Question

In: Physics

A 336-kg crate rests on a surface that is inclined above the horizontal at an angle...

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?

Expert Solution

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, F_p = W sin + F cos

The normal force, F_n = W cos - F sin

The coefficient of static friction, = F_p / F_n

= (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

genius_generous answered 8 months ago

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