Calculating Tension and Mass in a Hanging Spotlight System

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In summary, the person struggled to solve for the tensions in the second cable, and then found that they could if they found the angles of the triangle and solved for the tensions.
  • #1
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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  • #2
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.
 
  • #3
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:
 
  • #4
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:
  • #5
Looks OK.
 
  • #6
Alright, then the answers on the key must be wrong.

Thanks for your help.
 
  • #7
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.
 
  • #8
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.
 
  • #9
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.
 

FAQ: Calculating Tension and Mass in a Hanging Spotlight System

What is the mass of a hanging spotlight?

The mass of a hanging spotlight can vary depending on the type and size of the spotlight. Generally, a small spotlight can weigh around 5-10 pounds, while larger spotlights can weigh up to 50 pounds or more.

Why is the mass of a hanging spotlight important?

The mass of a hanging spotlight is important because it affects how the spotlight is hung and supported. A heavier spotlight may require stronger support and more secure hanging methods to prevent accidents or damage to the spotlight.

How is the mass of a hanging spotlight measured?

The mass of a hanging spotlight is typically measured using a scale or a balance. The spotlight is placed on the scale and the weight is recorded in either pounds or kilograms.

Does the mass of a hanging spotlight affect its brightness?

No, the mass of a hanging spotlight does not directly affect its brightness. The brightness of a spotlight is determined by the type and power of the light source, not its mass. However, a heavier spotlight may require more power to be hung and may affect the overall weight load of the hanging system.

Can the mass of a hanging spotlight change over time?

Yes, the mass of a hanging spotlight can change over time due to factors such as wear and tear, addition of accessories, or changes in the light source. It is important to regularly check the mass of a hanging spotlight to ensure proper support and safety.

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