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.
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.
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 \(T_1 = 2\,Mg ;\, \,T_2 =\dfrac{3\,Mg}{2}\)
A \(m= \dfrac{M}{2}\)
B \(m= \dfrac{M}{3}\)
C \(m= \dfrac{M}{4}\)
D \(m= \dfrac{M}{5}\)
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 \(3 \, mg , \, 2 \, mg\)
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 \(T= 3\,mg,\, N_x = mg, \,N_y = mg \)
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 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 \(0.5\)