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 acm

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.

Illustration Questions

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.

image image

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-

image
A image
B image
C image
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.

Illustration Questions

Which one of the following bodies is not in equilibrium?

A

B

C

D

×

Net force on body A 

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

image

Net force on body B 

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

\(\vec F_{net }= 0\)

image

Net force on body C 

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

\(\vec F_{net }= 0\)

image

Net force on body D 

as the body is at rest, 

\(\vec F_{net }= 0\)

image

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, \(ACW\rightarrow+Ve\)]

\(\tau_A = 10 \,\ell\)

image

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 

image

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. 

image

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.

image

Which one of the following bodies is not in equilibrium?

A image
B image
C image
D image

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.

Illustration Questions

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 \(\ell\)

image image

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.

image

For translational equilibrium

\(\vec F_{net} = 0\)

\(T= mg\)

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

image image

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)

image

As Rod is in rotational equilibrium

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

image

\(\vec \tau\) about point P

image image

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

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)

image

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

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

\(T_1 = \dfrac{3\,Mg}{2}\)

\(T_2 = 0\)

image

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.

image
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

Illustration Questions

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 

image image

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

image

Free Body Diagram at Center of mass

\(\therefore\) If  \(F_{net} =0\)  

\(Mg+mg =N\)

 

image image

For rotational equilibrium

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

where  \(\vec \tau_P\)  is to be calculated.

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

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

\(\Rightarrow mg\dfrac{\ell}{3}=Mg\dfrac{\ell}{6}\)

 \(\Rightarrow m = \dfrac{M}{2}\)

image

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? 

image
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.

Illustration Questions

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

image image

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)

image image

For the rod to be in rotational equilibrium; 

\(\overrightarrow\tau_{net} = 0\)

image

Torque about point B

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

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

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

Taking anticlockwise as positive

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

image image

If body is in equilibrium 

\(\vec \tau_B = 0\)

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

image

from (ii)

\(T_1 = \dfrac{3}{4} \, mg\)

Substituting the value of T1 in (i) 

\(T_2 = \dfrac{mg}{4}\)

image

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

image
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.

Illustration Questions

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.

image image

As the rod is at rest, 

\(\therefore\) Rod is in translational as well as in rotational equilibrium.

image

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.

image image

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)

image image

For the rod in rotational equilibrium.

\(\vec \tau_{net } = 0\)

image image

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.

image

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

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

As \(\tau_H = 0\)

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

image image

Solving equation (i),(ii) and (iii), 

\(T = \dfrac{mg}{2\sqrt3}\)

\(N_x = \dfrac{mg}{4\sqrt3}\)

\(N_y = \dfrac{3\,mg}{4}\)

image

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.

image
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)

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

 

 

Illustration Questions

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

×

Free body diagram of ladder

where N1 and N= Normal contact forces

f = friction force

image image

As rod is in translational as well as rotational equilibrium. 

image

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

image

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

image image

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 }

 

image

\(F_x = f-N_2\)

\(F_x= 0; \,\,f-N_2=0 \)

\(f=N_2\)      ........(i)

\(F_y = N_1 -mg\)

\(F_y=0; \,\,N_1-mg=0\)

\(N_1 = mg\)      ........(ii)

image

As body is in rotational equilibrium,

\(\vec \tau_{net} =0\)

image

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

image

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

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

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

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

image image

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)

image

From (i) and (iii)                                                                               

\(N_2 =f\)

\(f= \dfrac{mg}{2}\)

\(f= \dfrac{5×10}{2}\)

\(f= 25\,N\)

image

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: \(g= 10 \,m/s^2\))

image
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 \(\mu\,N_1\)

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

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

        \(\Rightarrow\dfrac{cot\,\theta}{2} \leq \mu\) 

Illustration Questions

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 

image image

If ladder is in translational as well as rotational equilibrium.

image

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

image

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

image image

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 }

image

\(F_x = f-N_2\)

\(F_x=0;\,\,f-N_2 =0\)

\(\Rightarrow f=N_2\) .......(i)

\(F_y = N_1 -mg\)

\(F_y =0; \,\, N_1 -mg =0\)

\(\Rightarrow N_1 =mg \)    .....(ii)

image

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

image

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

image

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

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

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

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

image image

As \(\tau_B =0 \)

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

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

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

\(\Rightarrow N_2 = \dfrac{mg}{2}\)

\(\Rightarrow N_2 = \dfrac{5× 10}{2}\)

\(\Rightarrow N_2 = 25 \,N\)

image

From (i) 

\(N_2 = f\)

\(f= 25 \,N\)

image

It is the value of friction required for equilibrium.

 

image

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.

image

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? 

image
A

\(0.4\)

.

B

\(0.2\)

C

\(0.25\)

D

\(0.5\)

Option D is Correct

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