Physics Help Forum Stretching of string (and tearing of string)
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 Jul 8th 2017, 04:28 AM #1 Junior Member   Join Date: Jul 2017 Posts: 27 Stretching of string (and tearing of string) Hello everbody! I found the following representation: F = 2ρdl(d²l/dt²) F - the maximum stretching force (force at which the string tears); ρ - density; dl - random differential section of the string; t - time. This is some kind of parallel between WAVE and STRING? I cannot find information in the world biggest search machine, so every single explaining comment will be highly appreciated... Thank you!!!
 Jul 8th 2017, 05:15 AM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 685 Well I don't recognise your equation, where does it come from? Force has dimensions $\displaystyle M{L^{ + 1}}{T^{ - 2}}$ but the right hand side of your equation has different dimensions since for $\displaystyle \rho (dL)\frac{{{d^2}L}}{{d{t^2}}}$ the dimensions are $\displaystyle \left( {M{L^{ - 3}}} \right)\left( L \right)\left( {L{T^{ - 2}}} \right) = M{L^{ - 1}}{T^{ - 2}}$
 Jul 8th 2017, 05:29 AM #3 Junior Member   Join Date: Jul 2017 Posts: 27 Hmm.. What do the different dimensions in the equation (which i quoted) imply?
 Jul 8th 2017, 05:41 AM #4 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 685 Well the missing quantity is L squared or area. So this must be acknowledging the fact that as you stretch a string it gets thinner to make your equation work. Sorry the mathml doesn't seem to have taken on this forum. It used to work.
 Jul 8th 2017, 05:47 AM #5 Junior Member   Join Date: Jul 2017 Posts: 27 So, the equation is right? Then, it is confusing why i cannot find information about it in the textbooks and google... I cannot grasp its derivation.
 Jul 8th 2017, 05:56 AM #6 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 685 No the equation must be missing a term equivalent to area. Density = mass / volume so the equation is basically Newton's Second Law when multiplied by area since the second derivative is acceleration. An alternative would be to transfer the area to the left hand side so it becomes Force / area or stress.
 Jul 8th 2017, 06:07 AM #7 Junior Member   Join Date: Jul 2017 Posts: 27 In the original text, it is mentioned for real that the equation has something to do with Newton's Second Law... but no more explanation. Maybe things are getting more clear for me... But.. Could you, please, write down the equation in the correct way (acknowledging the area)...
 Jul 8th 2017, 06:11 AM #8 Junior Member   Join Date: Jul 2017 Posts: 27 by the way: "ρ" is described as "mechanical" density. I have never heard of "mechanical density" and i don't understand what does it mean in the context of stretching (tearing) of a string.
 Jul 8th 2017, 06:19 AM #9 Physics Team     Join Date: Jun 2010 Location: Naperville, IL USA Posts: 2,269 Your equation should be: $\displaystyle F = \rho A dl (d^2l/dt^2)$ where A is the cross-sectional area of the string. This equation is simply a form of F=ma. It has nothing to do with waves, but rather is an equation that determines how a string stretches dynamically assuming it has zero stiffness (i.e. its spring constant = 0). I suspect that in the equation you gave p, density, is in units of Kg per meter of length of the string, not Kg per cubic meter which is how it's normally defined. With this understanding the units work out, but I have no idea why there is a factor of 2 in your equation.
 Jul 8th 2017, 06:19 AM #10 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 685 Consider a differential length of string, dL, with cross sectional area A. The force in the string = stress x A or stress = Force/A The mass of the differential length = density x volume = density x dL x A So, by Newton's Second Law Force = mass x acceleration = density x dL x A x acceleration. So stress in the string = Force/A = F/A = density x dL x acceleration. But force does not break a string, stress does. The string will break when the maximum stress reaches the breaking stress. I think the factor of 2 is to do with the fact that the maximum stress (or load) is twice the average stress (or load), but it is not clear from your translation what force (load) or stress we are talking about.

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