Two wooden crates rest on top of one another. The smaller top crate has a mass o
ID: 2060930 • Letter: T
Question
Two wooden crates rest on top of one another. The smaller top crate has a mass of m1=20kg and the larger bottom crate has a mass of m2=88kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crate is museS=0.78 and the coefficient of kinetic friction between the two crate is museK=0.61. 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). What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide. The tension in the rope to 1157N causing the boxes to accelerate faster and the top box to begin sliding. what is the acceleration of the upper crate and as the upper crate slides, what is the acceleration of the lower crate?Explanation / Answer
1)As the force is small enough the small krate doesnt slide on bigger crate so the two crates move with same accelaration
a = F/(M+m) (as the crate is moving as if they were a single crate)
a = 288/117 =2.461mts/sec^2
2)As the upper crate is moving only due to the friction force acting on it due to contact with the lower crate so friction force is equal to m*a
Fs = 23*2.461 = 56.615N
3)the maximum force is dependent on max accilaration the smaller crate can move
its calculated by a = g
a = 0.76*10 = 7.6mts/sec^2
as the crate is moving without the movement of upper crate as a single crate
F = (M+m)a = (117*7.6) = 889.2 N
4) as the crate is sliding on the lower block the friction is kinetic
the accilaration is equal to k*g = 0.6*10 = 6mts/sec^2
5 ) as the upper crate is sliding and force is acting on the opposite direction
force acting on lowwer crate is F- Fk Ma
=> 1213 - 23*6 = 94*a
acclelaration of lower block is 12.653 mts/sec^2
we can consider any direction as postive its doesnt change the answers
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