A 224-kg crate rests on a surface that is inclined above the horizontal at an angle of 20.1°. A horizontal force (magnitude = 543 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?
Force applied = 573
Force of Friction=(co-eff static friction)*(Normal force)
The Attempt at a Solution
So I calculated the normal force to be 2063.6N using N=m*g*cos(theta).
Then, I assumed that 573 was the limit for static friction. So the forces would have to equal up to 0 for there to be no acceleration.
Fapp - Ff = 0
573N - 2063N*(coeff. static) = 0
coeff. static = -0.278
I tried both this value and the absolute value and about 7 other values. I've only got three tries left and I have no clue what to do! Any help would be so great!! Thanks in advance.