Find the component of the Force along AC

[Northbysouth]

Homework Statement

The 350-lb force is to be broken into components along the
lines AB and AC.
2. The component along AC is most nearly
1. 175.0 lb
2. 186.2 lb
3. 202 lb
4. 225 lb
5. 239 lb
6. 268 lb
7. 294 lb
8. 303 lb
9. 404 lb
10. 457 lb

I have attached an image of the question

Homework Equations

The Attempt at a Solution

I had thought that I could do 350cos(30) = 303.1
But this answer is wrong. The answer is supposed to be 239 lb but I don't understand how to get it.

Help would be appreciated.

 

[gneill]

Suppose instead that there was a mass producing the same force but suspended by two cables with the same geometry (but inverted, of course). How would you approach the problem of finding the tensions?

 

[Northbysouth]

Now I get it. Redrawing the picture as you suggested helped me recognize that I'd dealt with this before.

ƩF_y=0
0 = -350 + ACsin(60) + ABsin(50)

ƩF_x=0
0= AC cos(60) - ABcos(50)

Rearranging this we get:

AB= ACcos(60)/cos(50)

Then I can plug this into the ƩF_y thus allowing me to solve for Ac which is 239 lb.

Thanks for your help. That change in perspective really opened the question up.