Young's Modulus Question

[rosi123]

A cyclist collides with a tree, and the collision creates an intense compressive stress on the cyclist’s left humerus (upper arm bone), causing a fracture. Which of the following changes in the initial conditions most likely could have prevented the fracture?

A. A decrease in his velocity by a factor of two
B. A decrease in his mass by a factor of two
C. An increase in his mass by a factor of two
D. A decrease in the cross-sectional area of his humerus by a factor of two

The answer is A.

I don't understand where why D isn't correct. Doesn't decreasing the cross-sectional area lead to a reduced stress/strain? Also, does the mass not play a role in the Force?

Sooo confused! Thanks in advance for anyone that can help clarify!

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[rosi123]

And this is the explanation. This question is from a princeton review practice test, btw. TPR #4.

A. The cyclist's kinetic energy just before the impact will be "used" to cause damage to the tree and to the cyclist. Decreasing this kinetic energy will result in a less violent collision. Because KE = (1/2)mv2, a decrease in the cyclist's velocity would have the greatest effect (since v is squared) in lowering his KE.

 

[The_Sunny_Doc]

Velocity is a squared term, so decreasing that number has the greatest impact on the change in energy the bone would have to withstand

KE = .5mv^2

8 = .5(1)(4)^2
2 = .5(1)(2)^2
.5 = .5(1)(1)^2

Not sure about Young's modulus in this problem, but there's no squared term in any of those formulas.

 

[monkeyvokes]

A cyclist collides with a tree, and the collision creates an intense compressive stress on the cyclist’s left humerus (upper arm bone), causing a fracture. Which of the following changes in the initial conditions most likely could have prevented the fracture?

A. A decrease in his velocity by a factor of two
B. A decrease in his mass by a factor of two
C. An increase in his mass by a factor of two
D. A decrease in the cross-sectional area of his humerus by a factor of two

The answer is A.

I don't understand where why D isn't correct. Doesn't decreasing the cross-sectional area lead to a reduced stress/strain? Also, does the mass not play a role in the Force?

Sooo confused! Thanks in advance for anyone that can help clarify!

As stated, velocity is squared so it has the biggest impact. mass is also directly proportional but not squared.

Strain is force/area. Area is inversely proportional to strain so decreasing the cross-sectional area of his humerus by a factor of two will double the strain on it and not help reduce the chances of a fracture.

 

[The_Sunny_Doc]

Strain is force/area. Area is inversely proportional to strain so decreasing the cross-sectional area of his humerus by a factor of two will double the strain on it and not help reduce the chances of a fracture.

👍

 

[eudovcic]

Is strain = F/A or stress?

 

[tuhtles]

Is strain = F/A or stress?

Stress = F/A

For young's modulus, Strain = change in length/original length

Young's Modulus (Y) is defined as stress/strain. It refers to tensile and compressive stress where as shear's modulus refers bending. The greater the shear's modulus, the harder it is to bend. Same goes with young's modulus.

 

[rosi123]

This was my reasoning for justifying why D would be correct. When reducing the area, the stress component of Young's Modulus would increase. Therefore, wouldn't the bone be less likely to fracture because it could withstand a greater amount of stress before causing that same amount of strain?

I understand why A is correct. I just don't think I fully understand Young's Modulus in relation to the likelihood of the bone fracturing.

 

[The_Sunny_Doc]

When you increase cross sectional area of the bone, you are reducing the stress, but you're also reducing the strain on the object, since YM is a constant. YM is an intrinsic property of the material, unaffected by the dimensions of the material. Only temperature changes it.

Perhaps the YM answer was only meant to be a distracter. It doesn't have a squared term, so that's the key piece of info. Being able to visualize the equations helps.

 

[Heday]

This was my reasoning for justifying why D would be correct. When reducing the area, the stress component of Young's Modulus would increase. Therefore, wouldn't the bone be less likely to fracture because it could withstand a greater amount of stress before causing that same amount of strain?

I understand why A is correct. I just don't think I fully understand Young's Modulus in relation to the likelihood of the bone fracturing.

What do you think is easier to break, a bone with the cross sectional area of a toothpick or a bone with the cross sectional area similar to that of a soda can? This is an extreme comparison, but it shows you how cross sectional area can affect the "fragility" of objects. Young's modulus is a constant (like specific heat).

The passage talks about how the elastic energy density under the entire curve would be the energy required to fracture the bone. So here we're relating how energy of the object can affect a fracture... more energy = more likely to fracture.

KE = (1/2)mv^2

A.) Decrease energy by a factor of 4
B.) Decrease energy by a factor of 2
C.) Increase energy by a factor of 2
D.) No affect in energy, but makes the bone more "fragile"--> a toothpick is more fragile than a similar wooden stick with the cross sectional area of a soda can

So it's apparent that only A and B can help reduce the chances of fracture.

 

[monkeyvokes]

👍

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[JetsFan]

Stress = F/A

For young's modulus, Strain = change in length/original length

Young's Modulus (Y) is defined as stress/strain. It refers to tensile and compressive stress where as shear's modulus refers bending. The greater the shear's modulus, the harder it is to bend. Same goes with young's modulus.

What's the difference between the three moduli, bulk, shear, and young?

 

[monkeyvokes]

What's the difference between the three moduli, bulk, shear, and young?

I think that question is really worth a google cos those are probably important to know. They are essentially stress/strain, but each of the three have slightly different formulas (same general idea though).

 

[JetsFan]

I think that question is really worth a google cos those are probably important to know. They are essentially stress/strain, but each of the three have slightly different formulas (same general idea though).

I tried searching it, however the results were fart too complex.

 

[monkeyvokes]

I tried searching it, however the results were fart too complex.

the youngs modulus is stress over strain. Y=stress/strain
stress: Force/Area (the amount of stress applied on an object by a force)
strain: deltaL/L (the deformation that occurs following application of said stress)

The shear modulus is also stress over strain, stress still force over area. This time though you have a shearing force (like sliding your hands flat across each other)
Strain: deltaX/L (the amount an object starts to bend over due to shearing stress, dividing by its height)


The bulk modulus is once again stress/strain. This is useful for fluids because pressure is applied perpendicularly to all surfaces around on object. The stress is the force over area, or more simply the change in pressure that occurs from the stress.
strain: delta V/Vo (The change in volume due to applied stress / original volume)

Hope that helps a bit