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

In summary, the conversation involves a student seeking help with a physics problem involving a massless rope and pulleys with no friction. The student found the tension in the rope to be 11.3N and the acceleration of the 10kg block to be 1.13 m/s2, but was told by another person that their answer was incorrect. The student then realized their mistake and corrected it, finding the correct answers to be 13.7N and 1.37 m/s2. The conversation also includes a discussion on the use of diagrams and equations in solving problems.
  • #1
SpacemanRich
26
2

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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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
 
  • #8
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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  • #9
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:
 

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

1. What is Newton's second law with pulleys, ropes, and boxes?

Newton's second law states that the net force acting on an object is equal to its mass multiplied by its acceleration. In the case of pulleys, ropes, and boxes, the mass includes the weight of the boxes and the tension in the ropes, while the acceleration is determined by the movement of the system.

2. How does friction impact Newton's second law in this scenario?

In a frictionless scenario, the tension in the ropes will be equal throughout the system, allowing the boxes to move with the same acceleration. However, if there is friction involved, the tension in the ropes will be higher on one side, causing the boxes to move with different accelerations and altering the overall net force on the system.

3. How do pulleys affect the masses and accelerations in this system?

Pulleys can change the direction of forces and can also help distribute the weight of the boxes among multiple ropes. This means that the masses and accelerations of the boxes may be different depending on the configuration of the pulleys.

4. Can the use of pulleys, ropes, and boxes help reduce the force needed to move an object?

Yes, using a pulley system can help distribute the weight of an object among multiple ropes, reducing the force needed to move it. However, this does not change the overall net force on the system, only how it is distributed.

5. How do the number of pulleys and ropes impact the overall force and acceleration in this scenario?

The more pulleys and ropes used, the more the weight of the boxes can be distributed, potentially decreasing the force needed to move them. However, the number of pulleys and ropes does not change the overall net force on the system, which is still determined by the masses and accelerations of the boxes.

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