Question

A uniform cube of mass $m$ and side $a$ is resting in equilibrium on a rough ${45}^o$ inclined surface. The distance of the point of application of normal reaction measured from the lower edge of the cube is:

• A. $zero$
• B. $\frac{a}{2}$
• C. $\frac{a}{\sqrt{2}}$
• D. $\frac{a}{4}$

Answer 1

Answer: (A)

$zero$

There are two forces acting on the block- gravitational force and the frictional force. Using rotational equilibrium we see that the moment is zero about the lower edge of the cube. Thus for equilibrium the normal force which is perpendicular to the plane should pass through the lower edge.

Assume Normal is at x distance from lower edge, taking moment about lower edge:

$Nx+mgsin\theta \frac{a}{2}=mgcos\theta \frac{a}{2}$

$Nx=0$

$N$ is non zero , hence $x=0$