Two wooden crates rest on top of one another. The smaller top crate has a mass o

Question

Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 25 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 crates is µs = 0.79 and the coefficient of kinetic friction between the two crates is µk = 0.62. 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 = 234 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 magnitude of 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 1187 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?m/s2
5)As the upper crate slides, what is the acceleration of the lower crate?

Explanation / Answer

1) T=234N

in this case, both the crates move together with same acceleration.
total mass, m = m1+m2 = 116kg
acceleration, a = 234/116 = 2.02 m/s2

2) the only horizontal force acting on upper body is the frictional force by the lower crate.

for the upper body, f = m1 * a = 25 * 2.02 = 50.5 N.

3) the maximum static frictional force is the maximum force that can be exerted on lower crate so that both the crates move together.

fmax = s * m1 * g = 0.79 * 25 * 9.8 = 193.55N.

a = f/m1 = 193.55 / 25 = 7.72 m/s2

Tmax = fmax + m2 * a = 193.55 + 91 * 7.72 = 895.52N

4)T=1187N

force on upper crate = fkinetic = k*m1*g = 151.9N

acceleration of upper crate, a1 = 151.9/25 = 6.076 m/s2

5)for lower crate,

T - f = m2*a2

1187 - 151.9 = 91*a2

acceleration of lower crate, a2 = 11.37 m/s2

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