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

ID: 2167297 • Letter: T

Question

Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 16.0 kg and the larger bottom crate has a mass of m2 = 90.0 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is ?s = 0.85 and the coefficient of kinetic friction between the two crates is ?k = 0.69. 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 = 389.0 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 1184.0 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

Answer is as following 1) T = 445 = (16 + 90) * acc acc of system mass = 3.938m/s2 acceleration of the small crate answer static friction force between m1 & m2 = 16 * 9.81 * 0.82 = 144.8 N kinetic friction force between m1 & m2 = 16 * 9.81 * 0.66 = 116.54 N 2)frictional force the lower crate exerts on the upper crate = 16 * 3.938 =70.88 N answer since frictional force the lower crate exerts on the upper crate