How Do You Calculate the Tension in a Chandelier's Cable?

[gb14]

Homework Statement

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 the forces in the x direction.

Find an expression for T₁, the tension in cable 1, that does not depend on T₂.

Homework Equations

F_T = ma

The Attempt at a Solution

Would I be correct in saying that the total F_tx = mgsin\theta1+mgsin\theta2? If so, I would take that and add it to mgcos\theta1+mgcos\theta2 for the total.

P.S This is my first post so I apologize if anything is in the wrong place.

 

[kuruman]

Hi gb14, welcome to PF.

Would I be correct in saying that the total F_tx = mgsin\theta1+mgsin\theta2?

No, you would not be correct. The sum of the vertical components of the two tensions must be equal to the weight of the chandelier. That's because there is no unbalanced force in the vertical direction and you need to say that with an equation.

If so, I would take that and add it to mgcos\theta1+mgcos\theta2 for the total.

Again, no. Forces are vectors and you cannot add the vertical components to the horizontal components in any meaningful way. You need to say that the two horizontal components of the tension are equal and opposite. That's because there is no unbalanced force in the horizontal direction either.

When you write the vector equation

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

that's really two equations in one. One equation for the horizontal direction and a separate equation for the vertical direction.