Physics: How much is this person's femur compressed?

In summary, the conversation is about calculating the area of a circle and using Hooke's law in a physics course. The person is stuck on converting mass to weight and calculating the correct units for the area.
  • #1
brittanyr
6
0
Homework Statement
A 89.0 kg person stands on one leg and 90% of the weight is supported by the upper leg connecting the knee and hip joint-the femur. Assuming the femur is 0.650 m long and has a radius of 2.00 cm, by how much is the bone compressed?
Relevant Equations
weight supported by femur=0.9(85.0kg)= 76.5kg
L= 0.650m
A=(pi)(0.02m)^2=0.1256m
Thats how far I got with my equation. Help me please.
 
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  • #2
brittanyr said:
weight supported by femur=0.9(85.0kg)= 76.5kg
That's a mass, not a weight. Weight is a force.
brittanyr said:
A=(pi)(0.02m)^2=0.1256m
What is 0.022? What is (m)2?
 
  • #3
In your course work, have they been talking to you about Hooke's law, stress, strain, deflection, Young's modulus and such? If not, we will need to start from first principles and notions of relevant proportionality.

Google can easily find a relevant parameter not given in the problem statement here.
 
  • #4
jbriggs444 said:
In your course work, have they been talking to you about Hooke's law, stress, strain, deflection, Young's modulus and such? If not, we will need to start from first principles and notions of relevant proportionality.

Google can easily find a relevant parameter not given in the problem statement here.
Yes. We are using stress, strain, and Young’s modulus
 
  • #5
haruspex said:
That's a mass, not a weight. Weight is a force.

What is 0.022? What is (m)2?
0.02^2 =0.4
 
  • #6
brittanyr said:
0.02^2 =0.4
0.4*3.14= 0.1256 ... 0.1255^2 =0.2512? Correct?
 
  • #9
0.2*0.2=0.04
The unit would be Meters
 
  • #10
brittanyr said:
0.2*0.2=0.04
Yes, but you want 0.02*0.02
brittanyr said:
The unit would be Meters
The metre is a unit of length. You are calculating an area. I gave you the hint in post #2. You had (0.02m)2. You (inaccurately ) squared the 0.02 but you left the m as just m.
 

1. How is the compression of a person's femur measured?

The compression of a person's femur is measured by comparing the original length of the femur to its length after a certain amount of force or weight is applied. This can be done using tools such as a caliper or a compression testing machine.

2. What factors can affect the compression of a person's femur?

The compression of a person's femur can be affected by various factors such as the person's weight, the angle and direction of the force applied, the strength and density of the bone, and any pre-existing conditions or injuries.

3. Is the compression of a person's femur the same for everyone?

No, the compression of a person's femur can vary depending on the individual's weight, bone density, and other factors. It is also important to note that the compression of a person's femur can change over time due to factors such as age and physical activity.

4. Can the compression of a person's femur be measured without using specialized tools?

While it is possible to estimate the compression of a person's femur without using specialized tools, it is not recommended as it may not provide an accurate measurement. It is best to use proper equipment and techniques to accurately measure the compression of a person's femur.

5. Can the compression of a person's femur be reversed?

In most cases, the compression of a person's femur cannot be reversed. However, in some cases where the compression is caused by factors such as muscle imbalance or poor posture, it may be possible to improve the compression through physical therapy and exercises.

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