Cummulative blocks on a rope, tension problem

  • #1
This isn't a joke, I really don't know if this is right or wrong. Please let me know if this is wrong so I can learn what I am not getting right.

Homework Statement


Three different sets of boxes are being pulled along frictionless surfaces attached to a rope that ends with the rightmost box, accelerating toward the left. All the boxes are identical, and the accelerations are as follows
1) 3a for one box (box is called a)
2) 2a for two boxes (from left to right, called b and c)
3) a for 3 boxes (from left to right, called d, e, and f)

Rank the magnitude of the tension in these ropes from largest to smallest.

Homework Equations


force is mass times acceleration.
magnitude means absolute value
tension=magnitude of the pulling force exerted by a string, cable, chain, or similar object on another object

This is my first point of confusion. What do they mean by magnitude of the tension? If tension is already a magnitude then what are they asking for?

The Attempt at a Solution



If the boxes are identical, then their masses should be the same. If that is correct, then it seems like the ropes would have the following tensions:
1) m(3a)
2) 2m(2a)
3) 3m (a)

It seems like the answer may be that 2 has the most and (1 and 3) are equal to each other but less than 2.

Is that right, can I just add the forces together like that and compare them? Or is this wrong?
 
Last edited:

Answers and Replies

  • #2
Redbelly98
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Welcome to Physics Forums.

You are correct.

The say "magnitude of tension" so that it's clear they mean the strongest tension force. Otherwise, because the tension forces act to the left, many people may assign a negative value to it -- and come up with the unintended result that a tension of -3ma is greater than a tension of -4ma, since -3 is greater than -4.
 
  • #3
PhanthomJay
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If I understand the problem correctly, your answer correctly gives the tension in the rope that is pulling the box(es), using Newton's 2nd law. It does not address the tension in any ropes that exist in between the boxes b and c, d and e, and e and f. A sketch would be helpful. But I don't know what you mean by "adding the forces together".
 
  • #4
Thank you both for your answers. It is so kind of you to spend time helping confused students. I truly appreciate what you are doing here.

I am attaching a scan of the question for clarification. I hope the question I asked earlier matches the picture. If not, could you please tell me how it does not match?
 

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  • #5
Redbelly98
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You're welcome, and thanks for posting the figure. That clarifies -- and changes -- things.

The Attempt at a Solution



If the boxes are identical, then their masses should be the same. If that is correct, then it seems like the ropes would have the following tensions:
1) m(3a)
2) 2m(2a)
3) 3m (a)

It seems like the answer may be that 2 has the most and (1 and 3) are equal to each other but less than 2.

Is that right, can I just add the forces together like that and compare them? Or is this wrong?
Your tension values (1), (2), and (3) above apply to ropes A, B, and D, respectively. That leaves ropes C, E and F to be considered. To get started with thinking about it, try drawing force diagrams for the furthest-to-the-right masses in both the 2-mass and the 3-mass systems.
 
  • #6
I'm sorry. I am so confused.

Trying again here--I editted the picture.
I have checked all the definitions again.


So in the gray box we have this rope. The acceleration of the 2block/1box unit is 2a. The mass of the 2 blocks is 2 each.

The force on the rope B is 4ma.
The mass of the second block is m. The acceleration of teh whole unit is 2a. So maybe the force on this second block is 2ma. Maybe the tension on the rope is C is opposite that of the block, but it can't be opposite because tension is always positive.
So that's wrong too.

Trying again--
The force on the rope B is 4ma.
The mass of the second block is m. The acceleration of the whole unit is 2a. The force on this second block may be 2ma. Since the answer is not 4ma and only positive values are allowed maybe I should add them. Could it be 6ma?

Thank you so much for helping me. I am so sorry to be asking such incredibly stupid questions.
 
  • #7
trying again to add teh picture
 

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  • #8
Redbelly98
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I'm sorry. I am so confused.

Trying again here--I editted the picture.
I have checked all the definitions again.


So in the gray box we have this rope. The acceleration of the 2block/1box unit is 2a. The mass of the 2 blocks is 2 each.

The force on the rope B is 4ma.
The mass of the second block is m. The acceleration of teh whole unit is 2a. So maybe the force on this second block is 2ma.
Yes. The net force on the second block must be 2ma, since
Net force = mass × acceleration = m × 2a = 2ma​
Maybe the tension on the rope is C is opposite that of the block, but it can't be opposite because tension is always positive.
So that's wrong too.
I'm not quite following your reasoning here, and don't understand your statement "the tension on the rope is C is opposite that of the block". As long as the rope is pulling on the block, it is okay -- ropes can pull, but they cannot push on objects.

(If you haven't already, draw yourself a force diagram for this second block. The only ropes that can exert a force on it are ropes that are directly attached to this block.)

Hope that helps. I know it gets confusing with multiple blocks and ropes, but this is good practice for understanding Newton's laws of motion.
 
  • #9
Thanks again. So it was the first way. I was able to do several more problems given that information.

One of my big mistakes was not realizing that one box could be involved in multiple action-reaction pairs. I was trying to balance the boxes on both sides as well as the ropes which obviously is completely wrong.
 

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