posted by Jul .
A 248-kg crate rests on a surface that is inclined above the horizontal at an angle of 19.5°. A horizontal force (magnitude = 537 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?
can some1 plz explain this to me?
To start the crate sliding down the incline, you must provide a force component in that direction that exceeds the maximum static friction force minus the weight component in that direction.
When the two forces are equal,
M g cos 19.5 * Us = M g sin 19.5 + 537 cos19.5
where Us is the static friction coefficient.
Solve for Us
could you clarify what M is? i know g is the gravity which will be 9.81 but im not sure about the M