Calculating Masses from Fnet: What Am I Missing?

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SUMMARY

The discussion centers on calculating the masses of two objects using net force (Fnet) and acceleration (a). The user initially struggles with the relationship between the two masses, m1 and m2, and their respective forces. The correct approach involves using the equations m1a = m1g - T and m2a = T - m2g to solve for tension (T) and subsequently derive the mass ratio as m1/m2 = (-a - g) / (a - g). The final answer is confirmed to be 2.32.

PREREQUISITES
  • Understanding of Newton's Second Law (Fnet = ma)
  • Knowledge of tension in systems involving multiple masses
  • Familiarity with algebraic manipulation of equations
  • Concept of gravitational force (g) in physics
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  • Study the application of Newton's laws in multi-mass systems
  • Learn about tension forces in connected objects
  • Practice solving problems involving mass ratios and net forces
  • Explore advanced topics in dynamics, such as friction and inclined planes
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Students in physics courses, educators teaching mechanics, and anyone involved in solving dynamics problems related to mass and force calculations.

JoeyBob
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Homework Statement
See attachwed
Relevant Equations
Fnet=ma, a1=-a2
The answer is supposed to be 2.32.

I've been trying to relate the two masses but am having troubles with it. Fnet for mass one would be m1a. Fnet for mass 2 would be -m2a.
The Fnets would be different though.

m2=-Fnet/a for mass two and m1=Fnet/a for mass one. I can't just divide m2 by m1 because the Fnets are different. Even if they did cancel, the ratio would just be 1 from it.

What am I missing here?
 

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JoeyBob said:
Homework Statement:: See attachwed
Relevant Equations:: Fnet=ma, a1=-a2

The answer is supposed to be 2.32.

I've been trying to relate the two masses but am having troubles with it. Fnet for mass one would be m1a. Fnet for mass 2 would be -m2a.
The Fnets would be different though.

m2=-Fnet/a for mass two and m1=Fnet/a for mass one. I can't just divide m2 by m1 because the Fnets are different. Even if they did cancel, the ratio would just be 1 from it.

What am I missing here?
Nothing is attached.
 
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Chestermiller said:
Nothing is attached.

Thanks. I think I just figured it out.

m1a=m1g-T and m2a=T-m2g. Solve for T and then subsitute into another equation. You can then do some algebra and get m1/m2=(-a-g)/(a-g).
 

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