Consider the situation shown in the figure. A | vedclass

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(C) First,calculate the tension $T$ in the string connecting the masses $M$ and $2M$ over the movable pulley. The tension $T$ is given by the formula

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$\frac{16}{3} M$

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(C) First,calculate the tension $T$ in the string connecting the masses $M$ and $2M$ over the movable pulley. The tension $T$ is given by the formula for an Atwood machine: $T = \frac{2 M_1 M_2 g}{M_1 + M_2}$.Substituting $M_1 = M$ and $M_2 = 2M$,we get $T = \frac{2(M)(2M)g}{M + 2M} = \frac{4 M^2 g}{3M} = \frac{4}{3} Mg$.The movable pulley is supported by a string connected to the end of the rod. The upward force exerted on the rod by this string is equal to the tension in the string supporting the movable pulley,which is $2T = 2 	imes (\frac{4}{3} Mg) = \frac{8}{3} Mg$.Let $M'$ be the mass of the rod. The rod is in rotational equilibrium about hinge $A$. The torque due to the weight of the rod $(M'g)$ acting at its center of mass $(L/2)$ must balance the torque due to the upward force $(8/3 Mg)$ acting at the end of the rod $(L)$.Taking torque about $A$: $M'g 	imes \frac{L}{2} = \frac{8}{3} Mg 	imes L$.Simplifying,$\frac{M'}{2} = \frac{8}{3} M$,which gives $M' = \frac{16}{3} M$.

Consider the situation shown in the figure. $A$ uniform rod of length $L$ can rotate freely about the hinge $A$ in a vertical plane. Pulleys and string are light and frictionless. If the rod remains horizontal at rest when the system is released,then the mass of the rod is:

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