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A tractor with 4000 cm² tyres and a car with 250 cm² tyres are in a field. Why can the tractor move easily while the car sinks?

Physics

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Answer

b

100000 Pa\text{100000 Pa}

400000 Pa\text{400000 Pa}

Step-by-step solution

3 steps

Here are the solutions to your tasks:

Part a

The physical features on the two vehicles leading to the observation in the scenario.

The tractor has a large mass (4 tonnes) but also very large tyres with a significant contact area (4000 cm²). The car has a smaller mass (1 tonne) but smaller tyres with a much smaller contact area (250 cm²). The large contact area of the tractor's tyres distributes its weight over a wider surface, reducing the pressure exerted on the soft ground, allowing it to move easily. Conversely, the car's smaller contact area concentrates its weight, leading to higher pressure and causing it to sink.

Part b

100000 Pa\text{100000 Pa}

The comparison of the pressure exerted by each of the two vehicles on the ground.

  1. Convert masses to kilograms and areas to square meters.

    Given: 1 tonne = 1000 kg, 1 m² = 10000 cm².
    For the tractor:
    Mass mT=4 tonnes=4×1000 kg=4000 kgm_T = 4 \, \text{tonnes} = 4 \times 1000 \, \text{kg} = 4000 \, \text{kg}
    Area AT=4000 cm2=4000×10−4 m2=0.4 m2A_T = 4000 \, \text{cm}^2 = 4000 \times 10^{-4} \, \text{m}^2 = 0.4 \, \text{m}^2

    For the car:
    Mass mC=1 tonne=1×1000 kg=1000 kgm_C = 1 \, \text{tonne} = 1 \times 1000 \, \text{kg} = 1000 \, \text{kg}
    Area AC=250 cm2=250×10−4 m2=0.025 m2A_C = 250 \, \text{cm}^2 = 250 \times 10^{-4} \, \text{m}^2 = 0.025 \, \text{m}^2

  2. Calculate the force (weight) exerted by each vehicle.

    Assume acceleration due to gravity g=10 m/s2g = 10 \, \text{m/s}^2.
    Force F=mgF = mg.
    For the tractor:
    FT=mTg=4000 kg×10 m/s2=40000 NF_T = m_T g = 4000 \, \text{kg} \times 10 \, \text{m/s}^2 = 40000 \, \text{N}

    For the car:
    FC=mCg=1000 kg×10 m/s2=10000 NF_C = m_C g = 1000 \, \text{kg} \times 10 \, \text{m/s}^2 = 10000 \, \text{N}

  3. Calculate the pressure exerted by each vehicle.

    Pressure P=FAP = \frac{F}{A}.
    For the tractor:

    PT=FTAT=40000 N0.4 m2=100000 PaP_T = \frac{F_T}{A_T} = \frac{40000 \, \text{N}}{0.4 \, \text{m}^2} = 100000 \, \text{Pa}

    For the car:

    PC=FCAC=10000 N0.025 m2=400000 PaP_C = \frac{F_C}{A_C} = \frac{10000 \, \text{N}}{0.025 \, \text{m}^2} = 400000 \, \text{Pa}

    Comparison:
    The pressure exerted by the tractor is 100000 Pa100000 \, \text{Pa}.
    The pressure exerted by the car is 400000 Pa400000 \, \text{Pa}.
    The car exerts four times more pressure on the ground than the tractor.
    The pressure exerted by the tractor is 100000 Pa\boxed{\text{100000 Pa}} and by the car is 400000 Pa\boxed{\text{400000 Pa}}.

Part c

The modifications that should be made to the father's car to enable it to move on soft ground with ease.

To enable the car to move on soft ground easily, the pressure it exerts on the ground must be reduced. This can be achieved by increasing the contact area between the car and the ground. Modifications could include:

  • Fitting wider tyres to the car.
  • Using more tyres (e.g., dual wheels) or tracks (like a tank) to spread the weight over a larger area.

Part d

One application in real life where knowledge on pressure is applied.

Snowshoes are designed with a large surface area to distribute a person's weight over a wider area, reducing the pressure on the snow and preventing them from sinking.

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