6.System of Particles and Rotational Motion
hard

A heavy iron bar, of weight $\mathrm{W}$ 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

$\frac{W}{2}$

B

$W$

C

$\mathrm{W} \cos \theta$

D

$\mathrm{W} \sin \theta$

(JEE MAIN-2024)

Solution

$R=$ net reaction force by shoulder

Balancing torque about pt of contact on ground:

$\mathrm{W}\left(\frac{\mathrm{L}}{2} \cos \theta\right)=\mathrm{R}(\mathrm{L} \cos \theta)$

$\Rightarrow \mathrm{R}=\frac{\mathrm{W}}{2}$

Standard 11
Physics

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