Tension Problem, with slope and 2 masses

[thoradicus]

Homework Statement

http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20%289709%29/9709_w10_qp_43.pdf number 3

Homework Equations

T=mg (no acceleration as equilibirum)
sin rule?

The Attempt at a Solution

Not sure where to start..the marking scheme shows that sin/cos rules are used but I don't understand why they are used.
At first i wrote that m1g=T, and m2gsin30=T. Therefore, resultant force on pulley is mqg+mwgsin30, but, this is a dead end

 

[Doc Al]

Set up an equation for the net force that the strings exert on the pulley. You'll need to add vectors, so that's where those rules would apply.

 

[thoradicus]

ok, so the resultant force is 2Tcos30 along AQ and AP? how can we say for sure the two angles are the same?

 

[Doc Al]

ok, so the resultant force is 2Tcos30 along AQ and AP?

How did you get that?

how can we say for sure the two angles are the same?

What two angles?

 

[thoradicus]

How did you get that?

What two angles?

i assumed AP and AQ are hypotenuses to the line of direction of 3sqrt3. So, Tcos30 +Tcos30=3sqrt3?

Since the angle is 30 degrees, the other angle on top is 60 degrees. I assumed that the angle is divided equally into 2?

 

[Doc Al]

i assumed AP and AQ are hypotenuses to the line of direction of 3sqrt3. So, Tcos30 +Tcos30=3sqrt3?

Since the angle is 30 degrees, the other angle on top is 60 degrees. I assumed that the angle is divided equally into 2?

Ah, I see what you're doing. Yes, that's a fine way to add them up. You took advantage of symmetry, taking their components along the bisector of the 60° angle. Good!

I think you meant: Tcos30 + Tcos30 = T√3

(You can also use the law of cosines to add them up. Same answer, of course.)