Acceleration of two masses on a pulley

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SUMMARY

The acceleration of the block on the left in a pulley system involving two masses, M and m, is derived using Newton's second law, F=ma. The correct formula for acceleration is a = mg / (m + 2M), which accounts for the tension in the string. The initial attempt incorrectly simplified the equation, neglecting the effect of tension on the system. The final equations used to derive the acceleration are (M + m)g - T = (M + m)a and T - mg = ma.

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  • Understanding of Newton's laws of motion
  • Basic knowledge of free body diagrams
  • Familiarity with tension in pulley systems
  • Ability to solve algebraic equations
NEXT STEPS
  • Study the concept of tension in pulley systems
  • Learn how to draw and analyze free body diagrams
  • Explore advanced applications of Newton's laws in multi-mass systems
  • Investigate the effects of friction in pulley systems
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Physics students, educators, and anyone interested in classical mechanics, particularly in understanding dynamics involving pulleys and multiple masses.

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Homework Statement


Find the acceleration of the block on the left in terms of M, m and g.
See attachment

Homework Equations


F=ma


The Attempt at a Solution


From the free body diagram,
-Mg+Mg+mg=(m+M)*a
a=m*g/(m+M)

I think this is correct, however the solution states the correct answer as
a=m*g/(m+2M), indicating a problem with the signs.
 

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  • pulleys.jpg
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While finding the acceleration of the masses, you have to take into account the tension in the string.

(M +m)g - T = (M+m)a...(1)

T - mg = ma. ...(2)

Now solve for a.
 

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