A heavy iron bar of weight W has one end on t | vedclass

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(A) Let $L$ be the length of the iron bar. The weight $W$ of the bar acts at its center of mass,which is at a distance of $L/2$ from the end on the gr

Vedclass · Accepted Answer

$\frac{W}{2}$

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(A) Let $L$ be the length of the iron bar. The weight $W$ of the bar acts at its center of mass,which is at a distance of $L/2$ from the end on the ground.Let $R$ be the reaction force exerted by the person's shoulder on the bar. The bar is in rotational equilibrium.Taking the torque about the point of contact on the ground:The torque due to the weight $W$ is $	au_W = W \cdot (L/2) \cos 	heta$ (acting in a clockwise direction).The torque due to the reaction force $R$ is $	au_R = R \cdot L \cos 	heta$ (acting in a counter-clockwise direction).For rotational equilibrium,the net torque about the point of contact on the ground must be zero:$\sum 	au = 0$$R \cdot L \cos 	heta = W \cdot (L/2) \cos 	heta$Dividing both sides by $L \cos 	heta$ (assuming $\cos 	heta 
eq 0$):$R = \frac{W}{2}$Thus,the weight experienced by the person is $\frac{W}{2}$.

$A$ heavy iron bar of weight $W$ has one end on the ground and the other on the shoulder of a person. The bar makes an angle $\theta$ with the horizontal. The weight experienced by the person is:

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