The discussion centers on the calculation of the force supported by a femur, emphasizing the distinction between mass and weight. The correct calculation involves using the formula for area, specifically A = π(0.02m)², which results in an area of approximately 0.001256 m². Participants highlight the importance of understanding fundamental concepts such as Hooke's law, stress, strain, and Young's modulus to accurately analyze the problem. Misinterpretations regarding units and calculations are addressed, reinforcing the need for precision in scientific computations.
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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