Physics
posted by Anonymous on .
A 19 kg crate is placed on an inclined ramp. When the angle the ramp makes with the horizontal is increased to 23.2 degrees the crate begins to slide downward. what is the coefficient of static friction between the crate and the ramp? At what angle does the crate begin to slide it its mass is doubled? I just need to know how to approach this problem because i'm lost.

Fc = 19kg * 9.8 = 186.2N @ 23.2Deg = Weight of crate.
The Wt. of crate is broken down into 2
components:
Fp = 186.2*sin(23.2) = 73.35N parallel
to plane downward.
Fv=186.2*cos(23.2)=71.1N.perpendicular
to the plane downward.
Since the crate just beginsw to
move at 23.2 deg, we can assume that
Fp is = to the force of friction(uFv):
uFv = Fp,
171.1u = 73.35,
u = 73.35 / 171.1 = 0.429 = coefficient
of friction.