Class 11PhysicsSystem of Particles and Rotational MotionMix Example - System of Particles and Rotational Motion
Difficult$A$ uniform cube of mass $m$ and side $a$ is placed on a frictionless horizontal surface. $A$ vertical force $F$ is applied to the edge as shown in the figure. Match the following (most appropriate choice):
$(a)$ $\frac{mg}{4} < F < \frac{mg}{2}$ $(i)$ Cube will move up $(b)$ $F > \frac{mg}{2}$ $(ii)$ Cube will not exhibit motion $(c)$ $F > mg$ $(iii)$ Cube will begin to rotate about $A$ $(d)$ $F = \frac{mg}{4}$ $(iv)$ Normal reaction effectively at $a/3$ from $A$,no motion
| $(a)$ $\frac{mg}{4} < F < \frac{mg}{2}$ | $(i)$ Cube will move up |
| $(b)$ $F > \frac{mg}{2}$ | $(ii)$ Cube will not exhibit motion |
| $(c)$ $F > mg$ | $(iii)$ Cube will begin to rotate about $A$ |
| $(d)$ $F = \frac{mg}{4}$ | $(iv)$ Normal reaction effectively at $a/3$ from $A$,no motion |
(A) Let the side of the cube be $a$. The weight $mg$ acts at the center of the cube,which is at a distance $a/2$ from point $A$. The force $F$ is applied at the edge,at a distance $a$ from point $A$.
$1$. For the cube to start rotating about point $A$,the torque due to $F$ must exceed the torque due to gravity about $A$:
$\tau_F > \tau_{mg} \implies F \times a > mg \times \frac{a}{2} \implies F > \frac{mg}{2}$. Thus,$(b) \rightarrow (iii)$.
$2$. If $F > mg$,the net upward force is positive,so the cube will move up. Thus,$(c) \rightarrow (i)$.
$3$. For no motion,the normal reaction $N$ must balance the forces,and the net torque about $A$ must be zero. Let $x$ be the distance of the normal reaction from $A$. Then $N = mg - F$ and $mg(a/2) - F(a) - N(x) = 0$. For $F = mg/4$,$N = 3mg/4$. Substituting: $mg(a/2) - (mg/4)a = (3mg/4)x \implies mg(a/4) = (3mg/4)x \implies x = a/3$. Thus,$(d) \rightarrow (iv)$.
$4$. For $\frac{mg}{4} < F < \frac{mg}{2}$,the torque of $F$ is less than the torque of gravity,so the cube remains in equilibrium on the surface. Thus,$(a) \rightarrow (ii)$.
$1$. For the cube to start rotating about point $A$,the torque due to $F$ must exceed the torque due to gravity about $A$:
$\tau_F > \tau_{mg} \implies F \times a > mg \times \frac{a}{2} \implies F > \frac{mg}{2}$. Thus,$(b) \rightarrow (iii)$.
$2$. If $F > mg$,the net upward force is positive,so the cube will move up. Thus,$(c) \rightarrow (i)$.
$3$. For no motion,the normal reaction $N$ must balance the forces,and the net torque about $A$ must be zero. Let $x$ be the distance of the normal reaction from $A$. Then $N = mg - F$ and $mg(a/2) - F(a) - N(x) = 0$. For $F = mg/4$,$N = 3mg/4$. Substituting: $mg(a/2) - (mg/4)a = (3mg/4)x \implies mg(a/4) = (3mg/4)x \implies x = a/3$. Thus,$(d) \rightarrow (iv)$.
$4$. For $\frac{mg}{4} < F < \frac{mg}{2}$,the torque of $F$ is less than the torque of gravity,so the cube remains in equilibrium on the surface. Thus,$(a) \rightarrow (ii)$.
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