# Homework Help: Tension direction in a string

1. Jan 23, 2015

### andyrk

Why is tension at a specific point in a string in 2 directions? For example in the first part of image 1, the string is pulling the two blocks upwards. And in the second part of image 1, the string is pulling the pulley down too. Does tension at a point in a string try to pull the point in 2 directions?

And in image 2, why is the bottom most pulley (which is not hinged) pulled up by a tension 2T? If it is, that means there is no net force acting on it but then how is it moving down if there is no force on it? No force means no acceleration but still it can have a constant velocity (no acceleration) with which it moves down. But why and how does it get that velocity?

#### Attached Files:

File size:
15.4 KB
Views:
292
• ###### Image-2.jpg
File size:
15.4 KB
Views:
198
2. Jan 23, 2015

### Bystander

Hold one end of a string in one hand. Apply a tension. How taut is the string? Hold one end of the same string in one hand, and the second end in the other and pull in opposite directions. How taut is the string? Can you apply tension by pulling on only one end?
The diagram shows two masses of 1 T hanging from the pulley by a string run over the pulley. Two times one is two.
It's not.

3. Jan 23, 2015

### andyrk

Right, for a string to have tension or remain taut, it needs to be connected or pulled from both ends with equal force.

But shouldn't it move down? Why should there be no forces on the unhinged pulley?

4. Jan 23, 2015

### haruspex

Your second diagram lacks important information about masses and friction. Is there some accompanying text you've omitted?

5. Jan 23, 2015

### Bystander

Two T down, and two T up --- no indication in the diagram that the block anchoring the upper string is moving/sliding on the table top. The string supporting the lower pulley is NOT wrapped around the pulley, or turning it --- it's not shown specifically in the diagram, but in the real world, it would be attached to a clevis which carries the shaft around which the pulley turns.

6. Jan 23, 2015

### andyrk

Nope. I just wanted to know that why is the unhinged pulley being pulled by a tension 2T upwards?

7. Jan 23, 2015

### haruspex

how do you know that it is, other than that someone has drawn it on the diagram?

8. Jan 23, 2015

### andyrk

Because that's what the solution I am looking at says.

I didn't mention some facts which were also included in the question-
1.No frictional forces
2. Massless pulleys and strings

Does that somehow help you to give a reasoning as to why the pulley is being pulled up by tension 2T?

9. Jan 23, 2015

### haruspex

If an object is massless then it doesn't matter whether it is accelerating or not. $\Sigma F = ma$ tells you that the net force must be zero.

10. Jan 23, 2015

### andyrk

Oops..sorry my bad. But how is it still moving down then?

11. Jan 23, 2015

### haruspex

I presume the three blocks have mass, so there should be the weights of the two smaller blocks acting down. It will turn out that these are less than 2T in sum, so those blocks will accelerate downwards. Since the strings do not break, they and the pulley must accelerate down too.

12. Jan 23, 2015

### andyrk

The 2T up and the two Ts down cancel out. So by the virtue of which force does the pulley move down?

13. Jan 23, 2015

### haruspex

Like I said, the pulley has no mass, so it needs no force to accelerate which ever way. Its motion is determined by whatever it is attached to.
To make it more realistic, you could let the pulley have a very small mass. When you work out the whole system you will now find that the tensions down slightly exceed the tension up, giving the pulley just the acceleration it needs to keep all the strings taut. The same applies to the strings themselves, etc.
The 'massless' fiction just means the mass is insignificant compared with the other masses in the system.

14. Jan 23, 2015

### andyrk

What? Why? The tensions down always equal the tension up. How can they be slightly greater than the upper tension?

15. Jan 23, 2015

### haruspex

Actually I got it backwards - forgot gravity. The tension up will exceed tension down.
The system is accelerating. If the pulley has mass then there must be a net force to accelerate it downwards. The acceleration will be less than g, so the net of the tensions must be upwards.

16. Jan 23, 2015

### andyrk

So if the net of the tensions is upwards then would the pulley start moving upwards instead of downwards?

17. Jan 23, 2015

### haruspex

No, you've forgotten gravity acting on the pulley (as I did). The net force on the pulley will be downwards, accelerating it down at the same rate as the rest of the system. But that rate is less than g, so the net force of the strings must be upwards.

18. Jan 23, 2015

### andyrk

But for the pulley to move down, why does need to have greater upper tension than lower? Even if they are same, pulley can still move down because of its mass. So why do the upper and lower tensions need to be different?

19. Jan 23, 2015

### haruspex

That's not what I said.
For the pulley to accelerate down, the net force must be downwards: tensions down + mg > tension up. But the system will accelerate at a rate less than g, and for that reason tensions down < tensions up.

20. Jan 23, 2015

### andyrk

If tensions down = tensions up still, tensions down + mg > tension up. So why is tensions down < tensions up?

21. Jan 23, 2015

### haruspex

The downward acceleration will be < g, so the net downward force must be < mg. Tdown+mg-Tup < mg. Tdown<Tup .

22. Jan 23, 2015

### andyrk

I didn't get the inequalities you just showed. Can you explain them once again?
Why should Tdown+mg-Tup be < mg? Is there any reason for it?

Last edited: Jan 23, 2015
23. Jan 23, 2015

### haruspex

Let the mass of the pulley be m and the acceleration of the system (i.e. all the bits that move) be a. Do you agree they will all have the same acceleration? Do you see why a < g?
For the pulley, Tdown+mg-Tup = ma < mg.

24. Jan 23, 2015

### Staff: Mentor

If the two masses are moving down, then T < mg. Then, if the pulley is massless, the 2T = T + T. So, the 2T < 2mg.

Is this what you were saying, andyrk? If so, then I agree.

Chet

25. Mar 21, 2015

### andyrk

But wait, why would the pulley be pulled by two forces T and T in the downward direction on either side of the pulley in the first place? Shouldn't the points at which the pulley is being pulled downwards by T also be pulled upwards by T (since tension at a point acts in two directions), such that effectively the pulley isn't being pulled down at all and so we don't need the 2T force to exist? But this is not the case as we see in the image in post #1. Can somebody explain this?

Last edited: Mar 22, 2015