Question

A 4.5-kg, three legged stool supports a 89-kg person. If each leg of the stool has a cross-sectional diameter of 2.8 cm and the weight of the person is evenly distributed, determine the pressure exerted on the floor by each leg.

Answer 1

Answer:

4.96 × 10⁵ Pa

Explanation:

F = mg

$F = (m_{person}+m_{stool})g\\\\F = (4.5 + 89)*9.8\\\\F = 916.3 N$

This force is evenly distributed on the three leg

radius, r = d/2

= 2.8 / 2

= 1.4 cm = 0.014 m

total cross sectional area of the three legs, A = 3*pi*r^2

$= 3\times\pi \times0.014^2\\\\= 1.847\times10^-^3m^2$

Pressure due to weight,

P = Weight/A

$P = / 1.847 × 10⁻³\\P = \frac{ 916.3N}{1.847\times10^-^3} \\\\P= 496032.9Pa$

P = 4.96 × 10⁵ Pa