# Pulley System- equation derivation

1. May 26, 2010

### hanlon

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

Determine a formula for acceleration of the system shown in Fig. 4-45 if the cord has a non-negligible mass mc. Specify in terms of lA and lB, the length of cord from the respective masses to the pulley ( The total cord length is l= lA + lB)

[PLAIN]http://img706.imageshack.us/img706/8575/3333v.png [Broken]

2. Relevant equations

F = ma

3. The attempt at a solution

Fby = Fax

Fby = mB *g

acceleration of system: a = (mB *g)/ma

acceleration of system with non-negligible cord

a = (mB + (mC * (lB/ l)))*g / (ma + (mC * (lA/ l)))

need an answer check, I can't tell if its right.

Last edited by a moderator: May 4, 2017
2. May 27, 2010

### ehild

This is not right. Draw the free-body diagram, showing the forces both at A and B.

3. May 27, 2010

### hanlon

Is my whole solution wrong, or just the forces I used.

I understand now that I didn't use tension force, but is the way I derived the equation wrong with

Fby = Fax

4. May 27, 2010

### ehild

If you do not specify what Fbx and Fax are the equation Fbx=Fax has no sense.

ehild