Question : A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor.
Doubt by Nevaeh
Solution:
m=50 kg
Weight (Force exerted) = mg
W=50×9.8
W= 490 N
Weight (Force exerted) = mg
W=50×9.8
W= 490 N
Diameter of the heel (d) = 1 cm = 10-2 m
Area of cross section of the heel (A) = πr2
=π(d/2)2
=π(d/2)2
= πd2/4
= π(10-2)2/4
=π(10-4)/4 m2
= π(10-2)2/4
=π(10-4)/4 m2
Pressure exerted on the heel (P)
P = F/A = W/A
= 490/[π(10-4)/4]
= (490×4×104)/π
=(490×4×104)/(22/7)
= (490×4×7×104)/22
=(490×4×104)/(22/7)
= (490×4×7×104)/22
= (490×28×104)/22
= (13720/22)×104
= 623.64×104
= 6.2364×106
= 623.64×104
= 6.2364×106
= 6.24×106 N/m2