Pulley System- equation derivation

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Homework Help Overview

The discussion revolves around deriving a formula for the acceleration of a pulley system that includes a cord with non-negligible mass. The problem involves analyzing forces acting on the masses and the cord's contribution to the system's dynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law and the role of tension in the system. There are attempts to derive the acceleration formula based on forces acting on the masses and the cord. Questions arise regarding the correctness of the initial assumptions and the need for a free-body diagram.

Discussion Status

Some participants are questioning the validity of the original poster's approach and whether the forces considered were appropriate. There is an indication that further clarification on the definitions of forces is needed, and the discussion is focused on refining the understanding of the problem rather than reaching a consensus.

Contextual Notes

There is a mention of needing to specify the forces involved in the equations, which suggests that assumptions about the system's setup may be under scrutiny. The original poster expresses uncertainty about the correctness of their derived equation.

hanlon
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Homework Statement



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

Homework Equations



F = ma

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:
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hanlon said:
3. The Attempt at a Solution [/b]

Fby = Fax

Fby = mB *g

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

.

This is not right. Draw the free-body diagram, showing the forces both at A and B.
 
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
 
If you do not specify what Fbx and Fax are the equation Fbx=Fax has no sense.

ehild
 

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