The weight of a 1200 kg car is supported equally by four tires which are inflated to the same gauge pressure what gauge pressure is required so the area of contact of each tire with the road is 0.001 m^2 to the
The Correct Answer and Explanation is:
Required gauge pressure Pgauge ≈ 2.9×106 Pa = 2 940 kPa = 4.3×102 psiP_{\text{gauge}} \;\approx\; 2.9 \times 10^{6}\ \text{Pa} \;=\; 2\,940\ \text{kPa} \;=\; 4.3 \times 10^{2}\ \text{psi}
Why that number?
- Weight on each tyre
The 1200 kg car’s weight is shared equally by the four tyres, so each supports one-quarter of the total mass: mper tyre=1200 kg4=300 kg.m_{\text{per tyre}} = \frac{1200\ \text{kg}}{4} = 300\ \text{kg}. Converting this mass to a force (weight) with g=9.8 m s−2g = 9.8\ \text{m s}^{-2}: F=mg=300×9.8 ≈ 2.94×103 N.F = mg = 300 \times 9.8 \;\approx\; 2.94 \times 10^{3}\ \text{N}. - Pressure–area relation
Pressure is force per unit area, P=F/AP = F/A. The specified contact patch for each tyre is only A=0.001 m2A = 0.001\ \text{m}^{2} (a square 3.2 cm on a side). Substituting: Pinside=FA=2.94×103 N0.001 m2=2.94×106 Pa.P_{\text{inside}} = \frac{F}{A} = \frac{2.94 \times 10^{3}\ \text{N}}{0.001\ \text{m}^{2}} = 2.94 \times 10^{6}\ \text{Pa}. - Gauge vs. absolute
“Gauge” pressure is the amount above atmospheric pressure. Because atmospheric pressure (≈ 1.01 × 10⁵ Pa) is less than 4 % of the value we have just calculated, adding or subtracting it changes the result only in the third significant figure. Thus the gauge pressure effectively equals the computed 2.94 MPa.
Interpreting the result
- Unusually high: 2.9 MPa is roughly 14 times higher than the 200 kPa (30 psi) found in everyday passenger-car tyres. To achieve a 0.001 m² contact patch the tyre would have to be inflated to racing-car or small-aircraft levels—far beyond safe limits for road tyres.
- Why the mismatch? In reality a typical tyre’s contact area is 0.025–0.030 m², not 0.001 m². That larger footprint lowers the required gauge pressure into the familiar 180–250 kPa range.
- Lesson: For any load-bearing surface, halving the contact area doubles the required pressure; reducing the area by a factor of 25 (as here) increases pressure twenty-fivefold. Engineers manipulate contact area (wider tyres, twin wheels, caterpillar tracks) precisely to keep working pressures within safe, efficient ranges.
Hence, to support the car on a 0.001 m² patch per tyre, a gauge pressure of about 2.9 MPa (430 psi) would be necessary—a value that highlights how crucial adequate contact area is for practical tyre design.
