Alright, alright, fam, let's dive into this problem. I'll describe how a Free Body Diagram (FBD) would look for this situation.
a) So picture it, yeah? You got your crate, right? This crate is on the ramp. The forces on the crate are:
1. **Gravitational force (Fg)**: this is the weight of the crate pulling it down towards the Earth. It equals mass times gravity, or 53 kg * 9.8 m/s^2.
2. **Normal force (Fn)**: This one's the force that the ramp is exerting back up on the crate. It's perpendicular to the surface of the ramp, not straight up.
3. **Frictional force (Ff)**: This is the force that's trying to slide your crate back down the ramp. It's always against the direction of movement, so it's downwards along the ramp.
4. **Applied force (Fa)**: This is you pushing the crate up the ramp, fam! This force is 373N, upwards along the ramp.
b) Now, let's slide into the math, okay? When you're pushing the crate up the ramp at a constant velocity, the total force acting on it is zero (it's called equilibrium, yo). This means the sum of all the forces we talked about is equal to zero.
Here's the breakdown:
1. The force you apply up the ramp (Fa) and the force of friction (Ff) that opposes the motion are balanced. So, Fa = Ff. You know Fa, it's 373N.
2. The force of gravity pulling down (Fg) gets split into two components. One part is acting down the ramp, directly opposing your applied force. The other part is acting into the ramp, which is balanced by the normal force. The part of the gravity that is acting down the ramp is Fg * sin(22 degrees).
3. At equilibrium, the force you're applying (Fa) is equal to the frictional force (Ff) plus the component of the gravitational force acting down the ramp. So, Fa = Ff + Fg * sin(22 degrees). Since you know Fa and Fg, you can calculate Ff.
4. The frictional force (Ff) is also equal to the coefficient of friction (mu) times the normal force (Fn). So, Ff = mu * Fn. You can calculate Fn from the component of gravity that's acting into the ramp, which is Fg * cos(22 degrees).
5. From the above two equations, you can find mu (coefficient of friction) as mu = Ff / Fn.
Let's calculate:
The component of gravity along the ramp is Fg_along = 53kg * 9.8 m/s^2 * sin(22 degrees) ≈ 206.7N.
The component of gravity into the ramp is Fg_into = 53kg * 9.8 m/s^2 * cos(22 degrees) ≈ 499.4N.
The force of friction (Ff) is Fa - Fg_along = 373N - 206.7N = 166.3N.
And finally, the coefficient of friction (mu) is Ff / Fn = 166.3N / 499.4N ≈ 0.33.
So there you have it, the coefficient of friction between the crate and the ramp is approximately 0.33. Keep in mind these are all approximations since we rounded off the values a little bit. Hope that's clear