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Physics

Mechanics

Easy

Question

A cubical block of mass m and edge length L is sliding down on an inclined plane with uniform velocity. Torque due to the normal force applied by the incline on the block about its centre of mass is

$m g \frac{L}{2} \cos θ$
$m g \frac{L}{2}$
$m g \frac{L}{2} \sin θ$
$m g \frac{L}{2} \cos^{2} θ$

The correct answer is: $m g L over 2 sin invisible function application theta$

$text Here end text m g sin invisible function application theta equals f subscript k equals mu subscript k m g cos invisible function application theta text About centre of mass torque due to normal contact force = Torque due to end text f subscript k text (equal and unlike) end text text Here end text straight N text will not act at centre of mass and end text N x equals f subscript k L over 2 equals left parenthesis m g sin invisible function application theta right parenthesis L over 2$

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