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?
Answers
Mass of the girl, m = 50 kg
Diameter of the heel, d = 1 cm = 0.01 m
Radius of the heel, r = d/2 = 0.005 m
Area of the heel = πr2
= π (0.005)2
= 7.85 × 10–5 m2
Force exerted by the heel on the floor:
F = mg
= 50 × 9.8
= 490 N
Pressure exerted by the heel on the floor:
P = Force / Area
= 490 / (7.85 × 10-5) = 6.24 × 106 Nm-2
Therefore, the pressure exerted by the heel on the horizontal floor is
6.24 × 106 Nm–2.
Answer:
6.24×10^6 N/m²
Explanation:
Given,
mass of the girl (M)=50kg
radius of the heel (r)=d/2
=0.01m/2
=0.005m
Area of the heel=πr²
=3.14×(0.005)²
=0.0000785
=7.85×10^-5
Now, force(F)=Mg
=50×3.14
=490N
Therefore,
Pressure =Force/Area
=490/(7.85×10^-5)
=(490×10^5)/7.85
=6.24×10^6 N/m²
So, the pressure exerted by the heel on the horizontal floor will be equal to (6.24×10^6 N/m²)