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प्रश्न
A cubical block of mass M and edge a slides down a rough inclined plane of inclination θ with a uniform velocity. The torque of the normal force on the block about its centre has a magnitude
विकल्प
zero
Mga
Mga sinθ
\[\frac{1}{2}\text{Mga }\sin\theta\]
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उत्तर
\[\frac{1}{2}\text{Mga }\sin\theta\]
Let N be the normal reaction on the block.
From the free body diagram of the block, it is clear that the forces N and mgcosθ pass through the same line. Therefore, there will be no torque due to N and mgcosθ. The only torque will be produced by mgsinθ.
\[\therefore \overrightarrow{\tau} = \overrightarrow{F} \times \overrightarrow{r}\]
Since a is the edge of the cube, \[r = \frac{a}{2}\]
Thus, we have
\[\tau = mg\sin\theta \times \frac{a}{2}\]
\[= \frac{1}{2}mga\sin\theta\]
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