Two wooden crates rest on top of one another. The smaller top crate has a mass o
ID: 2186883 • Letter: T
Question
Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24.0 kg and the larger bottom crate has a mass of m2 = 86.0 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is ?s = 0.82 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 = 416.0 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate?Explanation / Answer
The big trick in this sort of problem is keeping track of what is moving, respect to what else, and by how much. 1) We are given the tension (= force) exerted by the rope. We are given the masses of each crate -- 19 kg and 89 kg. The total mass of the two crates is 108 kg. We know there's no friction between the floor and the bottom crate, so there's no resisting force to the tension of the rope. We are also given that the top crate is not sliding, so the two crates are accelerating as a unit under the influence of the rope. We can regard the tension from the rope as acting on the combined mass of the crates. F = ma 356N = 108kg * a 396/108 m/s^2 = 3.296 m/s^2 = a (answer) 2) Since the upper crate is not moving *relative*to*the*lower*crate*, we know the force applied is exactly enough to accelerate it at the same rate as the two crate system we looked at in problem 1. F = ma = 19kg * 3.296 m/s^2 = 62.6N. (answer) We know the force can't be any more, or the small crate would be pulling ahead of the large one. And we know it can't be any less, or the small crate would fall behind. 3) This takes a little actual work. µs = 0.82. The amount of force parallel to the floor that it takes to cause the top crate to slide is equal to the normal force exerted by the top crate, times µs. The normal force is mg = 19kg * 9.8 m/s^2 = 186.2N. Multiply by µs and we get F = 186.2 * 0.82 = 152.7N The rope pulls on the lower crate only. The lower crate transfers force to the upper crate by friction. As long as the force being transmitted is less than 152.7N, the upper crate won't slip. This force corresponds to an acceleration of a = F/m = 152.7 / 19 = 8.037 m/s^2. When the acceleration exceeds this amount, the upper crate will slip. The amount of tension needed to accelerate the whole system at that rate is: F = ma = 108 * 8.037 = 868N. (answer)
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