Question

Calculate the pressure exerted on the floor by a boy standing on both feet if the weight of the boy is 40kg.(Assume that the area of contact of each shoe is 6 cm by 25 cm. Take 1 kg wt=10N)

Answer 1

Hint: In the given condition we are given details about a boy who simply stands on the floor. We can say that only the gravitational force acts on the boy as he is standing idle. No other force comes into place at that time. Hence, pressure can be evaluated from gravitational force.

Complete Step by step answer:
For the given question we can use the formula, $F=mg$ where F is the gravitational force, m is the mass of the body and g is the acceleration due to gravity.
We know that, $P={\displaystyle \frac{F}{A}}$ where P is the pressure exerted by a body, F is the force acting on the body and A is the area where the force acts.
We are given that a boy is standing on both feet on the floor. The weight of the boy is 40kg and the area of contact of each shoe is 6 cm by 25 cm .
We know that 1kgwt=10N
Hence, 40kg wt can be converted to $40\times 10=400N$
The area of contact of each shoe=$6cm\times 25cm=({\displaystyle \frac{6}{100}}\times {\displaystyle \frac{25}{100}})m^2={\displaystyle \frac{15}{1000}}{m}^2$
Hence, total area of contact is $({\displaystyle \frac{15}{1000}}\times 2)m^2={\displaystyle \frac{3}{100}}{m}^2$
We know that
$P={\displaystyle \frac{F}{A}}$
$\Rightarrow P={\displaystyle \frac{400}{3/100}}={\displaystyle \frac{40000}{3}}={\displaystyle \frac{40}{3}}=13.33kN{m}^2$

Therefore the pressure exerted by the boy is $13.33kN{m}^2$or $13.33kPa$.

Note: Pressure is directed normally towards the surface area. In such cases it is scalar it acts in a direction normal to the surface. But if the force and area have directions then it acts as a tensor.