# Homework Help: 2 masses, 2 pulleys, 2 ropes

1. Mar 17, 2014

### jbunniii

1. The problem statement, all variables and given/known data
I am teaching myself mechanics using Kleppner & Kolenkow, second edition. My background is in math, so I don't need assistance with that aspect, but my physical intuition is weak/nonexistent. :tongue2: This is problem 2.8.

Masses $M_1$ and $M_2$ are connected to a system of strings and pulleys as shown in the figure attached. The strings are massless and inextensible, and the pulleys are massless and frictionless. Find the acceleration of $M_1$.

http://postimg.org/image/hogs0e151/

2. Relevant equations

$F = ma$

3. The attempt at a solution

The hint given is that if $M_1 = M_2$, then the acceleration should be $g/5$. My answer is consistent with that hint, but I want to make sure it's right.

Referring to the attached image, let us call $L_1$ the length of the rope attached to $M_1$, and let us call $L_2$ the length of the rope attached to $M_2$. Assume also that the radii of the pulleys are $R_1$ and $R_2$, respectively. Then:
$$L_1 = (h_1 - y_1) + \pi R_1 + (h_1 - h_2)$$
$$L_2 = h_2 + \pi R_2 + h_2 - y_2$$
Differentiating each of these twice, and recognizing that $L_1$, $L_2$, $R_1$, $R_2$, and $h_1$ are constants, we obtain $\ddot h_2 = -\ddot y_1$ and $\ddot y_2 = 2\ddot h_2$. Combining these, we see that $\ddot y_2 = -2 \ddot y_1$.

Next, we consider the forces on the two masses. The mass $M_1$ is acted upon by tension $T_1$ pointing upward, and $W_1 = M_1 g$ pointing downward. Therefore,
$$T_1 - M_1 g = M_1 \ddot y_1$$
The mass $M_2$ is acted upon by tension $T_2$ pointing upward, and $W_2 = M_2 g$ pointing downward. Therefore,
$$T_2 - M_2 g = M_2 \ddot y_2 = -2M_2 \ddot y_1$$
where the last equality follows from the constraint we established above.

We now have two equations and three unknowns. We need to eliminate either $T_1$ or $T_2$. Here is where my confidence gets a bit shaky. I considered the middle pulley and argued as follows.

The pulley is acted upon by three forces: $T_1$ upward - clear enough because the first rope is attached to the pulley. Now consider the downward forces. Although the second rope is not actually attached to the pulley, it is certainly pulling downward on the pulley. The upward tension on $M_2$ is $T_2$. Since the other end of the rope is attached to the floor, there is also an upward force on the floor of $T_2$ at the connection point. There must be downward forces associated with both of these upward forces, by Newton's second law. The downward forces are acting on the pulley (what else can they act on?) Therefore the pulley experiences downward forces of $2T_2$ in addition to the upward force of $T_1$, for a net upward force of $T_1 - 2T_2$. If this argument is indeed correct, I would love to know how to make it clearer and more rigorous.

Assuming the above is correct, we have
$$T_1 - 2T_2 = M_{\text{pulley}} \ddot h_2 = 0$$
since the pulley is massless. Therefore, $T_1 = 2T_2$. Substituting this into $T_1 - M_1 g = M_1 \ddot y_1$, we obtain
$$2T_2 = M_1 g + M_1 \ddot y_1$$
Substituting our other force equation, $T_2 = M_2 g - 2M_2 \ddot y_1$, we end up with
$$2(M_2 g - 2M_2 \ddot y_1) = M_1 g + M_1 \ddot y_1$$
and assuming I did the algebra right, solving for $\ddot y_1$ yields
$$\ddot y_1 = g \left(\frac{2M_2 - M_1}{M_1 + 4M_2}\right)$$
We can verify that this matches the hint by observing that $M_1 = M_2$ indeed gives us $\ddot y_1 = g/5$.

#### Attached Files:

• ###### pulleys.png
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Last edited: Mar 17, 2014
2. Mar 17, 2014

### jbunniii

Made a few edits to fix typos, sorry for any inconvenience.

3. Mar 17, 2014

### jbunniii

It is interesting to consider two special cases.

Case 1: $M_2 = 0$. In this case, mass $M_1$ is the only object with any mass, so it is able to free-fall with gravity. We expect $\ddot y_1 = -g$ in this case. Indeed, the general solution reduces to $\ddot y_1 = g(-M_1 / M_1) = g$.

Case 2: $M_1 = 0$. In this case, mass $M_2$ is the only object with any mass, so it is able to free-fall with gravity. Thus $\ddot y_2 = -g$. We earlier found that $\ddot y_2 = -2 \ddot y_1$, so in this case, $\ddot y_1 = g/2$. This is once again consistent with the general solution: $\ddot y_1 = g(2M_2 / 4M_2) = g/2$.

However, in Case 2 it makes me a bit nervous that if the rightmost pulley had even a tiny bit of mass, it would free-fall with gravity instead of being pulled by $M_2$ at only half the acceleration of gravity ($\ddot h_2 = \ddot y_2 / 2$). But then $y_1$ would rise with an acceleration of $g$ instead of $g/2$. This kind of "discontinuity" makes me uneasy. But I guess it's no worse than the discontinuity between some massless object not falling at all versus something not-quite-massless falling with an acceleration of $g$...

Last edited: Mar 17, 2014
4. Mar 17, 2014

### ehild

Your derivation is correct, good job!

That is correct. Actually, the rope around the pulley acts at both ends of each line element of the circumference of the pulley, along the tangents of that piece. The resultant force exerted to every line element points to the centre. The sum of all these elementary forces is vertical and equal to $T_2$ when both the pulley and the rope are massless.

ehild

#### Attached Files:

• ###### pulleyrope.JPG
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5. Mar 17, 2014

### Intrastellar

The detailed proof is available at the next chapter of the book jbunniii, you can check it out if you want

6. Mar 17, 2014

### ehild

If the pulleys have mass, the problem is quite different. You can not assume that the tension is the same all along a rope, and you have to take the rotational inertia of the pulleys into account. You will get different result for the accelerations, and taking the limit for zero masses you should get the same result as you had now.

ehild

7. Mar 17, 2014

### jbunniii

Thanks guys, good info. I figured that there must be a way to find the force on the pulley by adding up (integrating) the contributions of small bits of the rope, but I couldn't see how to determine the the contribution of each bit, in such a way that the total adds up to $2T$ regardless of the radius of the pulley. Maybe the authors will get to this level of detail later in the book.

@montadhar - I didn't realize they worked this problem out in the next chapter. Will check it out when I get there. I'll try to do the rest of chapter 2 problems first. You will probably hear from me again.

8. Mar 17, 2014

### Intrastellar

Kleppner is an amazing text, enjoy working through it