Hanging Chandelier (A question about Tension Force)

AI Thread Summary
The discussion revolves around solving for the tension T1 in a chandelier supported by two cables at angles theta1 and theta2. The user initially set up the equations for net forces but struggled to eliminate T2 from their calculations. After some confusion, they realized their mistake in the vertical force equation, which led to an incorrect expression for T1. Correcting this, they derived the right formula for T1, which is T1 = mg / (sin(theta1) + cos(theta1) * tan(theta2)). The final expression successfully expresses T1 in terms of mass, angles, and gravitational acceleration without depending on T2.
BlakeGriffin
Messages
8
Reaction score
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
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?
 
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:
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top