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

• gb14
In summary, a chandelier with mass m is attached to the ceiling of a large concert hall by two cables. The cables are attached to the ceiling near the walls due to intricate architectural decorations. Cable 1 has tension T1 and makes an angle of \theta1 with the ceiling, while cable 2 has tension T2 and makes an angle of \theta2 with the ceiling. To find the forces in the x direction, we can use the equations FT = ma, where the sum of the vertical components of the two tensions must be equal to the weight of the chandelier, and the two horizontal components of the tension are equal and opposite. To find an expression for T1 that does not depend on T2, we can use
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 $$\theta$$1 with the ceiling. Cable 2 has tension T2 and makes an angle of $$\theta$$2 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.

FT = ma

## The Attempt at a Solution

Would I be correct in saying that the total Ftx = mgsin$$\theta$$1+mgsin$$\theta$$2? If so, I would take that and add it to mgcos$$\theta$$1+mgcos$$\theta$$2 for the total.

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

Hi gb14, welcome to PF.

gb14 said:
Would I be correct in saying that the total Ftx = mgsin$$\theta$$1+mgsin$$\theta$$2?
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$$\theta$$1+mgcos$$\theta$$2 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.

I would like to clarify that the equations used in this problem are based on the assumption that the chandelier is in equilibrium, meaning that the forces acting on it are balanced. The equations used are correct, but it is important to note that the forces in the x direction are equal and opposite, so the total force in the x direction would be 0. This means that the tension in cable 1, T1, is equal to the tension in cable 2, T2. Therefore, an expression for T1 that does not depend on T2 would simply be T1 = mg. This is because the weight of the chandelier, mg, is the only force acting downwards in this scenario.

## What is tension on two cables?

Tension on two cables refers to the force exerted by each cable in opposite directions, stretching or pulling the cables taut.

## How is tension on two cables calculated?

Tension on two cables can be calculated using the formula T = F * L, where T is tension, F is force, and L is the length of the cable.

## What factors affect the tension on two cables?

The tension on two cables is affected by the weight of the object being supported, the angle of the cables, and the distance between the attachment points of the cables.

## What happens if the tension on one cable is greater than the other?

If the tension on one cable is greater than the other, the object being supported may become unbalanced and may shift or tilt in one direction.

## How does tension on two cables impact the stability of an object?

The tension on two cables is crucial in maintaining the stability and balance of an object. If the tension is not evenly distributed, the object may become unstable and may even collapse.

• Introductory Physics Homework Help
Replies
8
Views
840
• Introductory Physics Homework Help
Replies
2
Views
5K
• Introductory Physics Homework Help
Replies
2
Views
26K
• Introductory Physics Homework Help
Replies
4
Views
3K
• Introductory Physics Homework Help
Replies
3
Views
9K
• Introductory Physics Homework Help
Replies
2
Views
12K
• Introductory Physics Homework Help
Replies
5
Views
13K
• Introductory Physics Homework Help
Replies
2
Views
4K
• Introductory Physics Homework Help
Replies
6
Views
10K
• Introductory Physics Homework Help
Replies
2
Views
6K