Question

The area of a man’s foot is $80\;cm^2$. How much pressure will the man exert on the ground standing, if his mass is $80\; kg$.

Answer 1

Hint: Try and recall the relation between pressure, force and area. In other words, we know that pressure is directly proportional to force and inversely proportional to the area. Also, keep in mind that the force exerted by the man is equivalent to his weight, which is nothing but the gravitational force, so proceed with your calculations accordingly.

Formula Used: Pressure exerted $P = \dfrac{F}{A}$, where F is the force exerted and A is the area over which this force is exerted.
Gravitational force (weight) exerted $F_{gravitational} = mg$, where m is the mass of the body and g is the acceleration due to gravity $g_{earth} = 9.8\;ms^{-2}$

Complete step-by-step solution:
Let us begin by looking at the different components of the question.
We are required to find the pressure exerted by the man on the ground when he is standing on his two feet.
Therefore, the area over which he will exert this pressure will be $2 \times 80 = 160\;cm^2 = 160 \times 10^{-4}\;m^2$
Now, the force that the man exerts on the ground will be his weight, which is equivalent to the gravitational force acting on him.
$F_{gravitational} = mg = 80 \times 9.8 = 784\;N$
Now, we know that the pressure exerted by the man will be directly proportional to the force exerted by him and inversely proportional to the area over which he exerts this force. This relation is given by:
Pressure $P = \dfrac{F_{gravitational}}{area} = \dfrac{784}{160 \times 10^{-4}} $
$\Rightarrow P = 4.875 \times 10^4\;Nm^{-2} = 4.875 \times 10^4\;Pa$
Now, since the standard atmospheric pressure is $101,325\; Pa$, the pressure exerted by the man is nearly $\dfrac{487,500}{101,325} = 4.8 \approx 5$ times the pressure exerted on the earth by the atmosphere.

Note: We found that the relative pressure exerted by the man was 4.8. This means that the pressure exerted by the man was 4.8 times stronger than the atmospheric pressure. This relative pressure that we found is also referred to as the gauge pressure, which the pressure relative to the ambient pressure.
Remember that pressure is ultimately a scalar quantity and though it is most often specified with a direction, this comes from the normal force vector that acts on the vector area. Thus, it is incorrect