Hanging Chandelier (A question about Tension Force)

In summary, the tension in cable 1, T1, can be expressed as T1 = mg / (sin(theta1) + cos(theta1) * tan(theta2)), where m is the mass of the chandelier, (theta1) is the angle made by cable 1 with the ceiling, (theta2) is the angle made by cable 2 with the ceiling, and g is the acceleration due to gravity. This does not depend on the tension in cable 2, T2.
  • #1
BlakeGriffin
8
0
Hanging Chandelier (Solved)

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 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 .

Homework Equations


No equation


The Attempt at a Solution


I figured that Net Force=0 and therefore
Net Force x = T2cos(theta2) - T1cos(theta1)=0
Net Force y = T1sin(theta1) + T2sin(theta2) - mg=0

I came up with these equations and I know that they are right. I just don't know how to eliminate T2.

Can anyone help me with this?

This is the figure : http://session.masteringphysics.com/problemAsset/1010934/37/MFS_1l_3_v1_a.jpg
 
Last edited:
Physics news on Phys.org
  • #2
I figured that T2 = (T1cos(theta1)/(cos(theta2)

Which makes the Net Force y equation T1sin(theta1)+(T1cos(theta1))/(cos(theta2))=mg

Then I just solved for T1 and I got T1=(mgcos(theta2)-T1sin(theta1))/(cos(theta2)

Which is not the right answer. It says my answer should not depend on T1.

Where is my mistake?
 
  • #3
I figured out my mistake. For some reason I thought :

Net Force y equation was T1sin(theta1)+(T1cos(theta1))/(cos(theta2))=mg instead of

T1sin(theta1)+(T1cos(theta1))(sin(theta2))/(cos(theta2))=mg

Then T1(Sin(theta1)+(cos(theta1)(tan(theta2)=mg

T1=mg/(Sin(theta1)+(cos(theta1)(tan(theta2)

I tried it. It's the right answer.
 
Last edited:

Related to Hanging Chandelier (A question about Tension Force)

1. What is the purpose of a hanging chandelier?

A hanging chandelier is a decorative light fixture that is suspended from the ceiling. It is often used to add elegance and ambiance to a room, and can also provide additional lighting.

2. How is the tension force of a hanging chandelier determined?

The tension force of a hanging chandelier is determined by the weight of the chandelier and the strength of the materials used to support it. The tension force must be strong enough to hold the weight of the chandelier without breaking or causing damage to the ceiling.

3. What factors can affect the tension force of a hanging chandelier?

The factors that can affect the tension force of a hanging chandelier include the weight and design of the chandelier, the strength and stability of the ceiling or support structure, and the type and quality of the hardware used to hang the chandelier.

4. Can a hanging chandelier fall due to tension force?

Yes, if the tension force is not properly calculated or the materials used are not strong enough, a hanging chandelier can fall. It is important to ensure that the chandelier is securely hung and that the tension force is within a safe range.

5. How can the tension force of a hanging chandelier be increased?

The tension force of a hanging chandelier can be increased by using stronger and more secure hardware, such as heavy-duty chains or cables, and by distributing the weight of the chandelier evenly across multiple support points rather than relying on a single point of attachment.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
501
Replies
2
Views
5K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
6K
  • Introductory Physics Homework Help
Replies
9
Views
736
  • Introductory Physics Homework Help
Replies
3
Views
9K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
12K
  • Introductory Physics Homework Help
Replies
9
Views
4K
  • Introductory Physics Homework Help
Replies
10
Views
4K
Back
Top