Question: A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter...

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?

Answer 1

Solution

Hint: Pressure exerted on the floor is equal to the force exerted by the weight of the girl per unit area of the heel. We are given the diameter of the heel from which we can calculate the area of the heel and from the mass of the girl we can calculate her weight.

Formula used:
The pressure can be defined as the force exerted per unit area which can be mathematically given as:
$P = \dfrac{F}{A}{\text{ }}...{\text{(i)}}$
where P is the pressure exerted, F is used to denote the force being exerted while A is the area over which the force is being applied.

Complete step by step answer:
We are given that the girl weighs 50 kg, therefore, we can write
$m = 50kg$
From this we can calculate the force exerted due to the weight of the girl by taking $g = 10m/{s^2}$
Force is given as
$F = mg$
Substituting the known values, we get
$F = 50 \times 10 = 500N$
The girl is wearing a heel which has diameter given as
$d = 1cm = 0.01m$
Therefore the radius of the heel is given as
$r = \dfrac{d}{2} = \dfrac{{0.01}}{2}m$
Therefore, the surface area of the base is given as
$A = \pi {r^2}$
Substituting the known values, we get
$A = \pi {\left( {\dfrac{{0.01}}{2}} \right)^2}$
Now the pressure exerted by the heel on the horizontal floor is equal to the force exerted by the weight of the girl per unit the surface area of the heel. Therefore, using equation (i), we get
$P = \dfrac{F}{A}$
Substituting the known values, we get
$P = \dfrac{{500}}{\pi } \times \dfrac{4}{{0.01 \times 0.01}} \\\ = 6.36 \times {10^6}N/m \\\$
This is the required answer to the given question.

Note: Here the girl is balancing on a single heel due to which the pressure is quite large. If we talk about both the heels at same time then the force due to the weight of the girl would have been equally distributed on both heels and pressure would become half of the calculated value above.