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Old Apr 11th 2019, 04:04 AM   #1
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Cool Young's Modulus: 'how much force can you apply before it breaks' help?

The question is: "A rope has tensile Young's modulus of 0.64 GPa and a breaking strain of 0.021.

The rope is 50m long and has a diameter of 0.015m. What is the maximum tensile force that can be applied to the rope before it breaks in N?"

My answer was wrong, and I don't know where to go from here.

This is my working out, there is an 'X' next to the wrong answer: https://imgur.com/a/zdDCkez

Is the breaking strain not tensile strain? If so, what is it? Or maybe the F= tensile stress x A is not the one to use?

Thanks for any help 😭😭😭

Last edited by astupidummydumstupid; Apr 11th 2019 at 04:07 AM.
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Old Apr 11th 2019, 05:19 AM   #2
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You are doing the arithmetic the hard way.
But I think you method seems basically sound, unless you have miseed out some information, the 50m rope length is irrelevent.

Youngs modulus = stress/strain

So 64 x 10^7 = stress/strain (note I have got rid of the decimals)

64 x 10^7 21 x 10^-3 = stress in Pascal. = 64 x 21 x 10^4

Force = stress x cross sectional area

F = 64 x 21 x 10^4 x pi x D^2 /4

F = 64 x 21 x 10^4 x pi x 3/2 x3/2 x 10^-2 x 10^-2 /4

= 4 x 21 x pi x 9

See you can almost do it in your head.
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Old Apr 11th 2019, 06:10 AM   #3
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Originally Posted by studiot View Post
You are doing the arithmetic the hard way.
But I think you method seems basically sound, unless you have miseed out some information, the 50m rope length is irrelevent.

Youngs modulus = stress/strain

So 64 x 10^7 = stress/strain (note I have got rid of the decimals)

64 x 10^7 21 x 10^-3 = stress in Pascal. = 64 x 21 x 10^4

Force = stress x cross sectional area

F = 64 x 21 x 10^4 x pi x D^2 /4

F = 64 x 21 x 10^4 x pi x 3/2 x3/2 x 10^-2 x 10^-2 /4

= 4 x 21 x pi x 9

See you can almost do it in your head.

Hi! Thank you for your reply. That is the correct answer!!

Although I have to say I don't know how those numbers work. I am confused where the pi x D^2 /4 comes in, and the 3/2s? And divide it by 4?

If it is stress x cross sectional area, why isn't it just the stress in pascal x cross sectional area:

13440000 x 2pix7.5x10^4

but this = 633345N so obviously that's no good haha.

How would you go about getting those other numbers for the 2400N, if you don't mind me asking? Furthermore, how did the above numbers get simplified for 9 in the final line?

Again, thank you for your comment
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