Proving angle equals 45° for beads on strings in equilibrium

[elle]

Mechanics help please!

Hi, can anyone help me with the following mechanics question:

http://tinypic.com/f1iogj.jpg"

I've attempted this question but I'm stuck on how i can prove that θ = 45°

I've constructed a diagram but I haven't scanned it but what I've got so far is from the diagram at point C, by resolving horizontally i find that the tensions, say T1 and T2 are equal to each other and I label this as T.

Resolving vertically I get 2T cos θ - w = 0

for the bead on the far left, by resolving horizontally I got:

F1 - T1 sin θ = 0 (F1 being the force acting towards the left)

vertically I get: N1 - T1 cos θ - w = 0


For the bead on the far right, by resolving horizontally I get:

F2 - t2 sin θ = 0

vertically:

N2 - T2 cos θ = 0

And this is where I'm stuck...I don't know where to go from here. Am I on the rght track? Can anyone help? Thanks! Sorry again that I haven't provided my diagram

 

[Physics Monkey]

Ok, a couple things here that you need to work on.

Sorry, I misread the question, ignore my first comment (which I have now deleted).

Second, you left out the weight in the vertical force equation for the point B on the far right.

Third, you need to figure out how the condition that the beads A and B be as far apart as possible plays into things. Here is a hint: what is the maximum horizontal force of friction that bead A can feel? What about bead B?

 

[elle]

hmm I still don't understand...

 

[Physics Monkey]

Where does the horizontal force on bead A that balances the tension come from? What is the maximum value this force can have? Answer the same thing for B.