By what amount does the 52-cm-long femur of an 73 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.

The ground's normal on the femur while running can reach several larger

than when standing.

Reasons:

The length of the bone, L = 52 cm = 0.52 m

Mass of the runner = 73 kg

Cross sectional area of the femur = 5.2 × 10⁻⁴ m²

Young's modulus = 1.6 × 10¹⁰ N/m²

Normal force = 3 × Body weight

Solution:

Compression, ΔL = L × Strain

Therefore;

ΔL = 0.052 m × 2.582 × 10⁻⁴ ≈ 1.343 × 10⁻⁴ m

The compression of the femur when running, ΔL ≈ 1.343 × 10⁻⁴ m.

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Question parameter obtained from a similar online question:

The normal force of the ground on the leg can be up to thrice the runners weight.

4.55×10^-5m

Explanation:

Young modulus of the material is equal to the ratio of the tensile stress to tensile strain of the elastic material.

Young modulus = Tensile stress/Tensile strain

Tensile stress = Force/cross sectional area

Give mass = 73kg

Force = mg = 73×10 = 730N

Cross sectional area = 5.2×10^-4m²

Tensile stress = 730/5.2×10^-4

Tensile stress = 1.4×10^6N/m

Strain = extension/original length

Given original length = 52cm = 0.52m

Tensile strain = extension(e)/0.52

Substituting the values given into the young modulus formula we have;

1.6×10^10 = 1.4×10^6/{e/0.52}

1.6×10^10 = 1.4×10^6×0.52/e

e = 1.4×10^6×0.52/1.6×10^10

e = 7.28×10^5/1.6×10^10

e = 4.55×10^-5m

This shows that the femur compresses by 4.55×10^-5m

b. lifting dumbbell weight at 20-degree angle from the body after a 45-degree angle

answer: a. same unit

explanation: