Newton's second with Pullys, Ropes, and Boxes (Frictionless)

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Homework Help Overview

The problem involves analyzing a system with pulleys and masses, specifically focusing on the tension in a rope and the acceleration of a block. The scenario is set in a frictionless environment with massless ropes and pulleys.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculations for tension and acceleration, with some questioning the original poster's results and suggesting a review of the problem's hint regarding the relationship between the movements of the masses.

Discussion Status

There is ongoing dialogue about the correctness of the original poster's calculations, with several participants providing feedback and prompting further examination of the equations used. The original poster has indicated a willingness to share their work for additional review.

Contextual Notes

Participants note the importance of accurately interpreting the hint provided in the problem statement, as well as the need for clarity in presenting work for review. There is also mention of the original poster's uncertainty regarding their understanding of the problem.

SpacemanRich
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Homework Statement


[/B]In the drawing, the rope and the pulleys are massless, and there is no friction. Find (a) the tension in the rope and (b) the acceleration of the 10.0-kg block. (Hint: The larger mass moves twice as far as the smaller mass.)

P91Pic.png


2. Formula's Used.
F = ma, W = Mg.

The Attempt at a Solution


Hi,
I need someone to check my answers to this problem. I feel pretty strongly that my answers are correct but they differ slightly from those in my text. Your help would be VERY appreciated. I've exhausted my local sources (Extra texts, and Scham's outlines)for another problem like this to confirm my approach, but was not able to find any. I would feel a whole better if someone else confirmed my answers.
I found the Tension in the rope to be 11.3N, and the acceleration of the 10kg block to be 1.13 m/s2
Thanks for your help.
 
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Your answer doesn't seem right. If the tension is 11N then the net force on the 3k mass is 30-22=8 N downwards. That makes its acceleration roughly double that of the 10k mass. Reread the hint.
 
haruspex said:
Your answer doesn't seem right. If the tension is 11N then the net force on the 3k mass is 30-22=8 N downwards. That makes its acceleration roughly double that of the 10k mass. Reread the hint.
I'm sure my answer is pretty close. They give the answer as 13.7N and 1.37 m/s2. Also I found the acceleration for the 10kg mass to be 2 times as fast as the 3kg mass. (2ax = ay.) That is in line with there 2x = y hint.
 
SpacemanRich said:
I'm sure my answer is pretty close. They give the answer as 13.7N and 1.37 m/s2. Also I found the acceleration for the 10kg mass to be 2 times as fast as the 3kg mass. (2ax = ay.) That is in line with there 2x = y hint.
Your answer may seem close to the given answer, but it makes a big difference to the net force on the 3k block. Your 11.3N gives 3*9.8-11.3*2=6.8; 3*9.8-13.7*2=2.
Please post all your working.
 
haruspex said:
Your answer may seem close to the given answer, but it makes a big difference to the net force on the 3k block. Your 11.3N gives 3*9.8-11.3*2=6.8; 3*9.8-13.7*2=2.
Please post all your working.

Ok, I'll post a scan of my work as soon as I can get it scanned in. That will most likely be sometime tomorrow.
 
SpacemanRich said:
Ok, I'll post a scan of my work as soon as I can get it scanned in. That will most likely be sometime tomorrow.
Scanning is ok for printed matter, such as textbooks, and your own diagrams, but for working it is much preferred that you take the trouble to type it in. It might take you a little more time, but it saves time for everyone reading it, and makes it much easier to reference specific steps when commenting.
 
haruspex said:
Scanning is ok for printed matter, such as textbooks, and your own diagrams, but for working it is much preferred that you take the trouble to type it in. It might take you a little more time, but it saves time for everyone reading it, and makes it much easier to reference specific steps when commenting.

Ok, Thanks for the information. I'll do that in the future. I have my work scanned already. My work equations are numbered for reference. That should help this time.
Thanks for taking a look.

P91text.png
 
As I hinted in post #2, it's your eqn 3 that's wrong. Reread the hint carefully (but you shouldn't need the hint since you can work this out from the diagram).
 
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haruspex said:
As I hinted in post #2, it's your eqn 3 that's wrong. Reread the hint carefully (but you shouldn't need the hint since you can work this out from the diagram).

From the diagram?? Really, I never would have guessed that. I'll take another look and see what I can see.
Just to check, if I can find the mistake in my eqn 3, will I get the answer the book has?? (13.7N, and 1.37m/s2)
Also I'm glad my FBDs are ok. I thought that may be the source of my error.
Thanks.
 
  • #10
SpacemanRich said:
Just to check, if I can find the mistake in my eqn 3, will I get the answer the book has?? (13.7N, and 1.37m/s2)
yes.
 
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  • #11
haruspex said:
yes.

Ok. I have seen the error! Boy do I feel silly.:oops:

My equation 3 should be ay = 1/2 ax

I swear, I think my reading comprehension is just off sometimes. o_O

Thanks a lot for the help, again. It is appreciated. :smile:
 

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