Young's Modulus Question

In summary, the maximum force that can be exerted on the femur bone is 3.90 X 10^5 N and the bone will shorten by 2.08 X 10^-5 m when a force of this magnitude is applied. The equations used are tensile stress = F/A and tensile strain = delta L / Initial L.
  • #1
yb1013
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Homework Statement



Assume that Young's Modulus for bone is 1.50 X 10^10 N/m2 and that the bone will fracture if more than 1.50 X 10^8 N/m2 is exerted.

(a) What is the maximum force that can be exerted on the femur bone in the leg if it has a minimum effective diameter of 2.60 cm?

(b) If a force of this magnitude is applied compressively, by how much does the 24.0 cm long bone shorten?

Homework Equations



Y = tensile stress / tensile strain

tensile stress = F/A
tensile strain = delta L / Initial L

The Attempt at a Solution



I understand the concept, but i can't figure out where to plug the numbers into. Since I don't have any lengths for part a, I guess you wouldn't use the entire young's equation, but what exactly would I use?

Any help would be great! Thanks!
 
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  • #2
never mind, I figured it out
 
  • #3


I would approach this problem by first identifying the relevant equations and variables. In this case, we are given Young's Modulus (Y) for bone, the maximum stress that will cause fracture, and the minimum effective diameter of the femur bone. The unknowns are the maximum force that can be exerted and the amount of compression that will occur.

To solve part (a), we can use the formula for tensile stress: stress = force/area. We know the maximum stress that will cause fracture, so we can set up the equation:

1.5 x 10^8 N/m^2 = F / (pi x 0.026 m)^2

Solving for F, we get a maximum force of approximately 1,508 N.

For part (b), we can use the formula for tensile strain: strain = (change in length) / (initial length). We are given the initial length of the bone (24 cm), so we just need to find the change in length.

Since the bone is being compressed, the change in length will be negative. We can use the formula for Young's Modulus: Y = stress/strain. Rearranging for strain, we get: strain = stress / Y.

Plugging in the values for stress and Young's Modulus, we get:

strain = (1.5 x 10^8 N/m^2) / (1.5 x 10^10 N/m^2) = 0.01

This means that the bone will shorten by 0.01 times its initial length. Multiplying this by the initial length (24 cm), we get a compression of 0.24 cm.

In summary, to solve this problem as a scientist, we first identify the relevant equations and variables, and then use algebra and unit conversions to solve for the unknowns. We also pay attention to units and make sure they are consistent throughout the problem.
 

What is Young's Modulus?

Young's Modulus, also known as the modulus of elasticity, is a measure of the stiffness of a material. It quantifies the relationship between stress (force per unit area) and strain (deformation or change in shape) of a material under applied force.

How is Young's Modulus calculated?

The formula for Young's Modulus is E = σ/ε, where E is the modulus of elasticity, σ is the stress applied to the material, and ε is the resulting strain. It is typically measured in units of Pascals (Pa) or Megapascals (MPa).

What are some examples of materials with high and low Young's Modulus?

Materials with high Young's Modulus, such as steel and diamond, are very stiff and difficult to bend or deform. Materials with low Young's Modulus, such as rubber and foam, are more flexible and easier to bend or deform.

How does Young's Modulus affect the behavior of a material?

The higher the Young's Modulus of a material, the stiffer and more rigid it will be. This means it will resist deformation and be able to withstand greater amounts of stress before breaking. Lower Young's Modulus materials will be more flexible and able to bend or stretch without breaking.

How is Young's Modulus used in engineering and manufacturing?

Young's Modulus is an important factor in designing and creating structures and products that can withstand the forces and stresses they will be subjected to. It is used to select appropriate materials for specific applications and to ensure the safety and reliability of structures and products.

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