Two men are carrying a uniform bar of length | vedclass

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Question

(A) Let $W$ be the total weight of the uniform bar acting at its center of mass,which is at a distance $L/2$ from both ends.Let $R_B$ be the load on t

Vedclass · Accepted Answer

$L / 3$

Vedclass · Answer

(A) Let $W$ be the total weight of the uniform bar acting at its center of mass,which is at a distance $L/2$ from both ends.Let $R_B$ be the load on the younger man at one end and $R_M$ be the load on the other man.Given $R_B = W/4$. Since the total load is $W$,$R_M = W - W/4 = 3W/4$.Let the other man be at a distance $d$ from the other end. The distance of the other man from the center of mass is $x = L/2 - d$.For rotational equilibrium,taking the torque about the center of mass:$R_B 	imes (L/2) = R_M 	imes x$$(W/4) 	imes (L/2) = (3W/4) 	imes x$$L/8 = 3x/4$$x = L/6$Since $x = L/2 - d$,we have $d = L/2 - L/6 = (3L - L)/6 = 2L/6 = L/3$.Thus,the distance of the other man from the other end is $L/3$.

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 total load. Suppose the younger man is at one end of the bar,what is the distance of the other man from the other end?

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