A friend asks you how much pressure is in your car tires. You know that the tire manufacturer recommends ${30}_{psi}$, but it's been a while since you've checked. You can't find a tire gauge in the car, but you do find the owner's manual and a ruler. Fortunately, you've just finished taking physics, so you tell your friend, "I don't know, but I can figure it out." From the owner's manual you find that the car's mass is$1500kg$. It seems reasonable to assume that each tire supports one-fourth of the weight. With the ruler you find that the tires are $15cm$ wide and the flattened segment of the tire in contact with the road is $13cm$ long. What answer-in psi-will you give your friend?

The weight is defined by the product of the mass and acceleration due to gravity, and is given by the following relation:

$W=mg$

Here, $W$is the weight, $m$is the mass, and $g$is the acceleration due to gravity.

The pressure is defined by the ration of force and the area of the surface, and is given by the following relation:

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

Here, $P$is the pressure, $F$is the force, and $A$is the area

Pressure on the tires is calculated from force acting on each tire to the area of each tire. Each tire supports one-fourth of the weight.

Here, the weight is the force

The force due to gravity is equal to the weight.

The force due to gravity of the car is,

$F_{\mathrm{g}}=mg$

Substitute$1500kg$for $m$and $9.8m/s^2$for $g$in the equation $F_g=mg$.

$F_{\mathrm{g}}=(1500\mathrm{kg})\left(10\mathrm{m}/{\mathrm{s}}^2\right)$

$=15000N$

The force due to gravity acting on each tire is,

$F_{forceoncachtire}=\frac{F_8}{4}$

Substitute$15000N$for $F_g$in the equation $F_{forceoncachtire=}\frac{F_8}{4}$

$F_{forceoncachtire}=\frac{15000N}{4}$

$=3750N$

$F_{forceoncachtire}=\frac{15000\mathrm{N}}{4}$

Calculate the pressure acting on each tire as follows:

$P=\frac{F_{forceoncachtire}}{A}$

The area of each tire is,$l$

$A=lb$

Substitute $13cm$for $l$and $15cm$for $b$in the equation $A=lb$.

$A=\left(13\mathrm{cm}\left(\frac{{10}^{-2}\mathrm{m}}{1\mathrm{cm}}\right)\right)\left(15\mathrm{cm}\left(\frac{{10}^{-2}\mathrm{m}}{1\mathrm{cm}}\right)\right)$

$=195\times {10}^{-4}{\mathrm{m}}^2$

Substitute $3750N$for $F_{forceoncachtire}$and$195\times {10}^{-4}{m}^2$for $A$in the equation $P=\frac{F_{forceoncachtire}}{A}$.

$P=\frac{3750\mathrm{N}}{195\times {10}^{-4}{\mathrm{m}}^2}$

$=192307.6\mathrm{Pa}\left(\frac{0.000145038\mathrm{psi}}{1\mathrm{Pa}}\right)$

$=27.8\mathrm{psi}$

$\approx 27.8\mathrm{psi}$

As a result, each tire is under ${27}_{psi}$pound of pressure.