# Forces and Tension

1. Sep 22, 2007

### aligass2004

1. The problem statement, all variables and given/known data
A 14 kg loudspeaker is suspended 2.0m below the ceiling by two 3.0m long cables that angle outward at equal angles. What is the tension in the cables?

2. Relevant equations

F=ma

3. The attempt at a solution

I drew the free body diagram, and I broke the tension of the two wires into their components. I'm unsure about how to tackle the problem.

2. Sep 22, 2007

### learningphysics

Where are you getting stuck? What are the forces acting on the suspended mass vertically?

3. Sep 22, 2007

### aligass2004

The forces acting are the tension of the two cables and the weight of the object. I just don't know how to find the angle so I can break up the tensions into components.

4. Sep 22, 2007

### learningphysics

Find the angle... you know the y displacement is 2.0m. You know the length of the cable is 3.0m. You can find the angle.

5. Sep 23, 2007

### aligass2004

Would the angle be 48.19 degrees? And if that's the case, where do I go from there?

6. Sep 23, 2007

### learningphysics

I'm getting 41.81 degrees. Well... if T is the magnitude of the tension... what's the vertical component of tension? What are the forces in the vertical direction?

7. Sep 23, 2007

### aligass2004

The forces in the vertical direction are T1x and T2x

8. Sep 23, 2007

### learningphysics

Oops... sorry, this angle is fine... I just got the other angle of the right triangle... 90-48.19 = 41.81

9. Sep 23, 2007

### learningphysics

The tensions are equal... So what is Tx in terms of T?

10. Sep 23, 2007

### aligass2004

Tx = Tcos(theta)

11. Sep 23, 2007

### learningphysics

So what is the F=ma equation in the vertical direction?

12. Sep 23, 2007

### aligass2004

Is it Tcos(theta) = m(ax)?

13. Sep 23, 2007

### learningphysics

accleretation is 0... you have 2 tensions... and also gravity.

14. Sep 23, 2007

### aligass2004

I put horizontal anyway. In the vertical direction it would be Fnety = T1y + T2y - mg = m(ay).

15. Sep 23, 2007

### learningphysics

Can you put in the angles, sin/cos etc... ay = 0... and T1=T2... so just use T.

16. Sep 23, 2007

### aligass2004

I don't understand.

17. Sep 23, 2007

### learningphysics

Did you find T?

18. Sep 23, 2007

### aligass2004

No I didn't. I don't understand how you can find T.

19. Sep 23, 2007

### learningphysics

Why are you using two different variables T1 and T2?

20. Sep 23, 2007

### aligass2004

Am I not breaking the tension into components?