Tensile stress, radius and Young's modulus

AI Thread Summary
To determine the radius of an aluminum cable under a tensile force of 15,000 N and a maximum stretch of 1 mm, the Young's modulus of aluminum (6.9 x 10^10 N/m^2) is essential. The relevant equation is Y = (F * L) / (A * ΔL), where A is the cross-sectional area. The tensile stress can be calculated using the formula F/A, which can be expressed as F/(π * R^2) for a circular cross-section. Understanding that ΔL represents the change in length (1 mm) is crucial for solving the problem. Proper application of these equations will yield the required radius of the wire.
Gewitter_05
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Homework Statement


An Aluminum cable of length 3.5m has 15,000 N tensile force acting on it if the wire is only allowed to be stretched by 1mm before it breaks,
What must be the radius of the wire if the Young's modulus of Al is 6.9 x 10^10 N/m^2?

I am also supposed to find the tensile stress, but I am sure I could find that myself after finding out the radius


Homework Equations


F/A or F/pi*R^2


The Attempt at a Solution


I haven't tried attempting the solution because I am not sure about the equation to use. We didn't learn about tensile stress in class or Young's modulus. I tried looking in my book and those are the equations I found, but most of the equations have a diameter or a radius already given.
 
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Did you read the section regarding Young's modulus or modulii of elasticity in your textbook? That will help you solve this problem.
 
I did look in my book and I my old notes. I just found my old notes because this is my second time taking Physics I. I thought there might be something in there. The book also shows that F/A = delta L/L. In my notes it mentions that delta L is the change in length over the original length. So would I say that delta L is the original length plus 1 mm (that I have to change to m) over the 3.5m? Then multiply that by the Young's modulus and that would be tensile stress? Although I still don't know how to find out the radius.
 
Delta represents the change. Do you think that Delta L represents L + 1 mm?
 
That's what I am thinking, but I don't know. I don't really have much else to go on. There might be more to go on and I just am not seeing it.
 
There's less going on. Delta L represents the change in the length L.
 
Gewitter_05 said:
I did look in my book and I my old notes. I just found my old notes because this is my second time taking Physics I. I thought there might be something in there. The book also shows that F/A = delta L/L. In my notes it mentions that delta L is the change in length over the original length. So would I say that delta L is the original length plus 1 mm (that I have to change to m) over the 3.5m? Then multiply that by the Young's modulus and that would be tensile stress? Although I still don't know how to find out the radius.

Have you come across the equation Y=\frac{FL}{AΔL}. If not, I suggest looking for it in your textbook or some other reliable book. This helps you solve the problem. (Y is the young's modulus here).

And as Steam King pointed out, ΔL is the change in length that the wire can tolerate here.
 
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