Question

A 100 kg person sits on a 5 kg bicycle. The total weight is borne equally by the two wheels of the bicycle. The tires are 2.0 cm wide and are inflated to a gauge pressure of 8.0 × 105 Pa. What length of each tire is in contact with the ground?

Answer 1

To solve this exercise it is necessary to take into account the concepts related to Pressure depending on the Force and the Area.

The pressure is defined as

$P =\frac{F}{A}$

Where,

$F_w = Force$

$A = Area$

Our values are given as,

$m_1 = 100Kg$

$m_2 = 5Kg$

Where $m_1$ is the mass of the person and $m_2$ is mass of cycle. Therefore the total mass at the system is 105Kg.

For Newton's second law we have:

$F = ma \rightarrow$ where a is the acceleration, in this case the same Gravitational acceleration.

We need to make two considerations, the first is to calculate the total weight and second to calculate the weight on each wheel (2), that is

$F_T = 105*9.8$

$F_T = 1030.05N$

Force in each wheel,

$F_{wheel} = \frac{1030.05}{2}$

$F_{wheel} = 515.025N$

In this way we can now calculate the length based on the pressure, that is

$P = \frac{F_T}{A}$

$P = \frac{F_T}{L*W}$

$L = \frac{F_T}{P*W}$

$L = \frac{515.025}{(8*10^{5})(2*10^{-2})}$

$L = 0.0321m$

Therefore each tire has 3.21cm