Masses on incline plane and pulley.

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SUMMARY

The discussion focuses on calculating the acceleration of a three-block system involving a 2.87 kg mass moving upward, a 1.5 kg mass sliding down a ramp, and an 8.4 kg mass moving downward. The solution utilizes Newton's second law (F=ma) and incorporates friction, with a coefficient of friction of 0.14. The final calculated acceleration is 3.43 m/s², confirming the correctness of the approach after considering all forces, including the parallel force to the driving force.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of friction calculations (Friction=μN)
  • Familiarity with forces on an incline (F parallel: mgsinφ)
  • Basic trigonometry for resolving forces (F perpendicular: mgcosineφ)
NEXT STEPS
  • Study advanced applications of Newton's laws in multi-body systems
  • Learn about friction coefficients and their impact on motion
  • Explore dynamics of pulleys in mechanical systems
  • Investigate the effects of varying angles on incline planes
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and tutors looking to enhance their understanding of forces in multi-body systems.

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Homework Statement



A suspended 2.87 kg mass is moving up, a 1.5 kg mass slides down the ramp, a suspended 8.4 kg mass on the left is moving down. There is friction between the block and the ramp. The acceleration of gravity is 9.8 m/s2 . The pulleys are massless and frictionless.

Find the acceleration of the 3 block system.
Here is a diagram

http://img8.yfrog.com/i/capturexki.jpg/


Homework Equations



F=ma
Friction=μN
F parallel: mgsinφ
F perpendicular: mgcosineφ

The Attempt at a Solution



Sumation equation I set up (mass 1= 8.4kg, mass 2=1.5kg, mass 3=2.87kg mt=total mass)

m1g-m2g(sinφ)-friction-m3g=mtg

after the numbers are plugged in:

(8.4)(9.8)-(1.5)(9.8)sin36-(0.14)(1.5)(9.8)cos36-2.87(9.8)=12.77a

acceleration would equal=3.43 m/s2

is this correct?
 
Last edited by a moderator:
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i think your attempt is exactly what the question required
 
I figured it out...

I was forgetting to add the parallel force to the driving force.

All is well now, thanks!
 

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