IIT JEE , AIEEE Forum - Physics , Mechanics

karthik2007

Really very simple. The main thing to remember here is that the normal reaction wont pass through the center of the cube, as there will be a shifting of normal reaction in order to maintain rotational equilibrium

Now, as it moves with constant velocity:
Here M stands for coefficient of friction.

mgsin@ - Mmgcos@ = 0

Or M = tan@

Now, balancing torques about center:

We have only 2 torques about center:

Clockwise torque due to friction, and anticlockwise torque due to Normal reaction

So, Mmgcos@xa/2 = T

T -> torque due to normal reaction.

(a/2 is the distance of force of friction from center)

So, Mmgacos@/2 = T

But M = tan@

So mgasin@/2 = T

which is the required torque.

Cheers!!!

Ask & Discuss Questions with Community & Experts