Question

A cube of side $l$ and mass $M$ placed on rough horizontal surface and the friction so that it will not move, if a constant force $F=Mg$ is applied horizontally $1/4$ above the surface. Then the torque due to normal force about center of the cube is equal to

• A. $\frac{Mgl}{2}$
• B. $\frac{Mgl}{4}$
• C. $\frac{Mgl}{8}$
• D. $Zero$

Answer 1

Answer: (A)

$\frac{Mgl}{2}$

answer is A.

torque = force * displacement

T = F * L /2

T = Mg * L/2