# Kleppner/Kolenkow Pulley Example

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1. Feb 5, 2016

### Cosmophile

1. The problem statement, all variables and given/known data

Hey, all. I am working through Kleppner/Kolenkow and encountered a problem when trying to follow the example pictured below. My issues comes when they say "Equations (1)-(3) are easily solved..."

As it turns out, they are not so easily solved for me! So, it appears I've found a hole in my mathematical training that needs to be filled (K&K have proved to be good at exposing these weak spots).

2. Relevant equations
All pictured above.

3. The attempt at a solution
Frankly, I'm not sure where to begin. I've tried expanding the $\ddot{y_1}$ and $\ddot{y_2}$ terms using $(3)$, but the equations quickly become quite long. I haven't had the time to really sit down and see if they eventually clean up nicely and give the desired result, but truth be told, I don't feel too confident going into this and would love some help.

2. Feb 5, 2016

### BvU

Build confidence by starting off with A = 0 !
Take your time to really sit down and try another tack if things still become too complicated.

3. Feb 5, 2016

### SammyS

Staff Emeritus
First of all, you may have noticed that the xp in the figure should have been yp .

Solve Eq.(3) for either $\ \ddot y_1\$ or $\ \ddot y_2\ .\$ Plug the result into Eq (1) or (2).

etc.

4. Feb 5, 2016

### Cosmophile

Update: Got it! I can type up my solution if you guys would like to see it.

5. Feb 5, 2016

### SammyS

Staff Emeritus
Sure. Why not?

6. Feb 5, 2016

### Cosmophile

I actually solved it right before you posted your advice (I solved it in the same way):

I solved Eq. (3) for $\ddot {y_2}$ and plugged that solution into Eq. (2). This gave $$T = W_2 + M_2 (2A - \ddot{y_1})$$
I then set the right-hand sides of Eq. (1) and Eq. (2) together: $$W_1 + M_1 \ddot {y_1} = W_2 + M_2 \ddot {y_2}$$
From here, it was just a matter for breaking down the $W$ terms into their corresponding $M_ig$ terms and rearranging:

$$M_1g+M_1 \ddot{y_1} = M_2g + 2A M_2 - M_2 \ddot {y_1}$$
$$\ddot{y_1} =(2A+g) \frac {M_2 - M_1g}{M_1 + M_2}$$

To solve for $T$, I simply plugged this result into Eq. (1), and the desired result followed immediately.