Young's Modulus: Calculate from Length, Radius, Elongation & Weight

  • Thread starter Thread starter monkeymak7
  • Start date Start date
  • Tags Tags
    Modulus
AI Thread Summary
To calculate Young's modulus for a wire, the formula E = (F/A₀)/(ΔL/L₀) is used, where F is the force applied, A₀ is the cross-sectional area, ΔL is the elongation, and L₀ is the original length. In this case, the force is calculated as F = mg, resulting in 117 N for a 12 kg weight. The cross-sectional area A₀ is determined using the radius, yielding approximately 2.8 x 10^-8 m². Substituting these values into the Young's modulus equation gives a result of 0.0165 GPa. This calculation illustrates the relationship between stress and strain in elastic materials.
monkeymak7
Messages
4
Reaction score
0

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
 
Physics news on Phys.org
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:

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

<br /> 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}}.<br />

I believe the answer to your question is:

<br /> Y = \frac{Y_{0}}{226}<br />
 
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...
 
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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top