Static equilibrium: calculating force of tension

[janelle1905]

Homework Statement

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)

Homework Equations

sin=opp/hyp
cos=adj/hyp

The Attempt at a Solution

a. Using T_y = T₁sin60 - mg = 0, T₁=34 N.

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

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

Eq'n 2: T_y = 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!

 

[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.

 

[janelle1905]

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

 

[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?

 

[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=49^o.
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.

 

[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=[T_3x²+T_3y²]^1/2 = [(m₂g)²+(T₂)²]^1/2

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

tanθ = T_3y / T_3x = m₂g / T₂

 

[janelle1905]

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

 

[kuruman]

Correct.