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Question
A cubical block of mass m and edge a slides down a rough inclined plane of inclination θ with a uniform speed. Find the torque of the normal force acting on the block about its centre.
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Solution
Let N be the normal reaction on the block.
From the free body diagram of the block, it is clear that forces N and mgcosθ pass through the same line. So, there will be no torque due to N and mg cosθ. The only torque will be produced by mg sinθ.
\[\therefore \overrightarrow{\tau} = \overrightarrow{F} \times \overrightarrow{r} \]
a is the edge of the cube.Therefore, we have
\[r = \frac{a}{2}\]
\[ \therefore \tau = mg\sin\theta \times \frac{a}{2}\]
\[= \frac{1}{2}mga\sin\theta\]
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