Two men are carrying a uniform bar of length | vedclass

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(C) Let the weight of the bar be $W$,acting at its center of gravity $(CG)$ at a distance $L/2$ from the end where the younger man is standing.Let $F_

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$2L/3$

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(C) Let the weight of the bar be $W$,acting at its center of gravity $(CG)$ at a distance $L/2$ from the end where the younger man is standing.Let $F_1 = W/4$ be the force exerted by the younger man at the end $(x=0)$.Let $F_2$ be the force exerted by the other man at a distance $d$ from the younger man.For vertical equilibrium: $F_1 + F_2 = W \implies W/4 + F_2 = W \implies F_2 = 3W/4$.For rotational equilibrium,taking torque about the center of gravity $(CG)$: $F_1 	imes (L/2) = F_2 	imes x$,where $x$ is the distance of the second man from the $CG$.$(W/4) 	imes (L/2) = (3W/4) 	imes x \implies x = L/6$.The distance of the second man from the younger man (at the end) is $d = L/2 + x = L/2 + L/6 = (3L+L)/6 = 4L/6 = 2L/3$.

Two men are carrying a uniform bar of length $L$ on their shoulders. The bar is held horizontally such that the younger man gets $(1/4)^{th}$ of the load. Suppose the younger man is at the end of the bar,what is the distance of the other man from that end?

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