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.