Question

A block of base $10cm\times 10cm$ and height $15cm$ is kept on an inclined plane. The coefficient of friction between them is $\sqrt{3}$ The inclination $\theta$ of this inclined plane from the horizontal plane is gradually increased from $0^o$.Then :

• A. at $\theta ={30}^o$, the block will start sliding down the plane.
• B. the block will remain at rest on the plane up to certain $\theta$ and then it will topple.
• C. at $\theta ={60}^o$, the block will start sliding down the plane and continue to do so at higher angles.
• D. at $\theta ={60}^o$, the block will start sliding down the plane and on further increasing $\theta$, it will topple at certain $\theta$.

Answer 1

The correct option is B the block will remain at rest on the plane up to certain $\theta$ and then it will topple.

As we increase the angle of inclination, the frictional force acts in upward direction on incline to prevent sliding of the block.

But this frictional force also causes torque about the CoM of the block, which is countered by the torque due to the Normal force on the block.

But as the angle of inclination $\theta$ increases, the normal force reduces and perpendicular distance of Normal force from CoM also reduces which decreases the counter-torque due to the normal force. So after an angle, the torque due to frictional force overcomes that due to normal force, and the block starts to topple.