Tensile stress, radius and Young's modulus

Click For Summary
SUMMARY

The discussion centers on calculating the radius of an aluminum cable subjected to a tensile force of 15,000 N, with a maximum allowable stretch of 1 mm before breaking. The Young's modulus for aluminum is given as 6.9 x 1010 N/m2. The relevant equation for tensile stress is F/A, where A is the cross-sectional area, calculated as πR2. Participants emphasize the importance of understanding the relationship between tensile stress, Young's modulus, and the change in length (ΔL) to solve the problem effectively.

PREREQUISITES
  • Understanding of Young's modulus (Y) and its application in tensile stress calculations.
  • Familiarity with the formula for tensile stress: F/A = Y(ΔL/L).
  • Basic knowledge of geometry, specifically the area of a circle (A = πR2).
  • Ability to convert units, particularly from millimeters to meters.
NEXT STEPS
  • Study the equation Y = FL/(AΔL) to understand its components and application.
  • Learn how to derive the radius (R) from the tensile stress formula.
  • Explore the concept of tensile stress in materials science to gain a deeper understanding.
  • Review textbook sections on Young's modulus and its significance in material properties.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and material properties, as well as educators seeking to clarify concepts related to tensile stress and Young's modulus.

Gewitter_05
Messages
9
Reaction score
0

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.
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 29 ·
Replies
29
Views
8K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
20
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
6K
  • · Replies 3 ·
Replies
3
Views
3K