Forces acting on a box-pulley system in an elevator?

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Homework Help Overview

The problem involves a box-pulley system within an elevator, where the elevator is initially moving at a constant velocity and later accelerates downwards. The system consists of two blocks, A and B, with given masses, and the task is to calculate various forces acting on the blocks and the pulley.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to calculate the tension in the rope and the forces acting on block A and the pulley. They express confusion about how to approach parts (b), (c), (d), and (e) of the problem.
  • Some participants suggest drawing Free Body Diagrams (FBDs) for the components of the system to clarify the forces involved.
  • There is discussion about the role of the tension in the rope and how it affects the pulley, with questions raised about the contributions of the rope to the forces acting on the pulley.
  • Participants also explore the relationship between the forces acting on the blocks and the tension in the rope, particularly in the context of the elevator's acceleration.

Discussion Status

The discussion is ongoing, with some participants providing hints and guidance on how to approach the problem. The original poster has made some attempts and is seeking further clarification on specific parts of the problem. There is recognition of correct answers for parts (a) and (b), but uncertainty remains regarding the remaining parts.

Contextual Notes

The original poster is encouraged to show attempts for each part of the problem to receive more targeted help. There is an emphasis on understanding the underlying concepts rather than simply obtaining answers.

  • #31
gneill said:
Was the mass of block A 3.4 kg or 3.8 kg? I seem to recall it being 3.8 kg.

Oh my goodness, you're absolutely right. I used the wrong value for mass. Thank you so much for catching that error. With that changed, my answer is now 20.1 N, which is the correct answer on my homework website!

Thank you so much for your help with this problem gneill. Again, your explanations helped clarify a lot. I will probably be back with more questions in the future. I hope I will receive your guidance again during those difficult times :wink: and I promise to have a free body diagram and my attempts written out beforehand when I come back.
 
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  • #32
Glad to be of help! See you next time. :smile:
 
  • #33
Sagrebella said:
Ok, that makes sense.

and thank you again for your clear explanation, I re-did my equation, taking into mind that the box is stationary on the floor. Also, I changed my overly-rounded answer to 11.8 N instead of 12 N. Hopefully that works.

Now,

FN + FT - FG = 0

FN + 11.8 - m(g-a) = 0

FN + 11.8 - 3.4(10-1.6) = 0

FN + 11.8 - 28.56 = 0

FN = 16.8 N
Why do we use "m(g-a)" instead of mg? IS it because we have to take the difference of the accelerations because theyre both going in the same direction?
 
  • #34
cjm said:
Why do we use "m(g-a)" instead of mg? IS it because we have to take the difference of the accelerations because theyre both going in the same direction?
Yes. Also, if the elevator were accelerating up, you would use "m(g+a)". If the elevator were not accelerating, you would use just "mg".
 
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