A 214-kg crate rests on a surface that is inclined above the horizontal at an angle of 19.4°. A horizontal force (magnitude = 541 N and parallel to the ground, not the incline) is required to start the crate moving down the incline. What is the coefficient of static friction between the crate and the incline?

Answer 1

forces down slope = Fd
Fd = m g sin 19.4 + 541 cos 19.4

forces normal to slope
= m g cos 19.4 - 541 sin 19.4
so force up slope =
Ff = mu ( m g cos 19.4 - 541 sin 19.4)

no acceleration so Ff=Fd
mu (m g cos 19.4 - 541 sin 19.4)=m g sin 19.4 + 541 cos 19.4