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Tension in a light inextensible string

  • Thread starter hms.tech
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  • #1
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Homework Statement



The attachment shows the string which is light and inextensible passing over two pulleys which are NOT massless. Hence they have some mass and some Moment of inertia.

There is no friction in the pulley thus the pulleys are free to rotate.

K ≥ 1

we know that the system will accelerate and the heavier pink object would accelerate downwards.

The red table is frictionless.

The thing that i am unable of grasp is finding the Tension in each part of the string.

I am not sure, if i have labelled them correctly.
Would someone please tell me the correct values.
Although i am quite sure that T1 and T2 are correctly labelled (ie showing that the tension in both of these parts of the string would be DIFFERENT)

What is the tension T3 and T4 equal to in terms of magnitude (is it the same as T1 or T2) and direction (will they be in the same direction as which i have drawn?)

Homework Equations



none

The Attempt at a Solution



the attachment is an attempt (and might contain errors )
 

Attachments

Answers and Replies

  • #2
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Basically i have to find the tension[magnitude and direction] (sown in orange) in the diagram.

The directions of T3 and T4 are just arbitrary, they are Assumed to be acting in that sense, but it's not necessary they are (as i mentioned before the directions might be wrong) therefore i need you guys to help me determine the direction of T3 and T4


Note : the yellow circles are Pulleys and the pink ovals are masses. The mass on the right is heavier than that on the left. The red line is the table.
 
  • #3
Doc Al
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I am not sure, if i have labelled them correctly.
Would someone please tell me the correct values.
Although i am quite sure that T1 and T2 are correctly labelled (ie showing that the tension in both of these parts of the string would be DIFFERENT)
You have 3 rope segments and thus three different tensions, not four. Label the tension in the horizontal segment T3.

What is the tension T3 and T4 equal to in terms of magnitude (is it the same as T1 or T2) and direction (will they be in the same direction as which i have drawn?)
When drawing forces it's good to keep in mind what those forces are acting on. It's usually best to draw the forces acting on the massive bodies.

Your drawing of the forces acting on the two pink bodies is correct. But you have to fix the drawing of the forces acting on the pulleys. The vertical rope segments pull down on the pulleys--you have that correct. Now fix the tension forces due to the horizontal rope segment, remembering that ropes can only pull.
 
  • #4
Doc Al
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Basically i have to find the tension[magnitude and direction] (sown in orange) in the diagram.
You'll need to know the masses and moments of inertia to find the tension forces.
The directions of T3 and T4 are just arbitrary, they are Assumed to be acting in that sense, but it's not necessary they are (as i mentioned before the directions might be wrong) therefore i need you guys to help me determine the direction of T3 and T4
See my last post. The ropes pull on the pulley. That should tell you the directions of the tension forces on the pulley.
 
  • #5
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Now fix the tension forces due to the horizontal rope segment, remembering that ropes can only pull.

You are implying that if i reverse the directions of [itex]T_{3}[/itex] and [itex]T_{4}[/itex], that would give me the correct directions of the Tensions.

Also that [itex]T_{3}[/itex] = -[itex]T_{4}[/itex]


Another thing which confuses me : Why is it not that [itex]T_{1}[/itex] = [itex]T_{2}[/itex] ?
Ans : Is it because the pulley has some mass (and is not mass-less or "light") ?



I want to learn more about strings and ropes and pulley systems, would provide some links ?
 
  • #6
Doc Al
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You are implying that if i reverse the directions of [itex]T_{3}[/itex] and [itex]T_{4}[/itex], that would give me the correct directions of the Tensions.
That would give you the correct directions of the tension forces exerted by the middle string segment on the pulleys.

Also that [itex]T_{3}[/itex] = -[itex]T_{4}[/itex]
I would let [itex]T_{3}[/itex] represent the magnitude of the tension force.


Another thing which confuses me : Why is it not that [itex]T_{1}[/itex] = [itex]T_{2}[/itex] ?
Ans : Is it because the pulley has some mass (and is not mass-less or "light") ?
Exactly. A continuous segment of massless rope has the same tension throughout, but not when it's divided over a massive pulley.

I want to learn more about strings and ropes and pulley systems, would provide some links ?
Don't know of anything off hand. I suggest getting a review book and solving as many problems as possible. You'll also find many, many problems discussed in great detail right here on PF. But you'll need to poke around to find them.
 
  • #7
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Exactly. A continuous segment of massless rope has the same tension throughout, but not when it's divided over a massive pulley.
I have done some more reading on this topic and found out (another reason) :

"If the pulley is smooth, the tension in the string is same throughout its length while on the other hand if the pulley is rough the tension in the strings on either side of the pulley is different"

Does this reasons add up to the one mentioned before ?
 
  • #8
Doc Al
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I have done some more reading on this topic and found out (another reason) :

"If the pulley is smooth, the tension in the string is same throughout its length while on the other hand if the pulley is rough the tension in the strings on either side of the pulley is different"

Does this reasons add up to the one mentioned before ?
I presume they mean that the pulley is so smooth that the string slides over it without rotating it. In that case, sure, the pulley would not change the tension in the string.

But usually it is assumed that the pulley does turn with the movement of the string. So the tension would be the same on both sides only if the pulley is massless and frictionless (on its axle). When mass (and thus moment of inertia) is involved or there is friction, there must be a difference in tension from one side to the other to create the net torque needed to rotate the pulley (at least when there is acceleration involved).
 

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