Question

A mass of 8.4 kg is placed on a incline with friction.   The coefficient static friction is 0.9 and the coefficient of kinetic friction is 0.44.   If the angle of inclination of the incline is 63.1 degrees, what is the amount of friction acting on it in Newtons?

Answer 1

Here ,

for the angle of inclination ,

theta = 63.1 degree

maximum static friction , fs = us * m * g * cos(theta)

fs = 0.9 * 8.4 * 9.8 * cos(63.1)

fs = 33.51 N

Now , gravtiy force down the plane,

Fg = mg * sin(theta) = 8.4 *9.8 * sin(63.1)

Fg = 73.4 N

as Fg is larger than fs . the block will move and kinetic friction will act on the box

hence , frictional force = uk * mg * cos(theta)

frictional force = 0.44 * 8.4 * 9.8 * cod(63.1)

frictional force = 16.4 N

the amount of friction acting is 16.4 N