How do I calculate the tension in cable 1 without using the tension in cable 2?

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In summary, because the chandelier was hung near the walls, instead of directly above the chandelier, there is a lesser amount of tension on the cables, which in turn reduces the amount of force that is needed to hold the chandelier in place.
  • #1
ijd5000
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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 θ1 with the ceiling. Cable 2 has tension T2 and makes an angle of θ2 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, θ1, and θ2, as well as the magnitude of the acceleration due to gravity g.


i have for my Forces:

F(x) T2 cos ( θ2)-T1cos(θ1)=0
F(y) T2 sin(θ2)+T1 sin(θ1)-gm=0

not sure how to describe T1 without using T2
 
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  • #2
These are two equations in two unknowns, T_1 and T_2.
If you write them as
T_1cos(v_1)=T_2cos(v_2)
T_1sin(v_1)-mg=-T_2sin(v_2)
then dividing the one equation with the other yields:
[tex]\frac{T_{1}\sin\theta_{1}-mg}{T_{1}\cos\theta_{1}}=-\tan\theta_{2}[/tex]

Hurray, T_2 has disappeared! :smile:
Now, rearrange the last equation, so as to solve for T_1
 
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  • #3
arildno said:
These are two equations in two unknowns, T_1 and T_2.
If you write them as
T_1cos(v_1)=T_2cos(v_2)
T_1sin(v_1)-mg=-T_2sin(v_2)
then dividing the one equation with the other yields:
[tex]\frac{T_{1}\sin\theta_{1}-mg}{T_{1}\cos\theta_{1}}=-\tan\theta_{2}[/tex]

Hurray, T_2 has disappeared! :smile:
Now, rearrange the last equation, so as to solve for T_1

It's ok to combine equations from different directions? I thought you have to keep x and y separate until you add the vectors.
 

Related to How do I calculate the tension in cable 1 without using the tension in cable 2?

1. What is a tension chandelier problem?

A tension chandelier problem is a physics problem that involves calculating the tension forces within the cables or wires supporting a chandelier or similar hanging object.

2. How do you solve a tension chandelier problem?

To solve a tension chandelier problem, you will need to use the principles of static equilibrium and apply them to the forces acting on the chandelier. This involves setting up equations and solving for the tension forces in each cable or wire.

3. What information is needed to solve a tension chandelier problem?

To solve a tension chandelier problem, you will need to know the weight of the chandelier, the angles at which the cables or wires are attached to the ceiling, and any other forces acting on the chandelier such as wind or movement.

4. What are some common mistakes when solving a tension chandelier problem?

Some common mistakes when solving a tension chandelier problem include forgetting to include all forces, using incorrect angles or measurements, and not properly setting up and solving the equations for static equilibrium.

5. How is a tension chandelier problem relevant to real-life situations?

Tension chandelier problems are relevant to real-life situations where objects are suspended by cables or wires, such as suspension bridges, cranes, and even carnival rides. Understanding the tension forces in these situations is crucial for ensuring the safety and stability of the structures.

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