A tractor weighing 4 tonnes has four tyres with a total contact area of 4000 cm2. A car weighing 1 tonne has four tyres with a total contact area of 250 cm2. Calculate the pressure exerted by each veh

Mass of tractor ($m_T$) = 4 tonnes = $4 \times 1000 \text{ kg} = 4000 \text{ kg}$
Weight of tractor ($F_T$) = $m_T \times g = 4000 \text{ kg} \times 10 \text{ m/s}^2 = 40000 \text{ N}$
Contact area of tractor ($A_T$) = 4000 cm$^2$ = $4000 \times 10^{-4} \text{ m}^2 = 0.4 \text{ m}^2$
Pressure exerted by tractor ($P_T$) is:

$$P_T = \frac{F_T}{A_T} = \frac{40000 \text{ N}}{0.4 \text{ m}^2} = 100000 \text{ Pa}$$

Mass of car ($m_C$) = 1 tonne = $1 \times 1000 \text{ kg} = 1000 \text{ kg}$
Weight of car ($F_C$) = $m_C \times g = 1000 \text{ kg} \times 10 \text{ m/s}^2 = 10000 \text{ N}$
Contact area of car ($A_C$) = 250 cm$^2$ = $250 \times 10^{-4} \text{ m}^2 = 0.025 \text{ m}^2$
Pressure exerted by car ($P_C$) is:

$$P_C = \frac{F_C}{A_C} = \frac{10000 \text{ N}}{0.025 \text{ m}^2} = 400000 \text{ Pa}$$

The pressure exerted by the tractor is $100000 \text{ Pa}$.
The pressure exerted by the car is $400000 \text{ Pa}$.
The car exerts a higher pressure on the ground than the tractor, which is why it sinks.

c) To enable the father's car to move on soft ground with ease, its contact area with the ground should be increased. This can be achieved by fitting the car with wider tyres or tracks (like those on a tractor). Increasing the area reduces the pressure exerted on the soft ground, preventing the car from sinking.

d) One application in real life where knowledge on pressure is applied is in the design of snowshoes. Snowshoes have a large surface area that distributes a person's weight over a wider area, reducing the pressure on the snow and preventing the person from sinking.