Question

The normal force of the ground on the foot can reach three times a runner's body weight when the foot strikes the pavement.By what amount does the 52-cm-long femur of an 79 kg runner compress at this moment? The cross-section area of the bone of the femur can be taken as 5.2×10−4m2 and its Young's modulus is 1.6×1010N/m2.Express your answer to two significant figures and include the appropriate units.

Answer 1

Using Young's modulus and the normal force exerted on a runner's foot, the femur of a 79 kg runner compresses by approximately 0.0145 mm when the foot strikes the pavement.

Step-by-step explanation:

The problem involves calculating the compression of a runner's femur using Young's modulus, the normal force on the runner's foot, and the dimensions of the femur.

Given that the normal force is three times the runner's body weight (W = mg), we can use the formula for stress (σ = F/A) and strain (ε = ΔL/L), where ΔL is the change in length, L is the original length, A is the area, and F is the force.

Firstly, the normal force (F) will be:

F = 3 * (mass of the runner * acceleration due to gravity)

F = 3 * (79 kg * 9.8 m/s2)

F = 2323.4 N (rounded to three significant figures)

Using Young's modulus (E = stress / strain), we have:

σ = F/A

ε = ΔL/L

E = σ / ε

So, ΔL = L * σ / E

ΔL = (0.52 m) * (2323.4 N / 5.2×10−4 m2) / (1.6×1010 N/m2)

ΔL = 1.45×10−5 m or 0.0145 mm

The femur will compress by approximately 0.0145 mm when the runner's foot strikes the pavement.