# Kleppner - Mass and Pulleys

1. Mar 17, 2014

### jbunniii

1. The problem statement, all variables and given/known data
This is problem 2.9 from Kleppner and Kolenkow, 2nd edition. I think I got it right, just want to check since I'm self-studying.

Masses on table
Two masses, $A$ and $B$, lie on a frictionless table, as shown (see my crudely drawn figure in the thumbnail :tongue:). They are attached to either end of a light rope of length $l$ which passes around a pulley of negligible mass. The pulley is attached to a rope connected to a hanging mass, $C$. Find the acceleration of each mass.

2. Relevant equations
$F = ma$

3. The attempt at a solution
First, I chose the coordinates as follows:
* $x$ points horizontally, away from and perpendicular to the table edge
* $y$ points downward vertically

Define the positions of the objects:
* $x_A$ = horizontal position of mass $A$
* $x_B$ = horizontal position of mass $B$
* $x_p$ = horizontal position of movable pulley (the one to the left in the figure)
* $y_C$ = vertical position of mass $C$
* Note that I didn't draw it very well, but the pulley to the right in the figure is fixed: it is attached rigidly to the table.

I obtained the following constraints due to the rope connections:
$$2 \ddot x_p = \ddot x_A + \ddot x_B$$
$$\ddot x_p + \ddot y_C = 0$$
Combining these, we get
$$-2 \ddot y_C = \ddot x_A + \ddot x_B$$

I then considered the forces on each mass, and on the movable pulley. Masses $A$ and $B$ each have tension $T_1$ pointing to the right (negative $x$) due to the rope. Thus,
$$T_1 = -M_B \ddot x_B = -M_A \ddot x_A$$
Mass $C$ is acted upon by gravity downward and by tension from the second rope upward. This gives us
$$M_C g - T_2 = M_C \ddot y_C$$
The two tensions are related by the movable pulley: tension $T_2$ pulls to the right, and $2T_1$ pulls to the left. Since the pulley is massless, this gives us $T_2 = 2T_1$.

Thus we have five equations and five unknowns. I won't go through all the algebra as it's rather ugly but routine. My answer, in case anyone has done this problem and has their solutions available to verify, is:
$$\ddot x_A = \frac{-2 M_C M_B g}{M_C M_B + M_C M_A + 4 M_B M_A}$$
$$\ddot x_B = \frac{M_A}{M_B} \ddot x_A = \frac{-2 M_C M_A g}{M_C M_B + M_C M_A + 4 M_B M_A}$$
$$\ddot y_C = -\frac{1}{2}(\ddot x_A + \ddot x_B) = \frac{(M_C M_A + M_C M_B)g}{M_C M_B + M_C M_A + 4 M_B M_A}$$

I checked several special cases:

Case 1 : $M_C = 0$. In this case, all the accelerations are zero, as expected since nothing is pulling the system.

Case 2 : $M_A = 0$. In this case, $\ddot x_B = 0$ (intuitively expected because the "path of least resistance" is to pull massless mass $A$ and leave mass $B$ alone), $\ddot x_C = (M_C M_B g)/(M_C M_B) = g$ (reasonable because $C$ is able to free-fall), and $\ddot x_A = -2 (M_C M_B g) / (M_C M_B) = -2g$ (reasonable since $C$ is free-falling and the movable pulley doubles the motion of $A$).

Case 3 $M_A = M_B = M_C$. Here the expressions reduce to $\ddot x_A = \ddot x_B = -g/3$ and $\ddot x_C = g/3$. I guess this makes sense because gravity's force gets distributed across the three masses equally, so the acceleration of each one is one third of $g$. I think.

Case 4 $M_A = M_B = 0$. Here the expressions are invalid because the denominators are zero. But if we step back to the earlier equations, we get: $T_1 = 0$, $M_c g = M_C \ddot y_C$, hence $\ddot y_C = g$ as expected since $C$ can free-fall. Also, $\ddot x_A + \ddot x_B = -2g$ and $\ddot x_A = \ddot x_B$, which forces $\ddot x_A = \ddot x_B = -g$, as expected since they simply follow $C$ down.

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

### TSny

:thumbs: It all looks good to me.

3. Mar 17, 2014

### jbunniii

Great, thanks!