Question

A 50 kg steel file cabinet is in the back of a dump truck. The truck's bed, also made of steel, is slowly tiled. What is the size of the static friction force on the cabinet when the bed is titled 20 degrees? At what angle will the file cabinet begin to slide?

Answer 1

I think I've figured this out, i'll post up my answers

Answer 2

For static firction
\[F _{?}=\mu_{s}F _{n}\]
\[\mu _{s}\sin(20)/\cos(20)=\tan(20) = .36\]
the mass and gravity for the x and y components cancel out.
And I'm guessing that the angle would just be the inverse of tangent right?
\[\tan ^{-1}(\theta)=87\]

Answer 3

To answer the first question:
Ffs = 50kg.9,81m/s².sin 20° = 167,76N
To answer the second question you need the static friction coefficient.