Solving Tension in 3 Cords: Find T_a in Terms of W

[dbakg00]

This is my first post...is there a way to add a picture from my own computer to a post?

Homework Statement

similar to this problem but with different numbers

In my problem, C=T3, B=T2, and A=T1. The angle on the left is 30 and the one on the right is 45

Homework Equations

$$\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}$$

The Attempt at a Solution

What I know...
1. T_c=w
2. T_a sin 30 + T_b sin 45 = w
3. T_a cos 30 - T_b cos 45 = 0
4. sin 45 = cos 45

T_a sin 30 + T_b sin 45 = w
T_a sin 30 + T_b cos 45 = w
becuse of # 3 above, I get
T_a sin 30 + T_a cos 30 = w
T_a(sin 30 + cos 30) = w
T_a(1.366) = w


...I'm stuck here. Please help direct me to find T_a in terms of W. As shown above I came up with "T_a=w/1.366" but the book came up with "T_a=.732w" Thanks is advance for the help.

 

[tiny-tim]

Welcome to PF!

Hi dbakg00! Welcome to PF!

(try using the X₂ tag just above the Reply box )

… I came up with "T_a=w/1.366" but the book came up with "T_a=.732w"

erm … they're the same!

 

[willem2]

w/1.366 is the same as 0.732w, so it is both correct

 

[dbakg00]

I'm not seeing how those two are the same, would someone mind clarifying for me?

 

[tiny-tim]

w/1.366 = w (1/1.366) = w(.732)

 

[dbakg00]

thank you...i've been staring at this problem for so long, even the obvious becomes oblivious to me!