Question

Two thigh bones, each of cross-sectional area 15cm2,support the upper part of a human body of mass 42kg.estimate the average pressure sustained by the femur?(g=10m/s2)

Answer 1

Pressure = Force / Area

Pressure exerted by femur = mg / (2A)
= (42 kg × 10 m/s²) / (2 × 15 × 10⁻⁴ m²)
= 140 × 10³ Pa
= 140 kPa

Average pressure exerted by one femur is 140 kPa

Answer 2

Answer:

1.4 × 10^5 Pa

Explanation:

Cross-sectional area of femur (A) = 15 cm^2 = 15 × 10^(-4) m^2

Mass of human body (m) = 42 kg

Acceleration due to gravity (g) = 10 m/(s^2)

Pressure sustained by femurs = p = (mg)/(2A)

$p = (42 \times 10) \div (2 \times 15 \times {10}^{ - 4} ) \\ = 420 \div (30 \times {10}^{ - 4} ) \\ = 14 \times {10}^{4} = 1.4 \times {10}^{5} kg \: {m}^{ - 1} {s}^{ - 2}$