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Problem 18 Medium Difficulty

A $975-\mathrm{kg}$ car has its tires each Problem 17 inflated to $"32.0$ pounds."
(a) What are the absolute and gauge pressures in these tires in $\mathrm{lb} / \mathrm{in.}^{2}$ , $\mathrm{Pa},$ and atm? (b) If the tires were perfectly round, could the tire pressure exert any force on the pavement? (Assume that the tire walls are flexible so that the pressure exerted by the tire on the pavement equals the air pressure inside the tire.) (c) If you examine a car's tires, it is obvious that there is some flattening at the bottom. What is the total contact area for all four tires of the flattened part of the tires at the pavement?

Answer

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Physics 101 Mechanics

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Video Transcript

can this problem? We're given the pressure and a car tyre as being 32 pounds per square inch. And this is equal to the gauge pressure because basically a gauge reads the pressure above the atmospheric pressure. So in part they were asked, Just tow, convert this Teo to pass calls and two out in spheres with the top here I've written with the conversion factors. For each of these are one atmosphere is 1.0 won three times into the five past calls, and that's equal to 14.7 pounds per square inch. So if we want to convert to pass calls, which I have in this middle equation, then we simply multiply by 1.13 10 cent of the five and divide by 14.7. And likewise, if we want to convert two atmospheres, we divide by 14.7. So it's for a in part. B were asked if the tire could exert any force if it were perfectly circular and the answer to this is no, because the force of the tire comes from the pressure that's inside it. And so you need some kind of contact area in order to exert a force. And if the tires perfectly circular, then it's just going to be hitting the road at a single point. It's going tohave on no area because it's just that tangent point and so mechanics or any force in part. See, we're given that the car is 970 kilograms and were asked what area? Um, the tyres must have in contact with road in order to support this weight, given the pressure that we were just told. And so the way of the car is going to be 975 times 9.8, which is the gravitational acceleration. This is just MG and we know that the tires have to support this week. So if we want to solve for the area, we can write a is equal tio force divided by pressure and the force. A CZ, we just determined, needs to be equal to the weight of the car in order for the tires to support it and um, for the pressure we need to include. Basically we want to give the pressure in units of Paschal's and so we would insert look p given in Paschal's, which comes from this middle equation

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