# Massless string pulled by a force

1. Mar 25, 2015

### andyrk

If we take one end of a mass-less string and pull it with force F, would the string have any tension in it? Would it have any tension when we pull it with force F from both the ends?

2. Mar 25, 2015

### Orodruin

Staff Emeritus
Yes.

3. Mar 25, 2015

### andyrk

Why? And what would it be? Can we derive it?

4. Mar 25, 2015

If I take a small element dx of the string, and if T1 acts on one side and T2 acts on the other, net force is $T_1-T_2=ma=0$. So T1=T2.
This means that net force on every element of string is zero right?

5. Mar 25, 2015

### andyrk

Yes, but what I am asking is that why would Tension arise in the first place?

6. Mar 25, 2015

### Orodruin

Staff Emeritus
Why would it not? What is your understanding of tension?

7. Mar 25, 2015

From the equation of in post 4, the force on every segment of the thread is zero. Then why does it need tension?

8. Mar 25, 2015

### andyrk

That tension arises only when something is pulling at some particle. So the point which is being pulled with F has some pulling force on it. But what is pulling on the particles behind that first point at which force is being applied?
The fact that you are using T1 and T2 on dx element itself means that tension is present.

9. Mar 25, 2015

### Orodruin

Staff Emeritus
The net force is zero, yes. This is not the definition of tension. The definition of tension is the force which one part of the string exerts on the other part of the string. What you have shown in #4 is that the tension is constant.

10. Mar 25, 2015

### Orodruin

Staff Emeritus
This is wrong. Tension (at point x) is the force that acts from one part of the string on the other part (for a given cut at the point x of the string).

11. Mar 25, 2015

### andyrk

So how would you explain that different parts of the string impart forces on other parts so that tension exists throughout the string?

12. Mar 25, 2015

### Orodruin

Staff Emeritus
It has to, otherwise the string would be accelerating as per Newton's 2nd law as shown in post #4 by Aditya. If you let one of the ends of the element you consider be an actual end of the string (where you are pulling with force F), the force on the other side of the element must also be F (but in the opposite direction). Thus, the tension in the string must be F.

13. Mar 25, 2015

@Orodruin , so tension is not the net force on each part of the string but the force that is exerted by a nearby segment of the same string on it.
also, when the string has some mass and when equal forces are applied on both ends, since acceleration of body is zero, again the tension in each segment of the string has to be F right?

14. Mar 25, 2015

### Orodruin

Staff Emeritus
Assuming that acceleration is zero, yes.

15. Mar 25, 2015

### andyrk

So how would you explain that this same tension exists at all points of the string besides the first? We know that F exists on the very first element. But what about the rest?

16. Mar 25, 2015

### Orodruin

Staff Emeritus
One part of the string pulls on the other part of string and vice versa. If it did not, the different parts of the string would accelerate, as post #4 clearly shows. This is the very definition of what tension is.

17. Mar 25, 2015

### andyrk

How do you know that one part pulls the other part with the same force as the first part was pulled by? Why can't they be different?

18. Mar 25, 2015

Third law.
when one part pulls the nearby part, then that nearby part will pull the first part with the same force.

19. Mar 25, 2015

### andyrk

There are 2 parts. One is right at the end of the string, which is being pulled by force F. The second part which is adjacent to this part but not in direct contact with the point at which the force is being applied. So how does this second part get pulley by force F?

20. Mar 25, 2015

### Orodruin

Staff Emeritus
By the internal forces of the string, how else?