# Box Problem Kinetic/Static Friction

Here's the question: Two boxes are placed on top of one another. The top box has a weight of 20 N and the bottom box has a weight of 50 N. You can neglect friction between the bottom box and the floor, but uS = 0.5 and uK = 0.4. You can assume g = 10m/s^2. Both boxes are initially at rest.

a) you tie a string to the top box and exert a force F to the right so both boxes accelerate together to the right with the same acceleration. What is the maximum value of F for which the boxes have the same acceleration?

I did F = uS(m1 * g)
F = 0.5 (20N * 10m/s^2) = 100N

Is this right??

b) what if you pull witha force twice as large as the maximum value you calculated? Determine the acceleration of the box now. You should get two different values.

I did: 20 N = m1*a - uK(m1*g) and got 24 m/s^2
and m2*a = -uK(m1*g) and got -1.6 m/s^2

## Answers and Replies

collinsmark
Homework Helper
Gold Member
Here's the question: Two boxes are placed on top of one another. The top box has a weight of 20 N and the bottom box has a weight of 50 N. You can neglect friction between the bottom box and the floor, but uS = 0.5 and uK = 0.4. You can assume g = 10m/s^2. Both boxes are initially at rest.

a) you tie a string to the top box and exert a force F to the right so both boxes accelerate together to the right with the same acceleration. What is the maximum value of F for which the boxes have the same acceleration?

I did F = uS(m1 * g)
F = 0.5 (20N * 10m/s^2) = 100N

Is this right??
Not quite. At least not the way the problem statement is written as it is now.

Hint: Does the problem statement give the weight in Newtons or the mass in Kilograms?
b) what if you pull witha force twice as large as the maximum value you calculated? Determine the acceleration of the box now. You should get two different values.

I did: 20 N = m1*a - uK(m1*g) and got 24 m/s^2
Okay, here you're back to 20 N (instead of 200 N) for the twice the force of the first part. But re-evaluate the mass of m1. Is it 20 N or 20 kg?
and m2*a = -uK(m1*g) and got -1.6 m/s^2
You might want to redo the above too, as long as your at it. But there's one more hint I'd like to give first. This part is special, and I think you might be missing an important idea.

Before, when you worked out the part for m1, you correctly made sure the direction of the frictional force was such to oppose the direction of the relative sliding. In other words, in this particular case of m1, the frictional force is in the opposite direction as the string's pulling force. [You are correct on the frictional force's direction for that part ]

But in the case of m2, what is the direction frictional force (on m2) such that it opposes the relative sliding? (By relative sliding, think of the direction of m2 relative to m1. What is the direction of the frictional force as seen by m2?)

There is yet another way to look at the same thing. As the top block m1 slides across the bottom block m2, m2 exerts a horizontal frictional force on m1 (which opposes the direction of the string's force). What does Newton's third law of motion have to say about this horizontal force (on m2)?