A heavy iron bar, of weight $W$ is having its one end on the ground and the other on the shoulder of a…
- $\mathrm{W} \cos \theta$
- $\frac{W}{2}$
- W
- $W \sin \theta$
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)