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western50 · Feb 14, 2011, 08:12 PM

Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24 kg and the larger bottom crate has a mass of m2 = 91 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.79 and the coefficient of kinetic friction between the two crates is μk = 0.63. 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 = 285 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate? 2.478m/s^2

2. In the previous situation, what is the frictional force the lower crate exerts on the upper crate? 59.478N

3. What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide? 891.2385N

4. The tension is increased in the rope to 1319 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?
how can I get this one?

jcaron2 · Feb 14, 2011, 08:31 PM

What's the force of the kinetic friction between the two boxes? Since that's the only horizontal force acting on the upper box, once you know it you should be able to easily calculate its acceleration using Newton's second law.

And by the way, I agree with your answers for the first three questions. :)

western50 · Feb 14, 2011, 08:59 PM

The kinetic friction is 0.63 between the blocks, but would you show me in detail on how to get the answer?

western50 · Feb 14, 2011, 09:03 PM

The kinetic friction is 0.63 between the blocks, but would you show me in detail on how to get the answer?

jcaron2 · Feb 14, 2011, 10:19 PM

That's the coefficient of kinetic friction. The actual force is that number times the normal (downward) force due to gravity.

$F_{friction} = k_{kinetic} \cdot F_{normal} = k_{kinetic} \cdot m \cdot g = 0.63\; \cdot\; 24 \;kg \;\cdot \;9.81 \;\frac{m}{s^2} = ???\;N$

And again, once you know the frictional force, you can compute the acceleration using Newton's second law, F=ma.

Can you take it from there?

western50 · Feb 14, 2011, 11:29 PM

But how can I take the 1319 N into account?

ebaines · Feb 15, 2011, 08:49 AM

F=ma. The 1319 N pulling force must equal the sum of the mass times acceleartion for each box:

$<br /> 1319N = m_1 a_1 + m_2 a_2<br />$

But you really don't need this. The acceleration of the upper crate is determined by the force of friction between it and the lower crate:

$<br /> \mu m_1 g = m_1 a_1<br />$

jcaron2 · Feb 15, 2011, 09:03 AM

Exactly as ebaines said. It's irrelevant whether it was 1319 N or 1000000 N. Either way, the force was great enough that the upper crate slips, so it's only being accelerated by the frictional force.

If the question was asking you about the acceleration of the LOWER box, then the 1319 N would be needed, but not for the upper box.

illinihawks · Feb 22, 2011, 09:10 PM

How would we use the data from the question and part 4 to find the acceleration for the lower box given a tension of 1319 N that will cause the upper crate to slip?

Thanks!

illinihawks · Feb 22, 2011, 09:30 PM

Nm just realized all I had to was ebaines said and set T = m1a1 + m2a2. Thanks!