Question:
A cubical block of mass $m$ and edge a slides down a rough inclined plane of inclination $\theta$ with a uniform speed. Find the torque of the normal force acting on the block about its centre.
Solution:
Since,
acceleration $=0$
ff $=m g^{\sin \theta}$
The normal contact will shift as block is not toppling.
So, $\tau_{\text {Normal }}=\tau_{f f}$
$=\mathrm{mg}^{\sin \theta\left(\frac{a}{2}\right)}$
HC VERMA Solutions for Class 11 Physics Chapter 10 Rotational Mechanics
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