In: Physics
A 336-kg crate rests on a surface that is inclined above the horizontal at an angle of 17.2°. A horizontal force (magnitude = 405 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?
Given,
The mass, m = 336 kg
An angle, = 17.2 o
The force, F = 405 N
The weight, W = mg
= 336 * 9.8
= 3292.8 N
The force on the crate parallel, Fp = W sin + F cos
The normal force, Fn = W cos - F sin
The coefficient of static friction, = Fp / Fn
= (W sin + F cos ) / (W cos - F sin )
= [3292.8 * sin (17.2) + 405 * cos(17.2 )] / [3292.8 * cos(17.2) - 405 * sin (17.2)]
= (973.71 + 386.89) / (3145.54 - 119.76)
= 1360.6 / 3025.78
= 0.45