# Further Mechanics 1: elastic strings problem

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#1
Bit stuck on this question because I think I haven't got my basic assumptions correct, or the question itself is flawed

'A light elastic string of natural length 0.2m has its ends attached to 2 fixed points A and B which are on the same horizontal level with AB = 0.2m. A particle mass 5kg attached to string at point P, AP = 0.15m. System released and P hangs in equilibrium below AB with APB = 90 degrees. If BAP = theta, show that ratio of extension of AP and BP is
[4cos(theta) - 3]/[4sin(theta) - 1]'

Link to solutions with diagram (QS 7 pg 7): https://activeteach-prod.resource.pe...fm1_ex3mix.pdf

I always thought if it's the same string, the tensions must be equal. But if you look at the second part of the question, it says
(c = cos(theta), s = sin(theta)):
show that c(4s-3) = 3s(4s-1) and this isn't true for 45 degrees, i.e. theta is not 45 degrees. This means the two tensions have to be different (resolving horizontally shows this), which is impossible?

EDIT think ive figured it out
can confirm I was making wrong assumptions. The reason we asume the tension is the same, is that for a molecule inside the string, it must also be in equilibrium. So the molecule should feel forces equally to left and right along the string, which is usually provided by tension in the string. But then it's not true if you attach a particle to the midpoint and let the string hang vertically, since the weight acts differently on each half of the string. I think that it's a bit circular but it should be the case that if there is symmetry, the tensions must be the same, just as if the tensions are the same, there is symmetry.
Last edited by DeBrevitateVitae; 1 year ago
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1 year ago
#2
(Original post by DeBrevitateVitae)
I always thought if it's the same string, the tensions must be equal. But if you look at the second part of the question, it says
(c = cos(theta), s = sin(theta)):
show that c(4s-3) = 3s(4s-1) and this isn't true for 45 degrees, i.e. theta is not 45 degrees. This means the two tensions have to be different (resolving horizontally shows this), which is impossible?
Tensions would be equal IF the particle was free to slide on the string and contact was smooth.

In Q7, the particle is attached to the string. It's not free to slide smoothly, and the tensions don't have to be the same.

Also the lack of symmetry, AP does not equal PB, means theta can't be 45 degrees.
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#3
(Original post by ghostwalker)
Tensions would be equal IF the particle was free to slide on the string and contact was smooth.

In Q7, the particle is attached to the string. It's not free to slide smoothly, and the tensions don't have to be the same.
Thank you, I didn't know that before! The new textbooks separate wording and assumptions in modelling in another chapter entirely and it's in single maths so it doesn't cover further stuff.
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