Question

A cubical block of mass $M$ and edge $a$ slides down a rough inclined plane of inclination $\theta$ with a uniform velocity. The torque of the normal force on the block about its centre has a magnitude

• A. zero
• B. $Mga$
• C. $Mga\sin \theta$
• D. $\frac{Mga\sin \theta }{2}$

Answer 1

The correct option is D $\frac{Mga\sin \theta }{2}$

Because the cubical block slides with a uniform velocity and does not topple, hence
Torque produced by weight = Torque due to normal force on the block
$\therefore$ Torque due to normal force
= Torque due to weight
= Component of weight parallel to plane $\times \perp$ distance from lower face
$=\left(Mg\sin \theta \right)\frac{a}{2}$
Hence, the correct answer is option (d).