Three-pulley system and coefficients of friction

[andyfeynman]

Homework Statement

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

Homework Equations

F_max = μ_sN

The Attempt at a Solution

First, I set up a coordinate system with m at (x_m, 0), 3m at (x_3m, 0) and the middle pulley at (0, y). (Note: downward positive)
I also took the length of the string x_m - x_3m + 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 μ_smg and 3μ_smg. After solving a bunch of linear equations, I find T = 0.5(μ_k + 2)mg. I found that for μ_s = 1.1, the corresponding value of μ_k must be between 0.2 and 1.1; for μ_s = 0.3, the value of μ_k would be negative. Therefore, I conclude that (1) is the only possible case for μ_s = 1.1.

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

 

[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).

 

[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.

 

[andyfeynman]

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.

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.

 

[haruspex]

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.

Yes, I agree with that, but i did not understand why you wrote

but 3m moves at uniform velocity

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.

 

[andyfeynman]

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.

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?

 

[haruspex]

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?

Yes.

 

[andyfeynman]

Yes.

Thank you so much