Question

A boy weighing $60\;kgf$ stands on a platform of dimensions $2.5\;cm \times 0.5\;cm$. What pressure in Pascal does he exert?

Answer 1

The formula for pressure is applied in terms of the force and cross-sectional area on which the pressure is applied in order to determine the required solution. The correct conversions must be done so that the correct units are used for the quantities that are involved so that the pressure is obtained in Pascal.

Formula used:
The pressure is given as:
$P = \dfrac{F}{A}$
Where, $F$ is the force and $A$ is the area.

Complete step by step answer:
The above problem revolves around the concept of pressure and its relation with force that is exerted on a certain area.When the boy stands on the platform he is said to thrust some force or exert some amount of force on the platform. Any quantity that has a mass or some amount of weight will exert a force on the surface area that is considered due to the pull of gravity. Force is quantity which is defined to be any push or pull that is exerted on an object.

When there is a force that is applied then there is a pressure associated with it. The force in this case is applied perpendicular to the platform since the gravitational pull or force is vertically downwards.Thus, pressure is said to be the force exerted by the boy per unit area and hence the formula for pressure is given by the following equation:
$P = \dfrac{F}{A}$ ----------($1$)

The SI unit of the quantity known as pressure is called Pascal. It is defined as the pressure which has its force as one newton and its area in square meters. This means that the force must be in the unit Newton and the area must be in meter square to obtain the unit Pascal which is what we are asked to determine. Let us first extract the data given in the question. We are given the force in terms of the unit kilogram-force and the area in terms of length and breadth values in the unit square centimeters.

Given, $F = 60\;kgf$ and $A = 2.5\;cm \times 0.5\;cm$.Thus, we must convert these quantities into the required units, that is, force into newton and area in terms of square meter. Let us first convert the force into newton. We are given the weight of the boy. The quantity known as force is not equivalent to the mass or the weight of the boy and hence we have a relation between them in order to convert the given force into newton.
Hence, the relation is as follows:
$1kgf = 9.8\;N$
We can round this off to a whole number to make calculations easier and hence we get:
$1kgf = 10\;N$
Hence, by unitary method, $60\;kgf = 60 \times 10\;N$
$\Rightarrow F = 600\;N$

Now we must convert the area from square centimeters to square meters. We do this from the relation:Since,
$1cm = {10^{ - 2}}{m^2} = \dfrac{1}{{100}}{m^2}$
We can say that,
$1c{m^2} = \dfrac{1}{{10000}}{m^2}$
Hence, by unitary method:
$\left( {2.5 \times 0.5} \right)c{m^2} = 2.5 \times 0.5 \times \dfrac{1}{{10000}}{m^2}$
$\Rightarrow A = 2.5 \times 0.5 \times \dfrac{1}{{10000}}{m^2}$
By simplifying further we get:
$\Rightarrow A = 1.25 \times \dfrac{1}{{10000}}{m^2}$
$\Rightarrow A = 1.25 \times {10^{ - 4}}{m^2}$

Thus, we now substitute the converted values into equation ($1$) to obtain the pressure in Pascal. Hence we get:
$P = \dfrac{{600\;N}}{{1.25 \times {{10}^{ - 4}}{m^2}}}$
We can also write this as:
$\Rightarrow P = \dfrac{{6 \times {{10}^2}N}}{{1.25 \times {{10}^{ - 4}}{m^2}}}$
We now simplify the above equation to get:
$\Rightarrow P = \dfrac{6}{{1.25}} \times {10^2} \times {10^4}$
We know that when the base numbers are equal the powers can be added and hence we get:
$\Rightarrow P = \dfrac{6}{{1.25}} \times {10^{2 + 4}}$
On further simplification we get:
$\therefore P = 4.8 \times {10^6}Pa$

Hence, $4.8 \times {10^6}Pa$ is the pressure exerted by the boy on the platform in Pascal.

Note: The thrust, force and pressure quantities are often confused together. The thrust is slightly different from pressure as thrust is actually the force or the net force acting on the given surface on a whole while pressure is the force acting per unit area. The common mistake that is done in the problem is that the conversions are not done properly and the units of force and area are not converted into newton and square meters respectively which is wrong since Pascal is Newton per square meters.