Pulley-Block-Spring System: Understanding the Forces Acting on a Hanging Block

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SUMMARY

The discussion focuses on the analysis of forces in a pulley-block-spring system, specifically addressing the misunderstanding of spring force in relation to a hanging block. Participants emphasize the importance of drawing accurate Free Body Diagrams (FBD) for each block to clarify the forces acting on them. The correct equation of motion for the second block is established as mg - (T + kx) = ma, where T represents tension and kx represents the spring force. The conversation highlights the need for clear problem statements and step-by-step analysis to derive solutions effectively.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Familiarity with Free Body Diagrams (FBD)
  • Knowledge of spring force and Hooke's Law (kx)
  • Basic principles of pulley systems
NEXT STEPS
  • Study the principles of Free Body Diagrams in mechanics
  • Learn about Hooke's Law and its applications in spring systems
  • Explore the dynamics of pulley systems and their equations of motion
  • Practice solving problems involving multiple forces acting on objects
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators teaching concepts related to forces in pulley and spring systems.

Abhimessi10
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Homework Statement


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Homework Equations


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The Attempt at a Solution


In the second question at 5:32,why doesn't the block have a spring force.In FBD of 2nd block it should have been mg-(T+kx)=ma right?
 
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Un-PF format and ambiguous:
"why doesn't the block have a spring force "
there are two blocks. The spring is not attached to the left block.

Abhimessi10 said:
In FBD of 2nd block it should have been T+kx-mg=ma right?
No. Think what it should be when the wire is attached to a fixed point.
 
BvU said:
Un-PF format and ambiguous:
"why doesn't the block have a spring force "
there are two blocks. The spring is not attached to the left block.

No. Think what it should be when the wire is attached to a fixed point.
Sorry but i don't understand what you mean
 
Draw the free body diagram of each block. What are the forces on the 15 kg block? What is the net force? Then use Newton's second law to calculate the unknown force. Now mov to the second block and do the same thin.
 
Chandra Prayaga said:
Draw the free body diagram of each block. What are the forces on the 15 kg block? What is the net force? Then use Newton's second law to calculate the unknown force. Now mov to the second block and do the same thin.

I am talking about the second question at 5:32
 
It is still a problem of not drawing a proper force diagram. In the equation which you gave, what is T? What is kx? What is mg, what is its direction? All these will become clear if you draw the FBD.
 
Chandra Prayaga said:
It is still a problem of not drawing a proper force diagram. In the equation which you gave, what is T? What is kx? What is mg, what is its direction? All these will become clear if you draw the FBD.
At 5:32 for me if the second block attached to spring is going down,then spring force and tension should act on the block
 
1. Why is the second block going down? The diagram shows it on the table.
2. The tension is coming from which object?
 
Chandra Prayaga said:
1. Why is the second block going down? The diagram shows it on the table.
2. The tension is coming from which object?
The question I'm talking about is at 5:32 which involves a pulley
 
  • #10
Sorry. I did not realize that you were talking about a different diagram later in the video. OK.
You are writing two separate forces, T and kx. What is exerting these two forces?
 
  • #11
Chandra Prayaga said:
Sorry. I did not realize that you were talking about a different diagram later in the video. OK.
You are writing two separate forces, T and kx. What is exerting these two forces?
The wire exerts tension ,spring exerts Kx in same direction so FBD for block 2 will be mg-(T+kx)=ma in my opinion
 
  • #12
The wire and the spring are not two separate things, and each one is not separately attached to the block. The tension on the block is because of the spring, so T is the same as kx. Both should not appear separately in your equation.
 
  • #13
Chandra Prayaga said:
The wire and the spring are not two separate things, and each one is not separately attached to the block. The tension on the block is because of the spring, so T is the same as kx. Both should not appear separately in your equation.
But if the block is going down what is causing the restoring force kx of spring
the weight of the block or the tension ?
 
  • #14
Go at it step-by-step. Draw separate FBD for each block and the spring. Then write the equation of motion for each object. Then start looking for answers for your questions. You will see the answers pop out one after the other.
 
  • #15
Asking about a question buried 5 minutes into a video is not a reasonable thing to do to our homework helpers. Why not draw your own diagram and spell out the question in the problem statement? What are the relevant equations?

Your attempt at solution should not be another question, but a demonstration of what you've tried so far.
 
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