Tension Problem

1. Nov 6, 2005

jbgibson

The problem states: a 0.13-kg meter stick is held perpendicular to a wall by a 2.5-m string going from the wall to the far end of the stick. (a) find the tension in the string; (b) if a shorter string is used, will its tension be greater than, less than, or the same as that found in part (a); (c) find the tension in a 2.0-m string.

I assume the meter stick is not moving or it's in equilibrium, so wouldn't the tension simply be T = mg? Any help is greatly appreciated. Thanks in advance.

2. Nov 6, 2005

kp

dont you have an angle in there some where?

3. Nov 6, 2005

lightgrav

The *vertical component* of the Tension must cancel mg.
The horizontal component of Tension pulls it in tight to the wall.

4. Nov 6, 2005

jbgibson

That's what I was wondering, but there is no angle!

5. Nov 6, 2005

jbgibson

Shouldn't there be a horizontal distance or some angle to allow for calculating tension?

6. Nov 6, 2005

kp

there is an angle in there...read carefully (hint: tang)

7. Nov 6, 2005

lightgrav

did you draw a diagram, and label its dimensions?
How long is the string? the stick? mass of the stick?

ALSO draw and label the Forces that act on the interesting object (stick)

8. Nov 7, 2005

jbgibson

Okay, I think I found the angle. If it is perpendicular to the wall, this would be 90 degrees. So, theta would be 45 degrees right. If that's the case then the length of the meter stick and the vertical distance to the point at which the string connects at the wall would be the same.

9. Nov 7, 2005

Fermat

You got the 90 degree angle correct, but the 45 degree angle is wrong.

Have you done what lightgrav suggested just above ? (post #7)

In particular, re-read the 2nd line.

10. Nov 7, 2005

jbgibson

I have the length of the string and the mass of the stick; that's it.

11. Nov 7, 2005

Fermat

You have a metre-stick. What is its length ?

12. Nov 7, 2005

jbgibson

Man, I feel dumb! Now that I know the length, would I use the formula:
T = mg(H/V)?

13. Nov 7, 2005

kp

now that you know theta, what is the y (vertical) component of the tension?

I knew once you reread the problem, you would say how stupid you felt. I once spent hours trying to figure out how long a yard stick was i a similiar type of problem.

Last edited: Nov 7, 2005
14. Nov 7, 2005

jbgibson

Is the y component of tension = mg? I seem to still be lost on this one. Now that I have solved for theta, I get 70.53 degrees, where do I go from here? Thanks for the help!

15. Nov 7, 2005

jbgibson

I think I got it! If I know the Tx = mg(H/V) and Ty = mg, I should be able to use pythagorean theorem right?

16. Nov 7, 2005

Staff: Mentor

Realize that there are several forces acting on the the meter stick:
(a) its weight (where does that force act?)
(b) the tension in the string (the angle of the string, and thus the angle of the tension force, depends on the length of the string)
(c) the force the wall exerts on the stick

The thing to do is calculate torques about the point where the stick touches the wall. (Of the three forces mentioned above, only two of them contribute to the torque about that point.) Since the stick is in equilibrium, the net torque must be zero.

17. Nov 7, 2005

jbgibson

Thanks for the quick response Doc! I think I got it. Here it is:
(T*sin theta)H - mg(H/2) and solve for T

18. Nov 8, 2005

You got it.