# Tension in string problem

1. Dec 30, 2007

### Joza

I have a ball. It is attached to a wall by a horizontal string to the left. it is attached to the ceiling by another string at an angle theta from the vertical to the right.

The ball is not moving. So I worked out that the tension in the upper string is w/(cos theta) , where w is weight. Is this correct before I go on?

2. Dec 30, 2007

I think I understand your description, but I am not sure. Try to post a pic through photobucket.com or something.

And yes, the vertical component of T in the upper string is equal in magnitude to the weight of the ball.

3. Dec 30, 2007

### Joza

Im not exactly sure how to get a picture up. But basically, I need to find the ratio of the tension at this point, in the upper string, to when it swings out to its farthest position when the horizontal rope is cut.

I am having trouble working out the tensions at these 2 points in the upper string. Any pointers?

4. Dec 30, 2007

### mda

Looks like you got the first one correct. For the second one, work out the angle... draw all the forces and work out the tension as you did before. It's probably easier than you think.

5. Dec 30, 2007

### Staff: Mentor

So far, so good.

How far does the ball swing out when the string is cut? Hint: What direction is the acceleration zero?

6. Dec 30, 2007

Go to Photobucket.com and register for free. Take a picture, scan, or screenshot of the problem. Log in to phototbucket and click browse and find the file you want to upload. After it uploads just cut and paste the image code into thread.

Try it sometime. It's easy and it helps us help you!

Casey

7. Dec 30, 2007

### Joza

Thanks Casey, I'll try it soon!

It swings out an angle of 2 theta....its acceleration is zero at the farthest point right?

8. Dec 30, 2007

### Staff: Mentor

What angle does it make with the vertical at its farthest point?
No it isn't. That's the big difference from the first case. But the component of the acceleration is zero in one particular direction. What direction would that be?

9. Dec 31, 2007

### Joza

It makes an angle of theta with the vertical at its farthest point. And the same before string is cut, so it swings out 2 theta.

Well, it swings out in an arc, so that's circular motion. At it's farthest point, it would have zero tangential acceleration right, because their is no change in its speed. But, its acceleration would be radial, that is it would be pointing towards the point of rotation?

10. Dec 31, 2007

### Staff: Mentor

Good.

Actually, just the opposite! The tangential velocity is zero, but not the tangential acceleration. (The velocity is changing--it's only zero for an instant.)

As far as radial acceleration goes, how does that depend on tangential speed?

11. Dec 31, 2007

### Joza

Ah, yes yes, I see, silly me. It will have maximum acceleration back towards the wall at that instant will it?

Anyhow, I need the tension in the string at the 2 instants. I think I have the first one right - the tension is a component of the balls weight (ignoring the strings weight).

But it is this second instant that I am just having a little trouble with. My intuition tells me that the tension will be greater because the ball's inertia (correct?)

I need a numerical answer. So, I have the component of the balls weight, which is the same as before right? But shouldn't there be some other force?

I'm thinking it has something to do with its acceleration, but that isn't a force, it's the result of a force. Hmmm....

12. Dec 31, 2007

### Staff: Mentor

Make use of the "hint" I tried to give you in post #8. Apply Newton's 2nd law along a direction where the acceleration is zero.

13. Jan 1, 2008

### Joza

I'm sorry guys, but I can't see my mistake.

I still get w/(cos theta) for both tensions.....a points 1 and 2.

Where am I going wrong? In both situations, is the only force acting on the strings the weight of the ball (ie. string tension)?

14. Jan 1, 2008

### Staff: Mentor

It's hard to say where you are going wrong, since you don't describe how you calculated the tension in the second case.

In the first case, there are three forces acting on the ball: its weight and the two string tensions. In this case the forces add to zero in all directions, since the ball is in equilibrium.

In the second case, only one string pulls on the ball. And the ball is not in equilibrium. (But its acceleration in one particular direction is surely zero. Find that direction and apply Newton's 2nd law.)

15. Jan 2, 2008

### Joza

Thanks for the input guys. I'm still having trouble with this one. Here is the picture:

http://i257.photobucket.com/albums/hh216/Megatallica/tension.jpg

I need to find the ratio of string in position B to position A. The horizontal string is cut and it swings out to B, max position.

I m confused. I calculated the tension in a as (mg)/(cos beta), thats mass of ball, string has no mass.

16. Jan 2, 2008

### Staff: Mentor

Did you do what I suggested?

17. Jan 2, 2008

### Joza

Yes, but the problem is I can only identify once force acting on the string causing the tension.....

18. Jan 2, 2008

### HallsofIvy

Staff Emeritus
Only one force? I see three. The obvious one is the force of gravity- the weight of the ball- that's vertical, downward. The horizontal string is exerting a force horizontally, The string attached to the ceiling is at an angle- it has two components, one upward, the other horizontal.

19. Jan 2, 2008

### Staff: Mentor

Just as before, all you need to consider are the forces on the ball. There are only two.

What about the acceleration? In which direction is it zero?