Assuming bicycle tires are perfectly flexible and support the weight of bicycle and rider by pressure alone, calculate the total area of the tires in contact with the ground. The bicycle plus rider has a mass of$80.0kg$, and the gauge pressure in the tires is$3.50\times {10}^{5}Pa$.

Fluid statics, often known as hydrostatics, is a branch of fluid mechanics that investigates the state of balance of a floating and submerged body, as well as the pressure in a fluid, or imposed by a fluid, on an immersed body.

The pressure is the rate of force appliedto the surface per unit area.

Here the open end is at height h so it will exert a force inside.

The total mass of the system is$80kg$.

So, the mass (m) acting on each tire will be$40kg$.

The gauge pressure (P) inside the tires is $3.5\times {10}^{5}Pa$.

We will consider the area of one tire and then we can double the area to find out the total area of the system.

$$\begin{gathered} A=\frac{F}{P} \\ =\frac{mg}{P} \\ =\frac{40\text{kg}\times 9.8{\text{m/s}}^2}{3.5\times {10}^5\text{Pa}} \\ =0.00112 {\text{m}}^2 \end{gathered}$$

So, the area exerting the pressure by one tire is$0.00112m^2$.

To calculate the total area, we need to double the area as,

$$\begin{gathered} A_{total}=2\times 0.00112 \\ =0.00224m^2 \end{gathered}$$

Hence, the total area on which pressure is exerted of the system of cycle is: $=0.00224m^2$.