A cubical block of mass m and edge a slides down a rough inclined plan

1.	A cubical block of mass m and edge a slides down a rough inclined plane of inclination θ with a uniform speed. Find the torque of the normal force acting on the block about its center.?
Answer» Given in the question . Total number of FORCE acting on the block is three. Weight= MG Friction= F Normal force= R Refer to the Attachment Here, Force mg acting on the two component i.e mg. cosθ and mg.sinθ Now, ∵ R and mg.cosθ passing by the CUBE CENTRE, hence no TORQUE will be working, only torque is produced by mg.sinθ Thertefore, i = F× r i = mgsinθ × (a/2) (r = a/2) $\frac{1}{2} mga sin\theta$ i = Hope it Helps :-)

A cubical block of mass m and edge a slides down a rough inclined plane of inclination θ with a uniform speed. Find the torque of the normal force acting on the block about its center.?

Answer»

Given in the question .

Total number of FORCE acting on the block is three.

Weight= MG

Friction= F

Normal force= R

Refer to the Attachment

Here, Force mg acting on the two component i.e mg. cosθ and mg.sinθ

Now,

∵ R and mg.cosθ passing by the CUBE CENTRE, hence no TORQUE will be working, only torque is produced by mg.sinθ

Thertefore,

i = F× r

i = mgsinθ × (a/2) (r = a/2)

$\frac{1}{2} mga sin\theta$ i =

Hope it Helps :-)

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A cubical block of mass m and edge a slides down a rough inclined plane of inclination θ with a uniform speed. Find the torque of the normal force acting on the block about its center.?

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