A 220-kg crate rests on a surface that is inclined above the horizontal at an angle of 19.4°. A horizontal force (magnitude = 525 N and parallel to the ground, not the incline) is required to start the crate moving down the incline. What is the coefficient of static friction between the crate and the incline?

To find the coefficient of static friction between the crate and the incline, we can use the equation:

Fs = μs * N

Here, Fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force.

First, let's calculate the normal force acting on the crate. The normal force is the perpendicular force exerted by the surface on the crate.

N = m * g * cosθ

where m is the mass of the crate, g is the acceleration due to gravity, and θ is the angle of inclination.

Given:
m = 220 kg
g = 9.8 m/s^2
θ = 19.4°

N = 220 kg * 9.8 m/s^2 * cos 19.4°

Next, we need to calculate the force of static friction (Fs). The force required to start the crate moving down the incline is equal in magnitude to the force of static friction.

Fs = F_applied

Given:
F_applied = 525 N

Now, we can substitute the values into the equation:

525 N = μs * N

We already found N in the previous step, so we can now solve for the coefficient of static friction (μs):

μs = Fs / N = 525 N / (220 kg * 9.8 m/s^2 * cos 19.4°)

Calculating this expression will give us the coefficient of static friction between the crate and the incline.