Question

1. the weight of your 1420 - kg car is supported equally by its four tires, each inflated to a gauge pressure of 35.0 lb/in². what is the area of contact each tire makes with the road? a. 719 cm² b. 823 cm² c. 145 cm² d. 524 cm² e. 237 cm²

Explanation:

Step1: Calculate the force on each tire

The weight of the car is $F_{total}=mg$, where $m = 1420\ kg$ and $g=9.8\ m/s^{2}$. The force on each tire $F=\frac{mg}{4}$. So $F=\frac{1420\times9.8}{4}=3479\ N$.

Step2: Convert pressure unit

The gauge - pressure is $P = 35.0\ lb/in^{2}$. We know that $1\ lb = 4.448\ N$ and $1\ in=2.54\ cm = 0.0254\ m$, so $1\ lb/in^{2}=\frac{4.448\ N}{(0.0254\ m)^{2}}\approx 6894.76\ Pa$. Then $P = 35.0\times6894.76\ Pa\approx241316.6\ Pa$.

Step3: Calculate the area of contact of each tire

From the pressure formula $P=\frac{F}{A}$, we can solve for the area $A=\frac{F}{P}$. Substituting $F = 3479\ N$ and $P=241316.6\ Pa$ into the formula, we get $A=\frac{3479}{241316.6}\ m^{2}$. Converting to $cm^{2}$, since $1\ m^{2}=10000\ cm^{2}$, $A=\frac{3479}{241316.6}\times10000\ cm^{2}\approx145\ cm^{2}$.

Answer:

C. $145\ cm^{2}$