Question

A cube is placed on an inclined plane of inclination $\theta$ as shown in figure. Coefficient of friction between the cube and the plane is $\mu$. As the angle $\theta$ is gradually increased, the cube slides before toppling if

• A. $\mu >1$
• B. $\mu >\frac{1}{2}$
• C. $\mu <1$
• D. None of these

Answer 1

The correct option is C $\mu <1$

Equilibrium of an object is broken when the object is on the point of toppling,
hence we draw the FBD and write the force and torque equation.


Let $m$ be the mass of the cube and $a$ the side of the cube. The cube will slide when: Here, $f=$ force of friction

$mg\sin \theta >f$
$mg\sin \theta >\mu mg\cos \theta$
or $\tan \theta >\mu ...\left(i\right)$

The cube will topple if torque of $mg\sin \theta$ about $P$ is greater than torque of $mg\cos \theta$ about $P$.

$\left(mg\sin \theta \right)\left(\frac{a}{2}\right)>\left(mg\cos \theta \right)\left(\frac{a}{2}\right)$
(By applying the torque balancing equation).
or $\tan \theta >1...\left(ii\right)$

From Eqs. $\left(i\right)$ and $\left(ii\right)$, we see that cube slides before toppling if:
$\mu <1$

Hence option C is the correct answer.