Find tension of a rope and kinetic force

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Homework Help Overview

The discussion revolves around a physics problem involving a rope lifting a block, where the tension in the rope varies along its length. Participants are tasked with finding the tension at a specific position y along the rope, which has a uniform mass and length. The problem involves concepts from dynamics and forces, particularly Newton's second law.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the meaning of finding tension at a position y and question which end of the rope this refers to. There is an exploration of drawing free body diagrams and identifying forces acting on segments of the rope. Some participants express uncertainty about how to incorporate acceleration into their reasoning.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and reasoning. Some guidance has been offered regarding the consideration of the entire system and the forces involved. Multiple interpretations of the problem setup are being explored, particularly concerning the role of acceleration.

Contextual Notes

Participants note confusion regarding the definitions of variables such as y and L-l, and there is mention of homework constraints and the importance of understanding the problem rather than just finding a solution.

ruskointhehizzy

Homework Statement


A constant force F pulls a rope vertically upward. The rope is lifting a block of mass M, to which it is attached. The rope is of uniform composition, and has mass m and length L. Find the tension in the rope at a position y along its length. (The tension will not be uniform in the rope.)

Homework Equations


kinetic force, Newtons second law

The Attempt at a Solution



attached file. I am not sure what it means by "Find the tension in the rope at a position y". I drew a FBD, and attempted to solve for the tension on the rope.
 

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ruskointhehizzy said:
Find the tension in the rope at a position y"
As it says, the tension is not the same all along the rope. You are to find the tension at distance y from one end of the rope. Unfortunately it is not made clear which end. I would take it as from the bottom of the rope.
 
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haruspex said:
As it says, the tension is not the same all along the rope. You are to find the tension at distance y from one end of the rope. Unfortunately it is not made clear which end. I would take it as from the bottom of the rope.
okay I think so too.

how would I go about finding the tension at a point y? I am not sure how to add that into the equations.
 
ruskointhehizzy said:
okay I think so too.

how would I go about finding the tension at a point y? I am not sure how to add that into the equations.
Draw a free body diagram for the portion of rope below y. What forces act on it?
 
haruspex said:
Draw a free body diagram for the portion of rope below y. What forces act on it?
This is what I worked out. I drew the FBD and found that the force F (tension) and weight of the rope from point y to the block.
I hope this is correct, but I am not sure.
 

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ruskointhehizzy said:
This is what I worked out. I drew the FBD and found that the force F (tension) and weight of the rope from point y to the block.
I hope this is correct, but I am not sure.
Two things...
Not sure why you switched from y to L-l. Aren't they the same? I would avoid using lowercase l for a variable since it can be confused with I in some fonts.
More importantly, you have assumed there is no acceleration.
 
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haruspex said:
Two things...
Not sure why you switched from y to L-l. Aren't they the same? I would avoid using lowercase l for a variable since it can be confused with I in some fonts.
More importantly, you have assumed there is no acceleration.
that's the part that was confusing to me - I am not sure what to do because of the acceleration. I was just assuming the velocity is constant, so that the tension = weight force - because the second law tells us the net force would be 0. Also, y is just the point where as L-l is the distance. I guess it should be L-y.

I am not sure what to do if the velocity is not constant.
 
ruskointhehizzy said:
I am not sure what to do because of the acceleration
Consider the whole rope and mass as a system. What forces act on it? What acceleration results?
Incorporate that into your force balance for the mass-plus-length-y subsystem.
ruskointhehizzy said:
y is just the point where as L-l is the distance
In post #2 I defined y as the distance.
 
haruspex said:
Two things...
Not sure why you switched from y to L-l. Aren't they the same? I would avoid using lowercase l for a variable since it can be confused with I in some fonts.
More importantly, you have assumed there is no acceleration.
haruspex said:
Consider the whole rope and mass as a system. What forces act on it? What acceleration results?
Incorporate that into your force balance for the mass-plus-length-y subsystem.

In post #2 I defined y as the distance.
I just saw this part so that would just change the L-y to y correct?
I just did that, I thought about it and came up with this result. I thought the Fnet of the system would have to include the total mass times acceleration. Then used that in the other equation I had from before.
I hope this is right because I really thought this through and I always hope that my thinking is correct. That is what is most important to me, is that I can think the problem through in the correct way - it really helped me through calculus.

Thank you for the guidance!
IMG_0112.jpg
 

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  • #10
Yes, Fnet=a(M+m), but you made a bit of a blunder getting to the next line.
 
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  • #11
haruspex said:
Yes, Fnet=a(M+m), but you made a bit of a blunder getting to the next line.
yeah I was thinking on paper :P sorry for the mess
 
  • #12
haruspex said:
Yes, Fnet=a(M+m), but you made a bit of a blunder getting to the next line.
I got all excited when it started to make sense I wanted to show you and thank you
 
  • #13
ruskointhehizzy said:
yeah I was thinking on paper :P sorry for the mess
I mean that your expression for a is wrong. It is not (M+m)/Fnet. That is not even dimensionally correct.
 
  • #14
aw man... I turned it in already but that's okay I have the solution now. The professor posts them afterwards. Thank you for the help. I have an exam on Friday I am pretty nervous for. I will get another problem set after that, so will be back here most likely lol :P thanks for the help. I will get this no matter what it takes.
 

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