bbrain
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#1
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Two elastic strings, A and B, stretch by 30 mm and 60 mm respectively when a weight of 4N is attached to each in turn. the strings are hung vertically from the same point, close together, so that when they are unstretched the lower end of A hangs 30mm below that of B. The lower ends are now attached together to a weight of 6N. Find the tension in each string when the weight is hanging in equilibrium.

I have found the two moduli of elasticity in terms of l_A, the natural length of string A. Can I consider A and B to be one string? If so how can I find the tension in each string?

\lambda_A = (2/15)l_A



\lambda_B = (1/15)l_A-2
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ghostwalker
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(Original post by bbrain)
...
You've not specified the stretch in the B string with the 4N attached, so can't check your working so far, and without seeing how you derived it, I don't know if it's correct or not.

It would be difficult to consider A and B as one string unless you really know what you're doing, and even then I don't think you'd gain much, if anything, as you want to find the tension in each.

I'd treat them as two strings. You know that the extension on B will be 30mm more than the extension on A,

And that the sum of their tensions is 6N.

That's enough to allow you to resolve the two tensions.
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bbrain
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(Original post by ghostwalker)
You've not specified the stretch in the B string with the 4N attached, so can't check your working so far, and without seeing how you derived it, I don't know if it's correct or not.

It would be difficult to consider A and B as one string unless you really know what you're doing, and even then I don't think you'd gain much, if anything, as you want to find the tension in each.

I'd treat them as two strings. You know that the extension on B will be 30mm more than the extension on A,

And that the sum of their tensions is 6N.

That's enough to allow you to resolve the two tensions.
I have changed my original post to 60 mm for 4N on B.
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ghostwalker
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(Original post by bbrain)
I have changed my original post to 60 mm for 4N on B.
Noted. I presume you can continue with the info I put in my last post. If you wish me to check anything, just post the working.
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bbrain
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(Original post by ghostwalker)
Noted. I presume you can continue with the info I put in my last post. If you wish me to check anything, just post the working.
T_A=(x_A/l_A)(2\lambda_A/15)



T_B=\frac{x_A+30}{l_A-30}(\frac{1}{15}l_A-2)

Add the two together to make six?
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ghostwalker
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(Original post by bbrain)
T_A=(x_A/l_A)(2\lambda_A/15)



T_B=\frac{x_A+30}{l_A-30}(\frac{1}{15}l_A-2)

Add the two together to make six?
Yep.
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bbrain
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(Original post by ghostwalker)
Yep.
But there are three unknowns.
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ghostwalker
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(Original post by bbrain)
But there are three unknowns.
Well you can substitute for lambda A. And you're left with 2 unknowns.

Not worked it through, but I'd think you should be able to work out the value of \frac{x_A}{l_A} and that's all you need to find the tensions.

Edit: Humm. Doesn't seem to be working out.
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bbrain
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(Original post by ghostwalker)
Well you can substitute for lambda A. And you're left with 2 unknowns.

Not worked it through, but I'd think you should be able to work out the value of \frac{x_A}{l_A} and that's all you need to find the tensions.

Edit: Humm. Doesn't seem to be working out.
How do I substitute for x_A/l_A ?
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ghostwalker
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(Original post by bbrain)
How do I substitute for x_A/l_A ?
I think we're going to have to abandon this method, as I can't see it working out. Sorry about that.

I have however thought of a much easier way to do things.

Consider the two strings hanging side by side.

If we have a weight of 2N on the B string alone then the B string is stretched by 30mm, and the two strings are now level.

If we have a weight of 8N, the B string is stretched by 60mm (carrying 4N), and the A string by 30mm (again carrying 4N), and the two ends are level.

Now since extension is proprotional to weight, as we go from 2N to 8N, the extensions increase linearly.

So, since the weight is 6N, we can work out the extension as a proportion.
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bbrain
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(Original post by ghostwalker)
I think we're going to have to abandon this method, as I can't see it working out. Sorry about that.

I have however thought of a much easier way to do things.

Consider the two strings hanging side by side.

If we have a weight of 2N on the B string alone then the B string is stretched by 30mm, and the two strings are now level.

If we have a weight of 8N, the B string is stretched by 60mm (carrying 4N), and the A string by 30mm (again carrying 4N), and the two ends are level.

And so using the previous moduli of elasticity, I have tension in B = 5N and therefore tension in A is 1N ?

Now since extension is proprotional to weight, as we go from 2N to 8N, the extensions increase linearly.

So, since the weight is 6N, we can work out the extension as a proportion.
So we can assume with a total weight of 6N, each string is carrying 3N, so extension of A=30*(3/4)=22.5mm and extension of B is 60*(3/4)+30=75mm ?

So using the lambda of B, tension in B is 5 N and therefore tension in A is 1 N?
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ghostwalker
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#12
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(Original post by bbrain)
So we can assume with a total weight of 6N, each string is carrying 3N.
'Fraid not.

As A extends from 0 to 30mm (and correspondingly B extends from 30mm to 60mm) the equilibrium load goes linearly from 2N to 8N

So a load of 6N will have a equilibrium position, where the extension of A is

(6-2)/(8-2) * 30 mm = 20mm.

And we can now work out the tensions.


This method works, but is a bit messy, I think a better method would be:

Spoiler:
Show

Since extension is proportional load, and the strings are joined at each end, then if we let x be the extension of A, we have:

load = 4x/30 + 4(x+30)/60

We now equate this to 6, and solve for x, the extension of A.


Sorry about all the different suggestions, the technique wasn't as obvious as I'd assumed.

But you now have two methods that work.
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