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Young's Modulus
Learning Goal: To understand the meaning of Young's modulus, to perform some real-life calculations related to stretching steel, a common construction material, and to introduce the concept of breaking stress.
Hooke's law states that for springs and other "elastic" objects

F=k\Delta x,

where F is the magnitude of the stretching force, \Delta x is the corresponding elongation of the spring from equilibrium, and k is a constant that depends on the geometry and the material of the spring. If the deformations are small enough, most materials, in fact, behave like springs: Their deformation is directly proportional to the external force. Therefore, it may be useful to operate with an expression that is similar to Hooke's law but describes the properties of various materials, as opposed to objects such as springs. Such an expression does exist. Consider, for instance, a bar of initial length L and cross-sectional area A stressed by a force of magnitude F. As a result, the bar stretches by \Delta L.

Let us define two new terms:

  • Tensile stress is the ratio of the stretching force to the cross-sectional area:

    {\rm stress}=\frac{F}{A}.

  • Tensile strain is the ratio of the elongation of the rod to the initial length of the bar:

    {\rm strain}=\frac{\Delta L}{L}.

It turns out that the ratio of the tensile stress to the tensile strain is a constant as long as the tensile stress is not too large. That constant, which is an inherent property of a material, is called Young's modulus and is given by

Y=\frac{F/A}{\Delta L/L}.

Part A  
What is the SI unit of Young's modulus?
Hint A.1 Look at the dimensions

Hint not displayed

ANSWER:
 Pa 
 pascal 
 Pascal 
 pascals 
 Pascals 
 kg/(m*s^2) 
 N/(m^2) 
Part B  
Consider a metal bar of initial length L and cross-sectional area A. The Young's modulus of the material of the bar is Y. Find the "spring constant" k of such a bar for low values of tensile strain.
Hint B.1 Use the definition of Young's modulus

Hint not displayed

Express your answer in terms of Y, L, and A.
ANSWER:
  k =  Y*A/L 
Part C  
Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are connected end to end, so that the resultant wire has length 10L. What is the "spring constant" of the resulting wire?
Hint C.1 The spring constant

Hint not displayed

ANSWER:
0.1k
k
10k
100k
Part D  
Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are slightly twisted together, so that the resultant wire has length L and is ten times as thick as each individual wire. What is the "spring constant" of the resulting wire?
Hint D.1  

Hint not displayed

ANSWER:
0.1k
k
10k
100k
Part E  
Ten identical steel wires have equal lengths L and equal "spring constants" k. The Young's modulus of each wire is Y. The wires are connected end to end, so that the resultant wire has length 10L. What is the Young's modulus of the resulting wire?
ANSWER:
0.1Y
Y
10Y
100Y
Part F  
Ten identical steel wires have equal lengths L and equal "spring constants" k. The Young's modulus of each wire is Y. The wires are slightly twisted together, so that the resultant wire has length L and is ten times as thick as the individual wire. What is the Young's modulus of the resulting wire?
ANSWER:
0.1Y
Y
10Y
100Y
By rearranging the wires, we create a new object with new mechanical properties. However, Young's modulus depends on the material, which remains unchanged. To change the Young's modulus, one would have to change the properties of the material itself, for instance by heating or cooling it.
Part G  
Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's modulus of the steel is Y=2.0 \times 10^{11} pascals. How far ( Delta L) would such a string stretch under a tension of 1500 newtons?
Use two significant figures in your answer. Express your answer in millimeters.
ANSWER:
  Delta L =  15   \rm {mm}
Steel is a very strong material. For these numeric values, you may assume that Hooke's law holds. However, for greater values of tensile strain, the material no longer behaves elastically. If the strain and stress are large enough, the material deteriorates. The final part of this problem illustrates this point and gives you a sense of the "stretching limit" of steel.
Part H  
Although human beings have been able to fly hundreds of thousands of miles into outer space, getting inside the earth has proven much more difficult. The deepest mines ever drilled are only about 10 miles deep. To illustrate the difficulties associated with such drilling, consider the following: The density of steel is about 7900 kilograms per cubic meter, and its breaking stress, defined as the maximum stress the material can bear without deteriorating, is about 2.0 \times 10^9 pascals. What is the maximum length of a steel cable that can be lowered into a mine? Assume that the magnitude of the acceleration due to gravity remains constant at 9.8 meters per second per second.
Hint H.1 Why does the cable break?
The cable breaks because of the stress exerted on it by its own weight. At the moment that the breaking stress is reached, the stress at the top of the cable reaches its maximum, and the material begins to deteriorate.

Introduce an arbitrary cross-sectional area of the cable (which will cancel out of the final answer). The mass of the cable below the top point can be found as the product of its volume and its density. Use this to find the force at the top that will lead to the breaking stress.

Part H.2 Find the stress in the cable

Part not displayed

Use two significant figures in your answer, expressed in kilometers.
ANSWER:
 26   \rm {km}
This is only about 16 miles, and we have assumed that no extra load is attached. By the way, this length is small enough to justify the assumption of virtually constant acceleration due to gravity. When making such assumptions, one should always check their validity after obtaining a result.
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