Question

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 = 93 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.87 and the coefficient of kinetic friction between the two crates is μk = 0.68. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem).

1) The rope is pulled with a tension T = 481 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate?

2) In the previous situation, what is the frictional force the lower crate exerts on the upper crate?

3) What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide?

4) The tension is increased in the rope to 1286 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?

5) As the upper crate slides, what is the acceleration of the lower crate?

Answer 1

1) F = ma
there is no sliding, so this force accelerates both crates equally.
a = F/m = 481N/(22 + 93)kg = 4.18 m/s^2

2) now we want to know the F=ma to accelerate the top crate.
F = 22kg(4.18m/s^2) = 91.96N
Note that frictional forces are usually working in the opposite direction, but here we need friction to have the top crate move so friction is pushing the top crate. Also note that the frictional force can be anything from zero to the maximum for that coefficient and normal force.

3) Maximum frictional force is us*Fn
Fn = mg = 22kg*9.8m/s^2 = 215.6N
Ffr max = 0.87*215.6N = 187.57N
to get this force on the top crate would require what acceleration?
F = ma
a = F/m = 187.57N/22kg = 8.53 m/s^2
To accelerate both crates with this acceleration requires what force?
F = ma = (22kg + 93kg)8.53m/s^2 = 980.95 N
The tension is 980.95N when the top crate is just about to slip.

4) Since the tension is bigger than the maximum (980.95N) the top crate will slip. Since it is slipping, the kinetic coefficient is used.
The maximum force that friction can provide is the coefficient times the normal force
F fr max kinetic = uk*Fn = 0.68*215.6N = 146.61N
This force accelerates the upper crate
F = ma
a = F/m = 146.61/22kg = 6.66 m/s^2

5) This is tricky. The 1286N is accelerating the upper crate, but not as much as the lower crate. Only 146.61N is going to accelerating the upper crate. This leaves
1286N - 146.61N = 1139.39N
The bottom crate is accelerated with this 1139.39N
F = ma
a = F/m = 1139.39N/93kg = 12.25 m/s^2