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

Question

Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 20 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 s = 0.77 and the coefficient of kinetic friction between the two crates is k = 0.62. 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 = 383 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate?

___________m/s2

2)In the previous situation, what is the frictional force the lower crate exerts on the upper crate?

_____________N

3)What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide?

_______________N

4)The tension is increased in the rope to 1175 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?

________________m/s2

5)As the upper crate slides, what is the acceleration of the lower crate?

__________________m/s2

Explanation / Answer

  1) F=ma
so
383=(20+88)a
so
acceleration = 383/108 = 3.546m/s²

Note: Since the the small crate will not slide(as per the question), they will both have the same acceleration, that is also why I used the mass of both crates in my calculation. I used the mass of both crates.


2) F=ma
so
F=20*3.546
so
F=7.92 newtons

Note: Since the upper crate is not sliding, we know its acceleration, and since its mass was given, the sum of all forces acting on it is easy to calculate. Since the only force that is mentioned as acting on it is friction, friction accounts for the entire 91.66 newtons.


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*9.81 = 20*9.81 = 196.2(assuming this is on earth).
Therefore the maximum force that can be exerted by static friction is:
F = 0.77*196.2 = 151.074 newtons

Acceleration max. for upper crate = (F/M), =151.074/20, = 7.553m/sec^2.
(88 + 20) = 108kg.
In order for it have that acceleration, the lower crate would also need to have that acceleration.
In order for the whole 108kg (both crates) to have that acceleration, 815.724 newtons have to be acting on them(by F=ma).


4)Since k = 0.62, the maximum force that can be exerted by friction is given by F=k = 0.62*N, where N is the same normal force we had before.
Weight of top crate = 245N. Friction when it is sliding = (245 x 0.62), = 151.9N.
The top crate will slide when the acceleration exceeds 7.595m/sec^2., after which the force levels at 169.05N. The acceleration then will be (F/M) = 169.05/25, = 7.595m/sec^2.

5) The total friction when the upper crate slides is 151.9N.
(1175 - 151.9) = 1023.1N. net accelerating force on the LOWER crate only.
Acceleration = (F/M), = 1023.1/88, = 11.626m/sec^2.

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