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.
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.
Option B is Correct
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.
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.
Option A is Correct
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 \(T_1 = \dfrac{3\,Mg}{2} ;\,T_2 = 0\)
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)
\(T_1 + 0 = \dfrac{3\,Mg}{2}\)
\(T_1 = \dfrac{3\,Mg}{2}\)
\(T_2 = 0\)
Option A is Correct
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.
\(\tau_P = mg\dfrac{\ell}{3} -Mg \dfrac{\ell}{6} \)
\(0 = mg\dfrac{\ell}{3} - Mg\dfrac{\ell}{6}\)
\(\Rightarrow mg\dfrac{\ell}{3}=Mg\dfrac{\ell}{6}\)
\(\Rightarrow m = \dfrac{M}{2}\)
Option A 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
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;
\(\overrightarrow\tau_{net} = 0\)
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
\(\tau_B = -\dfrac{2\,\ell}{3} \,T_1 + \dfrac{mg\,\ell}{2} \)
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)
\(T_2 = \dfrac{mg}{4}\)
Option A is Correct
when, \(\vec F_{net} =0\) {Translational equilibrium}
and \(\vec \tau_{net} =0\) {Rotational equilibrium}
Important Point
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),
\(T = \dfrac{mg}{2\sqrt3}\)
\(N_x = \dfrac{mg}{4\sqrt3}\)
\(N_y = \dfrac{3\,mg}{4}\)
Option A is Correct
A body is said to be in equilibrium,
when \(\vec F_{net} = 0\) {Translational equilibrium}
and \(\vec\tau_{net} = 0\) {Rotational equilibrium}
Important point
For example
\(\vec F_x=\) 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)
\(\vec \tau_{net} =0;\)
\((r_{\bot})_{N_2} = \ell\,sin\,\theta \)
\((r_{\bot})_{mg} = \dfrac{\ell}{2}\,cos\,\theta\)
\(\Rightarrow\tau_B = -(r_\bot)_{N_2}× N_2 +(r_\bot)_{mg} × mg\)
\(= - N_2 \,\ell \,sin\,\theta + mg \dfrac{\ell}{2}\,cos\theta\) (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 20 N
B 28 N
C 25 N
D 30 N
Free body diagram of ladder
where N1 and N2 = 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)
\(N_2 =f\)
\(f= \dfrac{mg}{2}\)
\(f= \dfrac{5×10}{2}\)
\(f= 25\,N\)
Option C is Correct
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.
\(\vec {F_x} =0 ; \,\,\vec{F_y}=0\) { \(F_x = \) force in \(x \) direction }
{ \(F_y = \) force in \(y\) direction }
\(\Rightarrow 0 = f-N_2\)
\(\Rightarrow f=N_2\) ........(i)
\(\Rightarrow 0= N_1 -mg\)
\(\Rightarrow N_1 = mg \) ......(ii)
\(\vec \tau_{net} =0\)
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)
\(N_2 = f\)
\(\Rightarrow f= \dfrac{mg}{2} cot\,\theta\)
\(\therefore \dfrac{mg}{2} cot \,\theta \leq \mu\,N_1\)
\(\Rightarrow\dfrac{mg}{2} cot \,\theta \leq \mu\,mg\) \((\because N_1 = mg)\)
\(\Rightarrow\dfrac{cot\,\theta}{2} \leq \mu\)
A \(0.4\)
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.
Option D is Correct
