- #1

- 5

- 1

A 250-kg crate is on a rough ramp, inclined at 30° above the horizontal. The coefficient of kinetic friction between the crate and ramp is 0.22. A horizontal force of 5000 N is applied to the crate, pushing it up the ramp. What is the acceleration of the crate?

Relevant Equations: F = ma, F_k=U_kn

I did it, but my answer was 13.23 m/s^2. This was from (5000 - 2450*sin(30) - 0.22(2450)cos(30))/250.

I think I messed up my normal force calculation, which was 250*9.8 cos(30). I looked online, and other people said normal force was equal to Fsin(30) +mgcos(30), which doesn't make sense to me. If possible could someone walk me through finding the normal force?

Thanks!