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(A) For the rod $(AB)$ to be in equilibrium,the net torque about any point must be zero. The forces acting on the rod are: $1$. The tension $(T)$ in t

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(A) For the rod $(AB)$ to be in equilibrium,the net torque about any point must be zero. The forces acting on the rod are: $1$. The tension $(T)$ in the rope $(AC)$ acting at point $(A)$. $2$. The weight $(Mg)$ of the rod acting at its center of mass. $3$. The normal force $(N)$ from the ice at point $(B)$. Since the ice is frictionless,the normal force $(N)$ must be vertical. For the rod to be in equilibrium,the lines of action of all three forces must intersect at a single point or be parallel. Since the weight $(Mg)$ acts vertically downwards and the normal force $(N)$ acts vertically upwards,these two forces are parallel. For the system to be in equilibrium,the tension $(T)$ must also be parallel to these forces. Therefore,the rope $(AC)$ must be vertical. This configuration is shown in option $(A)$.

$A$ rod $(AB)$ is attached to a fixed point $(C)$ using a light rope $(AC)$. The other end of the rod $(B)$ is resting on ice with negligible friction,and the system is in a stationary position. Which of the following can be the equilibrium configuration of this system?

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