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?
To find out by how much the bone is compressed, we need to calculate the change in length of the femur bone.
The change in length can be determined using Hooke's Law, which states that the change in length of a material is proportional to the force applied to it.
First, let's find the force applied on the femur bone. Since 90% of the person's weight is supported by the femur, we can calculate it as:
Force = 90% of person's weight
Force = (90 / 100) * 89.0 kg * 9.8 m/s^2 [Considering acceleration due to gravity is 9.8 m/s^2]
Force = 784.98 N
Next, we need to determine the Young's modulus of the bone. The Young's modulus (E) is a measure of the stiffness of a material. For human bones, it is approximately 18 GPa (Giga Pascals).
Now, we can use Hooke's Law to calculate the change in length (ΔL) of the femur bone:
ΔL = (Force * Length) / (π * Radius^2 * Young's Modulus)
Using the given values:
Force = 784.98 N
Length = 0.640 m
Radius = 1.90 cm = 0.019 m
Young's Modulus = 18 GPa = 18 * 10^9 Pa
ΔL = (784.98 N * 0.640 m) / (π * (0.019 m)^2 * (18 * 10^9 Pa))
Simplifying the equation further, we get:
ΔL = (784.98 N * 0.640 m) / (3.14 * 0.000361 m^2 * 18 * 10^9 Pa)
ΔL = 139.0 meters
Therefore, the bone is compressed by approximately 0.139 meters.