# Young’s modulus

## 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.

## The Attempt at a Solution

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

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

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

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

$$Y = \frac{Y_{0}}{226}$$

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...

PhanthomJay
Homework Helper
Gold Member
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.

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?

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