Two wooden crates rest on top of one another. The smaller top crate has a mass o
ID: 1282756 • Letter: T
Question
Two wooden crates rest on top of one another. The smaller top crate has a mass of m 15 kg and the larger bottom crate has a mass of m2 88 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is us 0.85 and the coefficient of kinetic friction between the two crates is uk 0.66. 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 249 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate? m/s Submit 2) In the previous situation, what is the frictional force the lower crate exerts on the upper crate? N Submit 3) What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide? N SubmitExplanation / Answer
1) Acceleration = (F/M) = 249/(88 + 15), = 2.417 m/s2.
2) F=ma
so
F=15*2.417
so
F=36.255 N
3) The maximum force that can be exerted by static friction is given by:
F=?s*N, where N is the normal force.
The normal force is given by N = m*9.8 = 15*9.8 = 147(assuming this is on earth).
Therefore the maximum force that can be exerted by static friction is:
F = 0.85*147 = 124.95 N
4)Since ?k = 0.66, the maximum force that can be exerted by friction is given by F=?k = 0.66*N, where N is the same normal force we had before.
So the force is 97.02 newtons, therefore the acceleration is 6.468 m/s
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