Solving for Tension: A chandelier is suspended by two chains

[ttpp1124]

Homework Statement

A 45 kg chandelier is suspended by two chains of lengths 5 m and 8 m attached to two points in the ceiling 11 m apart. Find the tension in the 5 m rope. Please include a neat diagram.

Relevant Equations

N/A

I've begun by drawing out the diagram. Since they want the tension in the 5m rope, I've broken it down into two vectors. I want to use T = mg + ma, but I don't think that's right because I don't have acceleration. Is there another formula I can use instead? Thanks!

 

[kuruman]

Homework Statement:: A 45 kg chandelier is suspended by two chains of lengths 5 m and 8 m attached to two points in the ceiling 11 m apart. Find the tension in the 5 m rope. Please include a neat diagram.
Relevant Equations:: N/A

View attachment 258498

I've begun by drawing out the diagram. Since they want the tension in the 5m rope, I've broken it down into two vectors. I want to use T = mg + ma, but I don't think that's right because I don't have acceleration. Is there another formula I can use instead? Thanks!

Is the system accelerating?

 

[ttpp1124]

Is the system accelerating?

No. I should substitute acceleration with 0, correct?

 

[kuruman]

Correct. What would Sir Isaac Newton have said when the acceleration is zero?

 

[ttpp1124]

An object with an acceleration of 0 has no net force on it but there may be forces acting on it that cancel out.
Is T = mg my formula then?

 

[kuruman]

An object with an acceleration of 0 has no net force on it but there may be forces acting on it that cancel out.
Is T = mg my formula then?

That's what Sir Isaac would have said. There are two tensions acting on the mass. Which one is T in your expression?

 

[ttpp1124]

Well, the question is asking for the tension in the 5m rope. So, T=(45)(9.8). That would work for both the 5m and 8m rope. How do I change it such that the equation applies to the 5m rope?

 

[kuruman]

Well, the question is asking for the tension in the 5m rope. So, T=(45)(9.8). That would work for both the 5m and 8m rope. How do I change it such that the equation applies to the 5m rope?

You didn't answer my question. There are two ropes which means that that there two tensions which are not the same. Call then T1 and T2. Which of these, if any, is equal to mg? You may have to review vector addition.

 

[ttpp1124]

You didn't answer my question. There are two ropes which means that that there two tensions which are not the same. Call then T1 and T2. Which of these, if any, is equal to mg? You may have to review vector addition.

Let T1 - 5
Let T2 - 8

T1 = mg

 

[kuruman]

You can't "let" the tensions be what you want. Why not "let" T1 = 12 and T2 = 1? Unlike some people these days, you cannot make up your own reality. This is science, man! The length of each chain is relevant only in so far as the angle they make with resprect to the vertical. If the 5 m chain is lengthened to 8 m but attached farther up the ceiling so that the angle with the vertical is the same, the tension in it will not change. You have to find what the tensions are noting that each one can have only one value given the circumstances of the problem. Hint: The two tensions add as vectors. What is the sum of their horizontal components? What is the sum of their vertical components?

 

[haruspex]

T = mg

Forces are vectors, i.e. they have magnitude and direction. You can only compare their magnitudes usefully if they are in parallel directions.
You need to consider components that are in the same or opposite directions.