Kinetic Friction Question

Question

A box of books that weighs 325N moves with a constant velocity across the floor when it is pushed with a force of 425N exerted downward at an angle of 35.2 below the horizontal. Can you help me find the kinetic friction(mu) between the box and the floor?

Answer 1

Here's a quasi sketch/free-body diagram:

We can resolve the applied force "F" into two components F_x and F_y as follows:

F_x = 425cos(35.2°) = 347.287 N

F_y = 425sin(35.2°) = 244.984 N

The motion in the x-direction has no acceleration (constant velocity) so the sum of the forces in the x-direction is zero:

∑F_x = F_x - F_F = 0

F_x = F_F

We know the F_F = μF_N so:

F_x = μF_N and we can plug in the value of F_x calculated above to get:

347.287 = μF_N or μ = 347.287/F_N but we don't know F_N so we need to look at the forces in the y-direction:

There is no acceleration (there's no motion at all) in the y-direction so the sum of the forces in the y-direction is zero:

∑F_y = F_N - F_y - F_g = 0 so F_N = F_y + F_g and we know that F_g = 325 N and Fy = 244.984 N so:

F_N = 325 + 244.984 = 569.984 N and we can plug that into our equation from ∑F_x to get:

μ = 347.287/569.984 = 0.609