Force: two pulley problem with 3 masses

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SUMMARY

The discussion focuses on solving a two-pulley problem involving three masses: a 0.840 kg glider and two hanging masses of 4.85 kg and 3.62 kg. The equations of motion were derived using Newton's second law, leading to the calculation of acceleration and tensions in the strings. The correct acceleration was determined to be 1.30 m/s², with tensions T1 and T2 calculated as 41.3 N and 40.2 N, respectively. The initial calculations contained errors, which were identified and corrected through the discussion.

PREREQUISITES
  • Understanding of Newton's second law (ΣF = ma)
  • Knowledge of tension in strings and forces acting on masses
  • Familiarity with solving systems of equations
  • Basic concepts of pulley systems in physics
NEXT STEPS
  • Review the principles of Newton's laws of motion
  • Study the mechanics of pulley systems in detail
  • Practice solving multi-mass problems with frictionless pulleys
  • Learn to derive equations of motion for connected objects
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in mastering mechanics, particularly in understanding forces and motion in systems involving pulleys and multiple masses.

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Homework Statement


A 0.840- kg glider on a level air track is joined by strings to two hanging masses. As seen in the figure, the mass on the left is 4.85 kg and the one on the right is 3.62 kg The strings have negligible mass and pass over light, frictionless pulleys. Find the acceleration of the masses and T1 and T2.

Problem and figure


Homework Equations



I assume all a's are equal because a of each object has to be equal to the a of the glider.

[itex]\Sigma[/itex]F =ma

The Attempt at a Solution


[itex]\Sigma[/itex]Fobject1=T1-W
m*-a=T1-(4.85*9.81)
T1= -4.85a+4.85*9.81 (equation 1)


[itex]\Sigma[/itex]Fobject2=T2-W
3.62a=T2-(3.62*9.81)
T2=3.62a+(3.62*9.81) (equation 2)


[itex]\Sigma[/itex]Fobject3=T2-T1
-0.840a=T2-T1 (equation 3)

plugging eq1 and 2 into eq 3
-0.840a=3.62a+35.51-(-4.85a+47.58)
-0.84a-3.62a-4.85a=-12.07
a=.771m/sˆ2
T2=38.3N, T1=38.9N

the answers are a=1.30m/sˆ2 T1=41.3N T2=40.2

I can't figure out what I did wrong!
 
Last edited:
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Check the file you uploaded - Doesn't look like a physics problem to me.
 
sorry! -__- the right problem is written in now.
 

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