Question

Friction: a 50-kg box is resting on a horizontal floor. A force of 250 n directed at an angle of 30. 0° below the horizontal is applied to the box. The coefficient of static friction between the box and the surface is 0. 40, and the coefficient of kinetic friction is 0. 30. What is the force of friction on the box?.

Answer 1

The net forces on the box parallel and perpendicular to the surface, respectively, are

∑ F[para] = (250 N) cos(-30.0°) - F[friction] = (50 kg) a

and

∑ F[perp] = F[normal] + (250 N) sin(-30.0°) - F[weight] = 0

where a is the acceleration of the box. (We ultimately don't care what this acceleration is, though.)

To decide whether the friction here is static or kinetic, consider that when the box is at rest, the net force perpendicular to the floor is

∑ F[perp] = F[normal] - F[weight] = 0

so that, while at rest,

F[normal] = (50 kg) g = 490 N

Then with µ[s] = 0.40, the maximum magnitude of static friction would be

F[s. friction] = 0.40 (490 N) = 196 N

so that the box will begin to slide if it's pushed by a force larger than this.

The horizontal component of our pushing force is

(250 N) cos(-30.0°) ≈ 217 N

so the box will move in our case, and we will have kinetic friction with µ[k] = 0.30.

Solve the ∑ F[perp] = 0 equation for F[normal] :

F[normal] + (250 N) sin(-30.0°) - F[weight] = 0

F[normal] - 125 N - 490 N = 0

F[normal] = 615 N

Then the kinetic friction felt by the box has magnitude

F[k. friction] = 0.30 (615 N) = 184.5 N ≈ 185 N