A heavy iron bar, of weight $W$ is having its one end on the ground and the other on the shoulder of a…

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 :
  1. $\mathrm{W} \cos \theta$
  2. $\frac{W}{2}$
  3. W
  4. $W \sin \theta$

Solution


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

Asked in: JEE Main 2024 (09 Apr Shift 1)

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