Acceleration of a system with three blocks

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duplaimp
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Homework Statement


Physics diagram of a pulley system: a 6 kg block on a 30° incline, connected over a pulley at the top to a 4 kg block on a horizontal surface, and a hanging 9 kg block on the other side.


What is the acceleration of this system?

Homework Equations


F=ma


The Attempt at a Solution


I tried the following:

Let blocks with mass 9kg, 4kg and 6kg be m9, m4, m6
This is the free body diagrams I drawn:

Three free-body diagrams labelled m9, m4 and m6, with red arrows for tension TR, TL, weight w and normal force, drawn on blocks on inclined and vertical surfaces.


Then I applied Newton's 2nd Law:
F = ma

(For m9) mg-TR=9a
(For m4) TR-TL=4a
(For m6) TL-w*sin(30)=6a

I solved for a and it didn't gave the right answer.
What I am doing wrong?
 
duplaimp said:
(For m9) mg-TR=9a
Good.

(For m4) TR-TL=4a
What about gravity?

(For m6) TL-w*sin(30)=6a
Good.

Gravity applies to m4 just like it does to m6.
 
Doc Al said:
Gravity applies to m4 just like it does to m6.

I didn't included that because as m4 is "in the middle" I thought it wasn't necessary but it seems I was wrong. Now it gives the right acceleration.
Thank you! :biggrin: