A force compresses a bone by 0.10mm. A second bone has the same length but twice the cross-sectional area as the first. By how much would the same force compress the bone?
To solve this problem, we can use the equation for compressive stress:
Stress = Force / Area
The two bones have the same length, but the second bone has twice the cross-sectional area of the first bone. So the second bone has a cross-sectional area of 2 times the area of the first bone.
Let's assign some variables:
- Stress1: Stress on the first bone
- Stress2: Stress on the second bone
- Force: Force applied (constant for both bones)
- Area1: Cross-sectional area of the first bone
- Area2: Cross-sectional area of the second bone
From the problem, we know that Δlength (change in length) for both bones is the same. Let's call it ΔL.
The stress on the first bone can be calculated as follows:
Stress1 = Force / Area1
The stress on the second bone can be calculated as follows:
Stress2 = Force / Area2
Since the force and ΔL are the same for both bones, we can set the stresses equal to each other and solve for ΔL on the second bone:
Stress1 = Stress2
Force / Area1 = Force / Area2
ΔL / Area1 = ΔL / Area2
ΔL = (ΔL / Area1) * Area2
Since we know that Area2 = 2 * Area1:
ΔL = (ΔL / Area1) * (2 * Area1)
Simplifying:
ΔL = 2 * ΔL
Therefore, the force would compress the second bone by the same amount (0.10mm).
To understand how a force compresses a bone, we need to apply Hooke's Law, which states that the force applied to an object is directly proportional to the displacement caused by that force.
In this case, we have two bones: the first bone and the second bone, which has double the cross-sectional area of the first bone. Let's assume the force applied is the same for both bones.
According to Hooke's Law, the force applied can be given by the equation:
F = k * Δx
Where F is the force applied, k is the stiffness constant of the material (in this case, the bone), and Δx is the displacement or compression of the bone.
Since we are trying to find the compression of the second bone, let's call it Δx2. We already know the compression of the first bone, which is given as Δx1 = 0.10 mm.
Now, let's consider the relationship between the cross-sectional area (A) and the stiffness constant (k) of the bone. For a given material, stiffer bones will have a higher value of k. The stiffness constant can also be related to the cross-sectional area, as follows:
k ~ 1/A
This means that the stiffness constant is inversely proportional to the cross-sectional area.
Since the second bone has twice the cross-sectional area (A2 = 2 * A1), its stiffness constant (k2) will be half that of the first bone (k2 = k1/2).
Now, we can calculate the compression of the second bone using the equation:
F = k2 * Δx2
Since the force applied is the same for both bones, we can equate the forces:
k1 * Δx1 = k2 * Δx2
Substituting the values, we get:
k1 * 0.10 mm = (k1/2) * Δx2
Simplifying the equation:
0.10 mm = (1/2) * Δx2
Multiplying both sides by 2:
Δx2 = 0.20 mm
Therefore, the same force would compress the second bone by 0.20 mm.