# Static equilibrium: calculating force of tension

1. Aug 30, 2009

### janelle1905

1. The problem statement, all variables and given/known data
The system (picture found in attachment) is in static equilibrium, and the string in the middle is exactly horizontal.
Find
a. Tension T1
b. Tension T2
c. Tension T3
d. angle (theta)

2. Relevant equations
sin=opp/hyp
cos=adj/hyp

3. The attempt at a solution
a. Using Ty = T1sin60 - mg = 0, T1=34 N.

b. Using Tx = T2 - T1cos60 = 0, T2 = 17 N

c. Eq'n 1: Tx = T3sin(theta) - mg = 0
Re-arranged to: theta = cos-1 T2/T3

Eq'n 2: Ty = T3sin(theta) - mg = 0

Then I substituted the re-arranged eq'n (1) into eq'n (2), and I got this:
0 = T3sin(cos-1 T2/T3) - mg

However, I don't know how to solve this ... so I think I may be doing something wrong.

Thank you in advance for any assistance!

2. Aug 31, 2009

### kuruman

Sorry, there is no attachment that I can see and without it I don't want to guess at what the picture looks like.

3. Aug 31, 2009

### janelle1905

Very sorry about that!
I have added the attachment to this post...hopefully it works this time!!

#### Attached Files:

• ###### staticequilib2.jpg
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Views:
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4. Aug 31, 2009

### kuruman

Parts (a) and (b) look OK.

Part (c)
If you draw the FBD for the mass m2, you get

T3sinθ - m2g = 0
- T2+T3cosθ = 0

So T3sinθ = m2g

and

T3cosθ = T2

These are the horizontal and vertical components of the tension. Can you find its magnitude?

5. Aug 31, 2009

### janelle1905

Thanks so much for your help :)

Using the following equations:

1. T3sinθ - m2g = 0
2. - T2+T3cosθ = 0

I re-arranged and substituted eq'n 1 into 2, and then calculated theta=49o.
The only thing I wasn't sure about was when I had cos(theta)/sin(theta), I simplified it to 1/tan(theta). Is this the correct was to use the trig identity tan=sin/cos ??

Then I substituted the calculation for theta to calculate T3=26 N.

6. Aug 31, 2009

### kuruman

Your method is fine. What I was hinting at is that if you know the x and y components of T3, then the magnitude is given by

T3=[T3x2+T3y2]1/2 = [(m2g)2+(T2)2]1/2

and, as you pointed out, you get the angle from

tanθ = T3y / T3x = m2g / T2

7. Aug 31, 2009

### janelle1905

Okay I see what you were saying - but you get the same answer either way, correct?

8. Aug 31, 2009

### kuruman

Correct.

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