The discussion revolves around calculating the compression of a femur under a specific weight, involving concepts from mechanics and material science, such as stress, strain, and Young's modulus.
The conversation is ongoing, with participants providing clarifications and corrections regarding calculations and concepts. Some guidance has been offered regarding the need to understand relevant physical principles, but no consensus has been reached on the specific calculations or interpretations.
There appears to be some missing information regarding parameters necessary for the calculations, and participants are referencing course material that may not have been fully covered.
That's a mass, not a weight. Weight is a force.brittanyr said:weight supported by femur=0.9(85.0kg)= 76.5kg
What is 0.022? What is (m)2?brittanyr said:A=(pi)(0.02m)^2=0.1256m
Yes. We are using stress, strain, and Young’s modulusjbriggs444 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.
0.02^2 =0.4haruspex said:That's a mass, not a weight. Weight is a force.
What is 0.022? What is (m)2?
0.4*3.14= 0.1256 ... 0.1255^2 =0.2512? Correct?brittanyr said:0.02^2 =0.4
No.brittanyr said:0.02^2 =0.4
haruspex said:No.
Consult http://www.math.com/school/subject1/lessons/S1U1L5GL.html
And what will the units of the area be?
Yes, but you want 0.02*0.02brittanyr said:0.2*0.2=0.04
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.brittanyr said:The unit would be Meters