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 0.640 m long and has a radius of 1.90 cm, by how much is the bone compressed?
weight supported by femur=0.9(85.0kg)= 76.5kg
L= 0.650m
A=(pi)(0.02m)^2=0.1256m
femur bone radius=2.0cm=0.020m
thats how far i got
To find out how much the bone is compressed, we need to use Hooke's law, which states that the displacement or compression of an elastic object is directly proportional to the force applied to it.
In this case, the force applied is the weight supported by the femur, which is equal to the weight of the person multiplied by the percentage supported by the femur:
Force = 76.5 kg × 9.8 m/s^2 = 749.7 N
Now, we need to find the stiffness or spring constant of the bone. The stiffness can be determined using the formula:
Stiffness (k) = (Young's modulus × Area) / Length
The Young's modulus of bone is approximately 17 GPa (gigapascals). Since we are working in SI units, we need to convert it to pascals:
Young's modulus = 17 × 10^9 Pa
The area of the circular cross-section of the femur can be found using the formula:
Area = π × radius^2
Area = π × (0.020 m)^2 = 0.00126 m^2
Substituting the values into the formula for stiffness:
Stiffness (k) = (17 × 10^9 Pa × 0.00126 m^2) / 0.640 m
Stiffness (k) ≈ 333,187.5 N/m
Finally, we can calculate the compression of the bone (displacement) using the formula:
Displacement (x) = Force / Stiffness
Displacement (x) = 749.7 N / 333,187.5 N/m
Displacement (x) ≈ 0.00225 m or 2.25 mm
Therefore, the bone is compressed by approximately 2.25 mm when supporting 90% of the person's weight.