Question

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º. What is the coefficient of static friction between the box and the plane?

Answer 1

Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

Vertical component of force :

`mg\ sin\ \theta = 120* 10 * sin\ 47^\circ{}=877.62 \ N`

Horizontal component of force(Normal reaction) :

`mg\ cos\ \theta = 120* 10 * cos\ 47^\circ{}=818.40 \ N`

Since, box is on the verge of slipping :

`mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07`

Therefore, the coefficient of static friction between the box and the plane is 1.07.

Hence, this is the required solution.