Calculating Tension and Mass in a Hanging Spotlight System

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Homework Help Overview

The discussion revolves around calculating the tension in cables supporting a spotlight and determining the mass of the spotlight. The problem involves analyzing forces acting on the spotlight, given specific angles and one known tension value.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss resolving tensions into x and y components and forming equilibrium equations. Some mention the importance of understanding the angles involved and how they relate to the forces acting on the spotlight. There is also mention of using the law of sines in a triangle formed by the forces.

Discussion Status

Participants have offered various methods for approaching the problem, including resolving forces and using geometric interpretations. There is acknowledgment of a potential misunderstanding regarding the angles involved, and some participants express uncertainty about the correctness of the provided answers.

Contextual Notes

There is a reference to an image that is crucial for understanding the problem setup, which some participants cannot view. Additionally, there are discussions about the angles provided in the problem statement and their implications for solving the problem.

Lancelot59
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Homework Statement



A spotlight hanging from two cables, given angles and other information, (see picture below)

Homework Equations



F=mg

The Attempt at a Solution



I re-wrote my initial equations on a clean sheet of paper.

Basically I just tried to isolate T2 from the X forces equation, and then substitute that into the Y forces equation. The questions asks for the tension in the other cable and the mass of the light.

The only given values are T1 being 140N and the angles of the cables.

The actual answers are supposed to be 18 kg and 91.4N. I'm lost on how to properly solve this.

attachment.php?attachmentid=19299&stc=1&d=1244869472.jpg

 

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Since I can't see the picture as yet, I will try to explain this best as I can.


You are given tensions at specific angles. The weight acts downwards. There is no net movement in the x-direction.


resolve the two tensions into x and y components. All the forces in the x and y directions are in equilibrium. If you are given one tension and the two angles, then when you resolve in the x-direction, you should be able to get the value of the second tension.

Then just form the equilibrium equation in the y-direction and substitute the the value of the second tension to find the unknown.
 
rock.freak667 said:
Since I can't see the picture as yet, I will try to explain this best as I can.


You are given tensions at specific angles. The weight acts downwards. There is no net movement in the x-direction.


resolve the two tensions into x and y components. All the forces in the x and y directions are in equilibrium. If you are given one tension and the two angles, then when you resolve in the x-direction, you should be able to get the value of the second tension.

Then just form the equilibrium equation in the y-direction and substitute the the value of the second tension to find the unknown.

Despite not being able to see the picture, he got it exactly. To the original poster: this is exactly how you do the question. :approve:
 
I added the image as an attached file:

https://www.physicsforums.com/attachment.php?attachmentid=19299&d=1244869468

Thanks, I'll try that solution and come back.EDIT:

It failed again. If my physics are right though then I don't need to worry about it.
The image is rather large so I just linked to it.

http://i35.photobucket.com/albums/d194/Lancelot59/Hosted%20Images/scanned_0006.jpg
 
Last edited by a moderator:
Looks OK.
 
Alright, then the answers on the key must be wrong.

Thanks for your help.
 
I did just double check the answers given and they are correct, if the angles of 30 and 50 are with the vertical and not the horizontal as shown in your drawing.
 
The initial problem gave the angles of 40 and 60 on the left and right respectively, and those angles were made by the cables with the roof.

I just solved for the angles at the horizontal for the free-body diagram...which I did incorrectly. That's what the problem was.

Late night physics apparently doesn't go too well. Thanks for helping me with that.
 
There's a really elegant way to solve this, and it's the same for any angles.
The spotlight is at equilibrium. That means that the net force acting on it is 0.
The net force, is a vectorial sum of the force vectors acting on the spotlight.

If a vector sum is 0, that means that if you add up the vectors, head to tail, you'll get an enclosed shape (The total translation is 0).
That enclosed shape is a triangle whose sides are the vectors T1, T2 and mg. Find the angles of the triangle, and apply the law of sines to them:
Where the sides of the triangle are a, b and c and the angles opposite those sides are A, B and C, then the law of sines states that:
a/sinA = b/sinB = c/sinC

From there, you can find the tension in the second rope and the weight of the spotlight.
 
  • #10
We were taught how to solve these problems geometrically, but I'm personally not big on it. I like my vector components :D.

The method works, but it's more of a personal preference type thing. This was just a result of me making a calculation error.
 

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