Question 108 of 280

A heavy iron bar, of weight $W$ is having its 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 :

A$\mathrm{W} \cos \theta$B$\frac{W}{2}$CWD$W \sin \theta$

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Answer: (B) $\frac{W}{2}$

$R=$ net reaction force by shoulder
Balancing torque about pt of contact on ground:
$\begin{aligned} & \mathrm{W}\left(\frac{\mathrm{L}}{2} \cos \theta\right)=\mathrm{R}(\mathrm{L} \cos \theta) \\ & \Rightarrow \mathrm{R}=\frac{\mathrm{W}}{2}\end{aligned}$