Question

How much force should be applied on an area of ${10^{ - 4}}{m^2}$ to get a pressure of $15Pa$?

Answer 1

Hint: We know that force is defined as the product of pressure and area of the surface. With the help of definition we can easily solve the above problem. We also know that all units are given in S.I. unit so we just need to use the formula directly.

Complete step-by-step solution:
According to initial data we have
Area = ${10^{ - 4}}{m^2}$
Pressure = 15 Pa
Also we know that,
$1Pa = 1N{m^{ - 2}}$
$\therefore 15Pa = 15N{m^{ - 2}}$
A force is a push or pull upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects. When the interaction ceases, the two objects no longer experience the force.
We know that force is defined as the product of pressure and area of the surface.
According to definition of force we have,
$F = P \times A$
Where, $F = $force applied
$P = $Pressure
$A = $Area of the surface
On substituting the values of pressure and area we have
$F = 15N{m^{ - 2}} \times {10^{ - 4}}{m^2}$
On solving we get
$F = 15 \times {10^{ - 4}}N({m^{ - 2}} \times {m^2})$
Finally we get
$F = 1.5 \times {10^{ - 3}}N$

Hence the required force, $F = 1.5 \times {10^{ - 3}}N$.

Note: While calculating pressure, normal force is taken into consideration, which means that only that component of force is taken into consideration which is perpendicular to the surface. Pressure is the amount of force applied at right angles to the surface of an object per unit area. The symbol for it is $P$. The IUPAC recommendation for pressure is a lower-case p. However, upper-case P is widely used. The usage of P vs p depends upon the field in which one is working, on the nearby presence of other symbols for quantities such as power and momentum, and on writing style.