$N_1 = mg$$      ........(ii)

As body is in rotational equilibrium,

$$\vec \tau_{net} =0$$

if $$\vec \tau_B$$ is calculated, more number of unknown forces can be eliminated.

$$\Rightarrow(r_\bot) _{N_2} = \ell \,sin\,\theta$$

$$\Rightarrow(r_\bot) _{mg} = \dfrac{\ell}{2} \,cos\,\theta$$

$$\Rightarrow\tau_B=-(r_\bot)_{N_2} × N_2 + (r_\bot)_{mg} × mg$$

$$= -\ell \,sin\,\theta . N_2 + mg \dfrac{\ell}{2} \,cos\,\theta$$   (Assuming Anticlockwise as positive)

As $$\tau_B =0$$

$$0 = \dfrac{mg\,cos\,\theta}{2}-N_2\,sin\,\theta$$

$$\Rightarrow N_2 = \dfrac{mg}{2} cot \,\theta$$

$$N_2 = \dfrac{mg}{2} cot \dfrac{\pi}{4}$$

$$\Rightarrow N_2 = \dfrac{mg}{2}$$ ......(iii)

From (i) and (iii)

$$N_2 =f$$

$$f= \dfrac{mg}{2}$$

$$f= \dfrac{5×10}{2}$$

$$f= 25\,N$$

### A ladder of mass m = 5 kg and length $$\ell = 2\,m$$, is kept inclined at an angle $$\theta=\dfrac{\pi}{4}$$ against a smooth wall, as shown in figure. Find the friction force on the ladder by the ground. (Given: $$g= 10 \,m/s^2$$)

A

20 N

.

B

28 N

C

25 N

D

30 N

Option C is Correct

# Condition for Ladder to Stay in Equilibrium

• A ladder of mass m and length $$\ell$$ is inclined against a smooth wall, as shown in figure.
• The coefficient of friction at the floor is $$\mu.$$
• We will determine the condition for the ladder to stay in equilibrium.

## Equilibrium

A body is said to be in equilibrium,

when $$\vec F_{net} =0$$ {translational equilibrium}

and $$\vec \tau_{net} =0$$ (Rotational equilibrium)

Important point

If  $$\vec F_{net} =0,\,\,\vec\tau_{net}$$ calculated about any point comes out to be the same.

• If the inclined ladder is in equilibrium, then friction is static, and the ladder adjusts itself according to the tendency of relative motion between the surfaces in contact.

### Free body diagram

• If ladder is in equilibrium, then it is in translational as well as rotational equilibrium.
• For $$\vec F_{net} =0 ;$$ free body diagram can be altered such that all the forces are drawn at the same point.

• For $$\vec F_{net} =0$$

$$\vec {F_x} =0 ; \,\,\vec{F_y}=0$$           { $$F_x =$$ force in $$x$$ direction }

{ $$F_y =$$ force in $$y$$ direction }

• $$F_x = f- N_2$$

$$\Rightarrow 0 = f-N_2$$

$$\Rightarrow f=N_2$$ ........(i)

• $$F_y = N_1 -mg$$

$$\Rightarrow 0= N_1 -mg$$

$$\Rightarrow N_1 = mg$$ ......(ii)

• For body to be in rotational equilibrium

$$\vec \tau_{net} =0$$

and point B is chosen such that $$\vec \tau_B$$ is calculated and more number of unknown forces can be eliminated.

$$\Rightarrow(r_\bot) _{N_2} = \ell \,sin \,\theta$$

$$\Rightarrow(r_\bot) _{mg} = \dfrac{\ell}{2} \,cos \,\theta$$

$$\tau_B = -(r_\bot) _{N_2} × N_2 + (r_\bot) _{mg} × mg$$

$$\Rightarrow 0 = -N_2 \,\ell \,sin \,\theta + mg \dfrac{\ell}{2} \,cos\,\theta$$       (Assuming Anticlockwise as positive)

$$\Rightarrow N_2 = \dfrac{mg}{2} \,cot\,\theta$$ .......(iii)

from (i) and (iii)

$$N_2 = f$$

$$\Rightarrow f= \dfrac{mg}{2} cot\,\theta$$

• It is the value of friction for the ladder to be in equilibrium.
• Maximum value of friction can be $$\mu\,N_1$$

$$\therefore \dfrac{mg}{2} cot \,\theta \leq \mu\,N_1$$

$$\Rightarrow\dfrac{mg}{2} cot \,\theta \leq \mu\,mg$$         $$(\because N_1 = mg)$$

$$\Rightarrow\dfrac{cot\,\theta}{2} \leq \mu$$

#### A ladder of mass m = 5 kg and length $$\ell =2\,m$$, is kept inclined against a smooth wall, making an angle $$\theta = \dfrac{\pi}{4}$$ with the horizontal as shown in figure. For which of the following value of $$\mu$$(coefficient of friction), ladder will not slip?

A $$0.4$$

B $$0.2$$

C $$0.25$$

D $$0.5$$

×

Free body diagram

$$N_1\; and \;N_2=$$ Normal contact forces

$$f=$$ friction

If ladder is in translational as well as rotational equilibrium.

Then $$\vec F_{net} =0;$$ free body diagram can be altered such that all forces can be drawn at the same point.

Free body diagram of all forces acting at a point on ladder

For $$\vec F_{net} =0$$       { $$\vec F_x =$$ forces in $$x$$ direction }

$$\vec F_{x } =0 ; \,\vec F_y =0$$    { $$\vec F_y =$$ forces in $$y$$ direction }

$$F_x = f-N_2$$

$$F_x=0;\,\,f-N_2 =0$$

$$\Rightarrow f=N_2$$ .......(i)

$$F_y = N_1 -mg$$

$$F_y =0; \,\, N_1 -mg =0$$

$$\Rightarrow N_1 =mg$$    .....(ii)

If body is in rotational equilibrium then  $$\vec \tau_{net } =0$$.

If $$\vec \tau_B$$ is calculated, then more number of unknown forces can be eliminated.

$$\Rightarrow(r_\bot ) _{N_2} = \ell \,sin \,\theta$$

$$\Rightarrow(r_\bot ) _{mg} =\dfrac{\ell}{2} \,cos \,\theta$$

$$\Rightarrow \tau_B = (r_\bot)_{N_2 } × N_2 + (r_\bot)_{mg } × mg$$

$$= -\ell sin\,\theta.N_2 +\dfrac{mg}{2} \,\ell \,cos\,\theta$$           (Assuming Anticlockwise as positive)

As $$\tau_B =0$$

$$0 = \dfrac{mg\,cos\,\theta}{2} - N_2 \,sin\,\theta$$

$$N_2 = \dfrac{mg \,cot \,\theta}{2}$$

$$\Rightarrow N_2 = \dfrac{mg}{2} \,cot \dfrac{\pi}{4}$$

$$\Rightarrow N_2 = \dfrac{mg}{2}$$

$$\Rightarrow N_2 = \dfrac{5× 10}{2}$$

$$\Rightarrow N_2 = 25 \,N$$

From (i)

$$N_2 = f$$

$$f= 25 \,N$$

It is the value of friction required for equilibrium.

Condition for ladder to be in equilibrium

$$f\leq\mu\,N_1$$

$$25\leq \mu\,mg$$

$$25\leq \mu× 5× 10$$

$$0.5\leq \mu$$

$$\therefore$$ Option D is correct.

### A ladder of mass m = 5 kg and length $$\ell =2\,m$$, is kept inclined against a smooth wall, making an angle $$\theta = \dfrac{\pi}{4}$$ with the horizontal as shown in figure. For which of the following value of $$\mu$$(coefficient of friction), ladder will not slip?

A

$$0.4$$

.

B

$$0.2$$

C

$$0.25$$

D

$$0.5$$

Option D is Correct