Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 22 kg and the larger bottom crate has a mass of m2 = 91 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two

1) The acceleration of the small crate can be calculated using the equation F = ma, where F is the net force acting on the system, m is the mass of the small crate, and a is the acceleration. The net force is equal to the tension in the rope minus the frictional force between the crates: Fnet = T - fs where fs is the static frictional force. To find fs, we first need to compare the force due to gravity on the top crate (Fg1) to the maximum force of static friction between the crates (fs,max). If Fg1 is less than fs,max, then the crates will not move relative to each other. If Fg1 is greater than fs,max, then there will be a static frictional force equal to fs,max opposing the motion of the top crate. We can find fs,max using the equation fs,max = sN, where s is the coefficient of static friction and N is the normal force. The normal force is equal to the weight of the top crate (mg1), which gives us: fs,max = s(mg1) Now we can compare Fg1 to fs,max: Fg1 = m1g = (22 kg)(9.81 m/s^2) = 216.42 N fs,max = (0.84)(22 kg)(9.81 m/s^2) = 181.49 N Since Fg1 is less than fs,max, there is no static frictional force and the acceleration of the small crate is equal to the net force divided by its mass: Fnet = T = (466 N) a = Fnet/m1 = (466 N)/(22 kg) = 21.18 m/s^2 2) In the previous situation, the frictional force the lower crate exerts on the upper crate is equal in magnitude and opposite in direction to the force the upper crate exerts on the lower crate. This force can be found using the same equation as in part 1, but with the roles of the crates reversed: Fnet = fs - T where fs is the kinetic frictional force. The kinetic frictional force is equal to the coefficient of kinetic friction (k) multiplied by the normal force, which again is just the weight of the top crate: fs = kN = k(mg1) fs = (0.66)(22 kg)(9.81 m/s^2) = 142.68 N Since the tension in the rope is less than the force of kinetic friction, the upper crate will not slide and the frictional force exerted by the lower crate on the upper crate is equal in magnitude and opposite in direction to the tension: Ffriction = T = 466 N 3) The maximum tension that the lower crate can be pulled at before the upper crate begins to slide can be found by comparing the force of gravity on the top crate (Fg1) to the maximum force of static friction between the crates (fs,max), just as in part 1. If the tension in the rope is greater than this maximum static frictional force, the top crate will begin to slide relative to the bottom crate. We have already calculated fs,max to be 181.49 N. Therefore, the maximum tension that can be applied to the lower crate without causing the upper crate to slide is: Tmax = fs,max = 181.49 N 4) When the tension in the rope is increased to 1322 N, the system experiences a net force in the positive direction: Fnet = T - fs where fs is the kinetic frictional force. We have already calculated fs to be 142.68 N. The net force is therefore: Fnet = (1322 N) - (142.68 N) = 1179.32 N This net force is then divided by the combined mass of the crates to find the acceleration of the system: a = Fnet/(m1 + m2) = (1179.32 N)/(22 kg + 91 kg) = 11.28 m/s^2 5) As the upper crate slides, it experiences a frictional force in the opposite direction of motion. This force can be found using the same equation as in part 2, but with the roles of the crates reversed: Fnet = T - fs where fs is the kinetic frictional force. The kinetic frictional force is equal to the coefficient of kinetic friction (k) multiplied by the normal force, which now includes both crates: fs = kN = k(m1 + m2)g fs = (0.66)(22 kg + 91 kg)(9.81 m/s^2) = 814.48 N The net force is then equal to the tension minus the frictional force: Fnet = T - fs = (1322 N) - (814.48 N) = 507.52 N This net force is then divided by the mass of the upper crate to find its acceleration: a = Fnet/m1 = (507.52 N)/(22 kg) = 23.07 m/s^2 As for the acceleration of the lower crate, it remains the same as in part 1 (21.18 m/s^2) since there is no net force acting on it in the horizontal direction.