Two wooden crates rest on top of one another. The smaller top crate has a mass o
ID: 1340017 • Letter: T
Question
Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 21 kg and the larger bottom crate has a mass of m2 = 90 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is s = 0.8 and the coefficient of kinetic friction between the two crates is k = 0.64. 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 = 270 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 1150 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?
Explanation / Answer
(1)
Total mass of crates: m1 + m2 = 90 + 21 = 111 kg
In this case, there's no friction between the floor and the bottom crate, thus there's no resisting force to the tension of the rope.
Use Newton' second law
F = ma
Solve for acceleration
a = F /m = 270 N / 111 kg = 2.43 m/s2
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(2)
As the upper crate is not moving relative to the lower crate, the force applied is exactly enough to accelerate it at the same rate as the two crate system as we solve in patt ( 1)
From Newton's law.
F = ma = ( 21 kg ) ( 2.43 m/s2 ) = 51 N
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(3)
Find normal force: F = mg = ( 21 kg ) ( 9.8 ) = 205.8 N
This the fricitonal force = sF = ( 0.8 ) ( 205.8 N ) = 164.64 N
This force corresponds to an acceleration of
a = F /m = 205.8 N / 21 kg = 7.84 m/s2
When the acceleration exceeds this amount, the upper crate will slip. The amount of tension needed to accelerate the whole system:
F = ma = ( 111 ) ( 7.84 m/s2 ) = 870 N
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