Question

The coefficient of static friction between the 3.00-kg crate and the 35.0° incline of Figure P4.31 is 0.300. What minimum force SF must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

Answer 1

$$N=mg\cos{35}+F$$

$$F_f=\mu N=\mu (mg\cos{35}+F)$$

For the equilibrium:

$$0.3 ((3)(9.81)\cos{35}+F)=(3)(9.81) \sin{35}$$

$$F=32.2\ N$$