Acceleration of masses in a pulley system

  • Thread starter jisbon
  • Start date
  • #1
467
30
Homework Statement:
2 masses on each side of the inclined plane. Coefficient of kinetic friction = 0.3, held by an inextensible string.
Masses of blocks as labelled on the diagram
Relevant Equations:
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1573733245458.png


So I figured out the equation, but it is probably wrong because the answer doesn't tally.
Since the string is inextensible, I can assume that tension is the same for both sides, and acceleration for both masses is the same too So:
I can say that the acceleration of 2kg block =acceleration of 7kg block

$$\dfrac {7g\sin 40-T}{2+7}=\dfrac {T+\left( 0.3\right) \left( 2g\cos 50\right) -2g\sin 50}{7+2} $$

I got T = 29.36N, and used one of the equation to get my acceleration. However, it seems to be wrong :/

Is there something wrong with my equation? Thanks
 

Answers and Replies

  • #2
21,488
4,864
The denominator of the first terms should be 7 and the denominator of the 2nd term should be 2, and the sign of the friction term should be negative.
 
  • #3
467
30
The denominator of the first terms should be 7 and the denominator of the 2nd term should be 2, and the sign of the friction term should be negative.
Mistaken it for static friction and the massess too. Thanks for the help
 

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