A crate of potatoes of mass 23.0 kg is on a ramp with angle of incline 30° to the horizontal. The coefficients of friction are μs = 0.78 and μk = 0.35. Find the frictional force (magnitude and direction) on the crate if the crate is at rest.
I don't understand why I am given both coefficients when it states the crate is at rest. I tried solving for static friction by doing normal force x the coefficient of static friction. That is wrong. I tried making it and equilibrium problem where the static friction and horizontal component of the weight = 0 but that just didn't make sense because I wouldnt even need the coefficient. Clearly I'm very lost. Please help!!!
Tricky little problem makers. You don't need the coefficients after all. I got it!
I can understand why you might be confused. Let's go step by step to clarify the problem and find the solution.
First, when the crate is at rest on the ramp, we know that the frictional force acting on it opposes the tendency for the crate to slide down the ramp. So, in this case, we need to consider static friction rather than kinetic friction.
To calculate the frictional force, we need to find the maximum static frictional force that can be exerted on the crate without it moving. The maximum static frictional force is given by the equation:
Fs (max) = μs * N
where Fs (max) is the maximum static frictional force, μs is the coefficient of static friction, and N is the normal force.
The normal force, in this case, is the component of the weight acting perpendicular to the incline. It can be calculated using the equation:
N = mg * cos(θ)
where m is the mass of the crate (23.0 kg), g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the incline (30°).
N = (23.0 kg) * (9.8 m/s^2) * cos(30°)
Simplifying this equation, we find:
N ≈ 199.94 N
Now we can calculate the maximum static frictional force:
Fs (max) = (0.78) * (199.94 N)
Fs (max) ≈ 155.96 N
Therefore, the magnitude of the static frictional force acting on the crate is approximately 155.96 N.
The direction of the frictional force will be opposite to the direction the crate would slide if it were to start moving. Since the crate is on an incline, the frictional force will be pointing up the ramp.
So, the frictional force on the crate, when it is at rest, has a magnitude of approximately 155.96 N and it acts in the upward direction along the ramp.