Solve Shear Stress Problem: Cartilage Deformed at 7.4°

In summary, the conversation discusses the calculation of deformation angle on a knee cartilage due to a force of 904N at an angle of 12 degrees. The shear modulus is given as 12MPa and the cartilage area is 11cm^2. The attempted solution involves using the equation for shear modulus and calculating stress and strain, but the final angle calculation is incorrect. The correct approach involves using the given shear strain and solving for the angle directly.
  • #1
M_J
5
0

Homework Statement



The force on a knee cartilage can cause a large force to deform it. The cartilage area is 11cm^2 and the shear modulus is 12MPa. Max force at impact is 904N at an angle of 12 degrees with the cartilage, through what angle (in degrees) will the cartilage be deformed?

Homework Equations



http://session.masteringphysics.com/problemAsset/1026529/6/yg.11.35.jpg

Shear modulus = Stress/Strain

Stress = F(perpendicular)/A

Strain = x/h

The Attempt at a Solution



Stress = 904sin(12)/11x10^-4

Strain = 0.0142

h=11x10-2 ??

angle = tan^-1(0.0142/11x10-2) = 7.4

This method appears a little wrong, is someone able to point out where I have gone wrong and how I should have got to the answer? Thanks! :-)
 
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  • #2
Your calculations are also units-free, which makes them even more difficult to follow.
 
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  • #3
All units are SI, stress has units of Pa, strain is a ratio so has no units and h is measured in metres. My angle is in degrees.

I would be grateful if you were now able to help me with my answer! :-)
 
  • #4
M_J said:
All units are SI, stress has units of Pa, strain is a ratio so has no units and h is measured in metres. My angle is in degrees.

I would be grateful if you were now able to help me with my answer! :-)
Your angle calculation is incorrect. You already have the shear strain, so why are you dividing by the area?

Chet
 
  • #5
Thanks, can you explain what I do need to do then please?
 
  • #6
Don't worry, I have done it now - thanks anyway!
 
  • #7
SteamKing said:
Your calculations are also units-free, which makes them even more difficult to follow.

Thanks!
 
  • #8
M_J said:
Don't worry, I have done it now - thanks anyway!

what exactly did you do to change your original setup? i can't figure this one out
 

Related to Solve Shear Stress Problem: Cartilage Deformed at 7.4°

What is shear stress and why is it important to cartilage deformation?

Shear stress is a type of stress that occurs when forces act parallel to each other but in opposite directions, causing one part of a material to slide past another. In the case of cartilage, shear stress can lead to deformation and damage, which can affect its function and overall health.

What does a cartilage deformed at 7.4° mean?

This means that the cartilage has undergone a deformation of 7.4° due to shear stress. This can happen due to various factors such as excessive weight bearing, repetitive movements, or injury.

How is shear stress problem in cartilage solved?

Solving shear stress problem in cartilage involves understanding the underlying causes and addressing them. This can include reducing weight bearing, modifying movements, and providing adequate rest and support for the affected cartilage. It may also involve physical therapy or surgery in severe cases.

What are the consequences of ignoring shear stress problem in cartilage?

If left untreated, shear stress problem in cartilage can lead to further deformation, pain, and reduced joint function. It can also increase the risk of developing conditions such as osteoarthritis, which can have a significant impact on a person's quality of life.

Are there any preventive measures for shear stress problem in cartilage?

Yes, there are some preventive measures that can help reduce the risk of shear stress problem in cartilage. These include maintaining a healthy weight, avoiding repetitive or strenuous activities, and using proper technique during physical activities. It is also important to listen to your body and stop any activity if you experience pain or discomfort.

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