Young’s modulus

  • Thread starter monkeymak7
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  • #1

Homework Statement



A wire, of length 250 centimeters, and radius 0.3 millimeters is stretched by hanging on a weight of 12 kilograms, and the elongation produced is 8 millimeters. Calculate the value of Young’s modulus for the wire.


Homework Equations





The Attempt at a Solution



I have no idea because we never went over this...please please guide me in the right direction
 

Answers and Replies

  • #2
2,981
5
In atomic units, where the unit of length is 1 bohr radius ([itex]a_{0} \equiv \hbar^{2}/(m_{e} \, k_{0} \, e^{2})[/itex], the unit of mass is the electron mass [itex]m_{e}[/itex] and the unit of energy is 1 hartree ([itex]E_{0} = m_{e} \, (k_{0} \, e^{2})^{2}/\hbar^{2}[/itex]), the unit for Young's modulus (pressure) is:

[tex]
[Y] = [F]/[A] = \mathrm{T}^{-2} \, \mathrm{L}^{-1} \, \mathrm{M} = \frac{[E_{0}]}{[a_{0}]^{3}}
[/tex]

[tex]
Y_{0} = \frac{27.2 \, \mathrm{eV} \times \frac{1.602 \times 10^{-19} \, \mathrm{J}}{1 \, \mathrm{eV}}}{(5.29 \times 10^{-11} \, \mathrm{m})^{3}}.
[/tex]

I believe the answer to your question is:

[tex]
Y = \frac{Y_{0}}{226}
[/tex]
 
  • #3
Where are you getting all the values for this equation? I apologize, but this is like reading a different language!! I'm just not understanding where 27.2, 1.602, and 5.29 are taken from...
 
  • #4
PhanthomJay
Science Advisor
Homework Helper
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Look up the stress strain relationships for elastic materials where the stress (P/A) is linearly proportional to the strain (elongation/length). Watch your units.
 
  • #5
I think I've figured it out!! Thanks for your help. I posted another question and still haven't had any responses. I've been trying to research this question for a few days and still haven't a clue as to what needs to be done. Do any of you know what the heck this question is asking and how I go about solving it. (The book I have for my class has nothing like this in it, nor have I seen anything like this throughout the chapters I have studied).


A balance, the beam of which is 30 centimeters long and weighs 40 grams, is deflected through 1° by an excess of 1 milligram in one of the pans. What is the distance of the center of gravity of the beam below the central pivot?
 
  • #6
18
0
Young’s Modulus, E= tensile stress/tensile strain= (F/Ao)/(∆L/Lo)
F= force applied to object= ma= mg= (12kg)(9.81m/s2)= 117 N
Ao= cross-sectional area through which force is applied= πr2= 2.8 x 10^-8
∆L= amount by which length of object changes= 0.008m
Lo= original length of the object= 0.25m

So, E= (mg/Ao)/(∆L/Lo)= (117N/2.8x10^-8)^2/(0.008m)(0.25m)= 0.0165 Gpa,

Gpa=gigapascals
 

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