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

Question

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 (A) Zero (B) Mga (C) Mga sin θ (D) (1/2) Mga sin θ

Tardigrade Admin · Accepted Answer

τN=τf τN=f ⋅ (a/2)=m g sin θ (a/2) τN=(1/2) m g a sin θ

Q. 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

Zero

Mga

$M g a sin θ$

$\frac{1}{2} M g a sin θ$

$τ_{N} = τ_{f}$
$τ_{N} = f \cdot \frac{a}{2} = m g sin θ \frac{a}{2}$
$τ_{N} = \frac{1}{2} m g a sin θ$

Q. 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

Zero

Mga

Mgasinθ

21​Mgasinθ

τN​=τf​ τN​=f⋅2a​=mgsinθ2a​ τN​=21​mgasinθ

Q. 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

$M g a sin θ$

$\frac{1}{2} M g a sin θ$

$τ_{N} = τ_{f}$
$τ_{N} = f \cdot \frac{a}{2} = m g sin θ \frac{a}{2}$
$τ_{N} = \frac{1}{2} m g a sin θ$