Force Equilibrium: Find T1 Expression w/ m, theta1, theta2, g

[enantiomer1]

Homework Statement

(Intro 1 figure) A chandelier with mass m is attached to the ceiling of a large concert hall by two cables. Because the ceiling is covered with intricate architectural decorations (not indicated in the figure, which uses a humbler depiction), the workers who hung the chandelier couldn't attach the cables to the ceiling directly above the chandelier. Instead, they attached the cables to the ceiling near the walls. Cable 1 has tension T₁ and makes an angle of theta₁ with the ceiling. Cable 2 has tension T₂ and makes an angle of theta₂ with the ceiling.

Find an expression for T₁, the tension in cable 1, that does not depend on T₂.
Express your answer in terms of some or all of the variables m, theta₁, and theta₂, as well as the magnitude of the acceleration due to gravity g.

Homework Equations

T₁ SHOULD equal: mg/ (sin theta 1 + (cos theta 1/cos theta 2)* sin theta 2), but any time I put in sin they say that I haven't formatted the equation right.
what am I missing that can only be computed by thetas 1 & 2 as well as m and g?

 

[tiny-tim]

Hi enantiomer1!

(have a theta: θ )

T₁ SHOULD equal: mg/ (sin theta 1 + (cos theta 1/cos theta 2)* sin theta 2), but any time I put in sin they say that I haven't formatted the equation right.
what am I missing that can only be computed by thetas 1 & 2 as well as m and g?

hmm … looks ok to me …

maybe they want you to keep (θ₁ + θ₂) as it is?

 

[Carid]

Maybe there's a shortage of brackets to make the expression clear?

mg/ (sin theta 1 + ((cos theta 1/cos theta 2)* sin theta 2))

BTW I can't see how you got your expression..;={

Since the chandelier isn't moving, it must be that the horizontal component of T1 is equal to the horizontal component of T2. So we can derive an expression for T2 in terms of T1.

Then we equate mg to the vertical components of T1 and T2, and then substitute for T2.

That gives me...

T1 = mg*(cos theta2)/((sin theta1)*(cos theta2) + (sin theta2)*(cos theta1))

NO it's the same, just prettier!