The femur is a bone in the leg whose minimum cross-sectional area is about 3.60 10-4 m2. A compressional force in excess of 6.90 104 N will fracture this bone.
(a) Find the maximum stress that this bone can withstand.
(b) What is the strain that exists under a maximum-stress condition
i tried to solve for part a and i got 1.92 and it was worng using thre equation force divided by area plaease help me. what would the right answer be
Your formula is correct, but you forgot to take care of the exponents (10^4=10000, etc.)
Someone has asked the same question, and the response is here.
http://www.jiskha.com/display.cgi?id=1289886124
i tried to answer part b but not 100 percent sure what the young modulus is for bone i know part a is 1.92E8. please help how to solve part b
If you read the article given in the link, you will realize that Young's modulus for bone is variable, depending on many factors, one of which is bone density.
Such information for a physics question would normally be supplied, but for completeness, the answer to part b had been completed, using an average value of 20 GPa.
To solve this problem, we need to use the equation for stress, which is given by:
Stress = Force / Area
(a) To find the maximum stress that the bone can withstand, we need to divide the compressional force by the minimum cross-sectional area:
Stress = (6.90 * 10^4 N) / (3.60 * 10^-4 m^2)
Calculating this, we get:
Stress = 1.92 * 10^8 N/m^2
So, the maximum stress that the bone can withstand is 1.92 * 10^8 N/m^2.
(b) The strain is a measure of the deformation of a material under stress and is given by:
Strain = Change in length / Original length
Unfortunately, the given information does not provide sufficient details to calculate the strain. The strain depends on the change in length of the bone, which is not given in the problem. Therefore, we cannot determine the strain without this information.
In summary, the correct answer for part (a) should be 1.92 * 10^8 N/m^2, and for part (b), the strain cannot be determined without additional information.