Question

A 4.5-kg, three legged stool supports a 83-kg person. If each leg of the stool has a cross-sectional diameter of 1.4 cm and the weight of the person is evenly distributed, determine the pressure exerted on the floor by each leg. answer is not 1.76*10^4 pa

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the person = M = 83 kg

Mass of the stool = m = 4.5 kg

Total weight supported by the three legs of the stool = W

W = (M + m)g

W = (83 + 4.5)(9.81)

W = 858.375 N

Force supported by each leg of the stool = F

F = W/3

F = 858.375/3

F = 286.125 N

Diameter of the stool leg = D = 1.4 cm = 0.014 m

Cross-sectional area of the stool leg = A

A = $\pi$D²/4

A = $\pi$(0.014)²/4

A = 1.54 x 10^-4 m²

Pressure exerted on the floor by each leg = P

$P = \frac{F}{A}$

$P = \frac{286.125}{1.54\ast 10^{-4}}$

P = 1.858 x 10⁶ Pa

Pressure exerted by each leg on the floor = 1.858 x 10⁶ Pa