# 2 masses, 2 pulleys, 2 ropes

**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##.

**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##.

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