Homework Help: Three-pulley system and coefficients of friction

1. Sep 16, 2015

andyfeynman

1. The problem statement, all variables and given/known data
In the machine shown in the attached picture, all pulleys have negligible mass and rotational inertia. The coefficient of static friction μs between the table and either mass is larger than the coefficient of kinetic friction μk. The two masses 3m and m originally were held stationary and then released. After release, find the tension of the string in terms of μk, m and acceleration due to gravity g for the following two cases:
(a) μs = 1.1
(b) μs = 0.3

2. Relevant equations
Fmax = μsN

3. The attempt at a solution
First, I set up a coordinate system with m at (xm, 0), 3m at (x3m, 0) and the middle pulley at (0, y). (Note: downward positive)
I also took the length of the string xm - x3m + 2y = constant as a constraint.

Okay, my approach to this problem goes like this:
I consider two possible motion:
(1) only m accelerates but 3m moves at uniform velocity,
(2) both m and 3m accelerate.

For case (1), I assume that for this to happen the tension must lie between μsmg and 3μsmg. After solving a bunch of linear equations, I find T = 0.5(μk + 2)mg. I found that for μs = 1.1, the corresponding value of μk must be between 0.2 and 1.1; for μs = 0.3, the value of μk would be negative. Therefore, I conclude that (1) is the only possible case for μs = 1.1.

Similarly, for case (2) the tension must be larger than 3μsmg. I find T = 6/7(μk + 1)mg, and for μs = 0.3, the value of μk should be greater 0.05; for μs = 1.1, this value would be greater than 2.85, which is impossible (since the question states that μs > μk). Therefore, I conclude that (2) is the case for μs = 0.3.

The problem is that my approach looks terribly cumbersome, and I'm not even sure whether it is right. (Are there really only two possible cases which depend on the value of μs?)

Attached Files:

• test.png
File size:
8.9 KB
Views:
101
2. Sep 16, 2015

andyfeynman

Wow! I just notice how similar this problem is to Kleppner & Kolenkow 2.15.
At least now I can be sure T is indeed 6/7(μk + 1)mg for case (2).

3. Sep 16, 2015

haruspex

For (1), why do you think 3m moves at all? Since kinetic friction is less than or equal to static friction, if one of the masses slides at all it will then accelerate.
I was not able to follow your reasoning, but I agree with the answer. Maybe it would be clearer if I saw all your working.
To answer your question, I would start by treating the general case: allow for each of the two side masses to have an acceleration (possibly 0), and assign a friction coefficient to the 3 mass of $\mu_3$, which may later be either static or kinetic. Having established the generic equations and some inequalities, you can get into individual cases as necessary.

4. Sep 16, 2015

andyfeynman

This is what I thought:
If the tension is larger than the maximum static friction of m but not of 3m, then only m would accelerate but 3m would stay unaccelerated.

Last edited: Sep 16, 2015
5. Sep 16, 2015

haruspex

Yes, I agree with that, but i did not understand why you wrote
If it moves at all then it will be subject to kinetic friction, not static, and it will accelerate. It cannot move at a nonzero uniform velocity.

6. Sep 16, 2015

andyfeynman

Yes, I forgot that anything moving should be subject to kinetic friction, not static.
So it should be m accelerates while 3m stays at rest?

7. Sep 16, 2015

haruspex

Yes.

8. Sep 16, 2015

andyfeynman

Thank you so much