A 4.5-kg, three legged stool supports a 65-kg person. If each leg of the stool has a cross-sectional diameter of 2.6 cm and the weight of the person is evenly distributed, determine the pressure exerted on the floor by each leg.

Answer :

The pressure exerted on the floor by each leg is 427832.0079 Pa.

F = mg

F = (4.5 + 65) × 9.8

F= 681.1 N

This force is evenly distributed on the three leg

radius, r = d/2

= 2.6 / 2

= 1.3 cm = 0.013 m

Total cross sectional area of the three legs, A = 3×π×[tex]r^{2}[/tex]

A = 3 × 3.14 × [tex]0.013^{2}[/tex]

A = 0.00159 [tex]m^{2}[/tex]

Pressure due to weight,

P = Weight/A

P = 681.1/ 0.00159

P = 427832.0079 Pa

Hence, the pressure exerted on the floor by each leg is 427832.0079 Pa.

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