# Homework Help: Tension in Strings HELP thanks!-very simple.

1. Feb 9, 2008

### physicsbhelp

[SOLVED] Tension in Strings HELP thanks!--very simple.

1. The problem statement, all variables and given/known data

A 16 kg object hangs in equilibrium from a string with a total length of L=5m and a linear mass density of u= 0.0010 kg/m. The string is wrapped around two light, frictionless pulleys that are separated by a distance of d=2.0m.

A) Determine the tension in the string.
B) At what frequency must the string between the pulleys vibrate in order to form the standing-wave pattern shown in Figure b)?

Picture is attached!!

2. Relevant equations

3. The attempt at a solution

If there was an axis at the bottom of the inverted triangle and the angles were measured from the axis to the strings;

I have also tried using;
v = Square root of (tension/u), but I don't have v,
So i tried to find v using f = v/ wavelength, but i don't have frequency.

I have tried so many things and I can't seem to get a start. Any hints?

#### Attached Files:

• ###### physics diagram.doc
File size:
24.5 KB
Views:
198
Last edited: Feb 9, 2008
2. Feb 9, 2008

### Hootenanny

Staff Emeritus
Since the 12kg mass is stationary, what can you say about the sum of the forces acting on it?

3. Feb 9, 2008

### rsala

since its in equilibrium the net force must be 0.

acknowledge all the forces in the y direction.
tension upwards
gravities pull downward..

since their total combined force is 0 due to mr.newton

then tension must be equal to gravities pull, besides direction

(assuming the weight is connected to one string)

4. Feb 9, 2008

### Hootenanny

Staff Emeritus
Wow physicsbhelp! You said that without even moving your lips...

5. Feb 9, 2008

### physicsbhelp

wait so is the tension equal to gravity? i typed in 9.81 and it said it was wrong so i typed in -9.81 and it said that was wrong too! i am still in need of help!

6. Feb 9, 2008

### Staff: Mentor

9.81 m/s^2 is the acceleration due to gravity (for an object in freefall). What's the weight of the 16 kg mass?

Last edited: Feb 9, 2008
7. Feb 9, 2008

### physicsbhelp

ooo the weight would be 16*9.81? = 156.96. i typed that in and it was also wrong.

8. Feb 9, 2008

### physicsbhelp

so what else could it be?

9. Feb 9, 2008

### Staff: Mentor

If I understand the setup correctly, then that would be the tension in the string (in N). How fussy is the system about significant figures? Round it off. (You certainly don't have 5 significant figure accuracy!)

10. Feb 9, 2008

### physicsbhelp

no that is not the answer. i didnt even round up on the real answer.

11. Feb 9, 2008

### physicsbhelp

okay i attached the diagram. i think that will help. part a) has to do with picture a).

12. Feb 9, 2008

### Staff: Mentor

Providing a diagram makes a huge difference! Now it's clear.

Analyze the forces acting on the hanging mass. You know it must be in equilibrium--that will allow you to figure out the tension. (First figure out the angle the strings make.)

13. Feb 9, 2008

### physicsbhelp

what angle are you talking about? like the angle that the string makes with the object? i thought it was a 45 45 90 triangle. but i still don't see how angles will help find tension. is there an equation for that.

14. Feb 9, 2008

### Staff: Mentor

How did you figure out those angles? Don't guess--they gave you the length of the string for a reason. Once you have the angles, then you can draw yourself a free-body diagram of the forces acting on the mass. (The tension acts along the strings--that's why you need the angles.) The equation you need is the equilibrium condition, specifically: The sum of the vertical force components must equal zero.

15. Feb 9, 2008

### physicsbhelp

i am still not quite understanding this equilibrium angle equation.
so the base of the inverted trianlge is 2 meters and the two sides are 1.5 meters long each. then i still dont get the angles. or basically how to find them.

16. Feb 9, 2008

### Staff: Mentor

Draw your own diagram, marking the lengths. Then use a little trig to find any angle you might need. Hint: Draw a vertical line passing through the mass. Use that as a reference.

17. Feb 9, 2008

### physicsbhelp

i drew my triangle the way u said. but i still don't seem to get it. like the vertical line passing throught the mass is 1.118 meters. i figured that out. but i cant tell how to do trig? like do i do cos (1/1.5)?

18. Feb 9, 2008

### Staff: Mentor

OK.
What's the cosine of the angle that the string tension makes with the vertical? You'll need that when you write your force equation. (Write that equation!)

19. Feb 9, 2008

### physicsbhelp

ooo wait i think i just got it!!! the whole time my calculator was in radian mode, that is why i kept getting wierd answers. haha.
okay so for the ablge between the side that equals 1meter and 1.5 meter i got 48.187 degrees and for the other angle i got 41.81 degrees.
now i really don't know how i can find out the tension.

20. Feb 9, 2008

### Staff: Mentor

Call the unknown tension in each string T. Draw a diagram showing the forces. Add up the vertical components. Solve for T.

21. Feb 9, 2008

### physicsbhelp

i did that but i have no numerical values. i got that 2Ft=Fg? is that right?

22. Feb 9, 2008

### physicsbhelp

or is it not?

23. Feb 10, 2008

### Staff: Mentor

That would be true if the strings were vertical, but they are not. If the tension in each string is T, and the angle that the string makes with the vertical is $\theta$, what is the vertical component of the force that each string exerts on the mass?

24. Feb 10, 2008

### physicsbhelp

Ft(theda)
or
0

25. Feb 10, 2008