First look at the washer in equilibrium:
Tsin62 = mg and Tcos62 + mgsin40 = ma
eliminate T:
mg/sin62 * cos62 +mg sin40 = ma
eliminate m
a = g cot62 + g sin40
We'll need this in part two
Now the crate:
Normal force
mg cos40 = Fn = 232*9.8*cos40
And mg sin40 - Ff = ma
Use the a you found above and solve for Ff.
mu = Ff/Fn
A steel washer is suspended inside an empty shipping crate from a light string attached to the top of the crate. The crate slides down a long ramp that is inclined at an angle of 40 ∘ above the horizontal. The crate has mass 162 kg . You are sitting inside the crate (with a flashlight); your mass is 70 kg . As the crate is sliding down the ramp, you find the washer is at rest with respect to the crate when the string makes an angle of 62 ∘ with the top of the crate. What is the coefficient of kinetic friction between the ramp and the crate?
2 answers
Tsin62 = mg ?????