# A 50 N force is applied to the box at a 37 degrees. The coefficient of kinetic friction between the box and the plane is .4. ?

I started doing this problem but I don’t think I’m doing it right. Please help.

### 1 Answer

- oldschoolLv 72 weeks agoFavorite Answer
Barely legible.

50N pulled at 37° resolves to 30N up and 40N right.

Assume g = 10m/s²

Fn = 10kg * 10m/s² - 30N = 70N

Friction force = 70*0.4 = 28N

Fx = 50*cos37 - 28 = 12N

a = Fx/m = 12/10 = 1.2m/s²

50N pushed at -37° resolves to 30N down and 40N right.

Assume g = 10m/s²

Fn = 10kg * 10m/s² + 30N = 130N

Friction force as high as 130*0.4 = 52N

Fx = 50*cos(-37) - 40 = 0

There is no movement which means Fnet = 0 which means the friction force exactly counters the applied force: Fnet = 50*cos37 - Ffriction = 0 so Ffriction = 50*cos37

a = 0 because the horizontal force is canceled by the friction force.

We understand 52N will be required to counter the friction force so there will be no movement until the horizontal component exceeds the friction force (F = 72N at 37° approx).