A block of mass m has dimension ( ℓ × b × h ) slides downward on fixed rough inclined plane (whose inclination angle θ ) with a constant velocity. The torque of normal contact force about the centre o

Question

A block of mass m has dimension ( ℓ × b × h ) slides downward on fixed rough inclined plane (whose inclination angle θ ) with a constant velocity. The torque of normal contact force about the centre of mass of the block will be : [Friction coefficient between block and inclined is μ ]

TestHub · Accepted Answer

Block slides with constant velocity, so it is in rotational equilibrium

Torque about the centre of mass of the block is zero:

$τ_{g} + τ_{N} + τ_{f r} = 0$

Weight acts at the centre of mass, so $τ_{g} = 0$

Constant velocity gives friction $f = m g sin θ$ , acting at a distance h/2 from the centre

$τ_{N} + m g sin θ \times \frac{h}{2} = 0$

$∣ τ_{N} ∣ = m g sin θ \times \frac{h}{2}$

Answer: C

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