Calculating Masses from Fnet: What Am I Missing?

  • Thread starter Thread starter JoeyBob
  • Start date Start date
  • Tags Tags
    Second law
AI Thread Summary
The discussion revolves around calculating the masses of two objects using their net forces, with the expected answer being 2.32. The user initially struggles to relate the two masses due to differing net forces, expressing confusion over how to derive a ratio between them. They clarify that Fnet for mass one is m1a and for mass two is -m2a, leading to complications in direct division. Ultimately, the user resolves their issue by establishing equations for tension and acceleration, allowing them to derive a mass ratio of m1/m2=(-a-g)/(a-g). This algebraic approach provides clarity in solving the problem.
JoeyBob
Messages
256
Reaction score
29
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?
 

Attachments

  • Question.PNG
    Question.PNG
    8.5 KB · Views: 156
Physics news on Phys.org
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.
 
  • Like
Likes JoeyBob
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).
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...

Similar threads

Back
Top