Question

The weight of a car of mass 1.20 × 103 kg is supported equally by the four tires, which are inflated to the same gauge pressure. What gauge pressure in the tires is required so the area of contact of each tire with the road is 1.00 × 102 cm2? (1 atm = 1.01 × 105 Pa.)

Answer 1

Answer:

P = 294300Pa or 42.67psi by conversion.

Explanation:

Since Four tyres were inflated, we have that area of the four tyres are

4×1×10²cm²

Pressure is given as:

P = f/a but f = mg

P = m×g/a

Therefore,

P = 1.20x10³kg × 9.81m/s² / (4 ×1x10² cm²)

P = 1.20x10³kg×9.81m/s² / (0.04m²)

P = 294300Pa or 42.67psi by conversion.

Answer 2

Answer:

$P = 2.94 \times 10^5 Pa$

Explanation:

Normal force due to four tires is counter balancing the weight of the car

So here we will have

$4F_n = mg$

$F_n = \frac{mg}{4}$

$F_n = \frac{1.20 \times 10^3 \times 9.81}{4}$

$F_n = 2943 N$

now we know that pressure in each tire is given by

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

Here we know that

$A = 1.00 \times 10^2 cm^2 = 1.00 \times 10^{-2} m^2$

$P = \frac{2943}{1.00 \times 10^{-2}}$

$P = 2.94 \times 10^5 Pa$