Question

A uniform cube of mass m =3 kg and side a = 9 cm is resting in equilibrium on a rough ${45}^o$ inclined surface. What is the distance (in cm) of the point of application of normal reaction measured from the lower edge of the cube?

Answer 1

Given

M= 3kg

a= 9cm

$\theta$=45

Distance of normal from lower edge= ?

Solution

Let the distance for normal reaction from lower edge be 'x'.

As body is in equilibrium therefore moment of center will be 0

This means couple forces at center is 0

$f\times \frac{a}{2}-N\times x=0$

$f$ is friction force

$f=mgsin\theta$

$N=mgcos\theta$

putting the values

$mgsin45\times \frac{a}{2}=mgcos45\times x$

$x=a/2$

Therefore Normal reaction will be $a/2$ distance from center

This means Normal reaction Act on lower edge of cube

Hence distance from lower edge will be 0

The correct option is 0