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 1500 kg. It seems reasonable to assume that each tire supports one-fourth of the weight. With the ruler you find that the tires are 15 cm wide and the flattened segment of the tire in contact with the road is 13 cm long. What answer—in psi—will you give your friend?

Question

Accepted Answer

Hi, everyone in this practice problem. We're being asked to determine the pressure of air within the tires of a truck. We will have a truck with an owner's manual stating that the truck will have a mass of 4300 kg and a transporter have just placed a 38 100 kg load on the truck. The loaded truck will have six tires that will bear 1/6 or one divided by six of the truck's mass and will have equal air pressure. A way bridge measures the flattened section of the tire that is in contact with the ground to be 27.9 centimeters by 9.2 centimeter. And we're being asked to determine the pressure of air within the tires and the options given R A 39.7 P SI B 76.3 P S IC, 35.1 P S ID, 74.8 P SI and lastly D 31.2 P si. So this is a great way of reasoning, pressure within the tires to support the weight of a vehicle. So it is uh given in the problem statement that the expression of B equals F divided by A can formulate an indirect way of measuring pressure in the vehicle's tire. The normal force will push on the flattened section and the air pressure in tires will push on the opposite side of the flat and section. Otherwise the flat or the flat in section will just move up or down. So in this case, you using Newton second law based on that fact, earlier, where we will have the normal force bushes downwards and the air pressure force bushes upwards to uh keep the tire at equilibrium. Using Newton's second law, we can get Sigma F Y to be equals to zero because it is at equilibrium. Next, as we have uh discussed previously, the only two forces I think is going to be first, the gravitational acceleration F G and second, going to be the normal force F N. In this case, F G is the gravitational force for the overall to pro the load. And F N is divided equally into six tires. So in this case, F G will equals to six F N where Sigma F Y will then equals to six F N minus F G to be equals to zero. So F G will equals to six F N just like. So F N is the normal force on a single tire. And six F N will also equals to six F B where F B is the force due to the pressure in a single tire. So therefore, F G will equal six F B and F B can be calculated using the expression of B equals to F divided by A where F through rearrangement will equals to P multiplied by A. So we want to substitute that equation right here into our Newton Second law. Final expression to get F G equals six multiplied by P multiplied by a rearranging this because we are interested to find the pressure. We wanna rearrange this to get an equation for pressure where we'll have P equals F G divided by six A. And in this case can actually start plugging everything into this formula right here. So F G will be, as I have discussed previously, the mass uh or the weight of the truck plus the load. So there will be M of the truck plus M of the load in parentheses, all multiplied by the gravitational acceleration. When I then define that with six A and A is going to be given by the um 27.9 centimeters by 9.2 centimeters. So it's essentially just length multiplied by its width. So L multiplied by W and then from there, we can substitute all the given information in the problem statement and start calculating for our pressure. So first, what we have is the mass of the truck, which is given to be 4300 kg in the problem statement. So open parenthesis is 4300 kg plus the mass of the load which is 3800 kg, 3800 kg close parenthesis multiplied by G or the gravitational acceleration which is 9.81 m per second squared. All of that divided by six multiplied by a length which is 27.9 centimeter. You wanna convert that in two m. So 27.9 times 10 to the power of negative two m and multiply that by the width as well, which is 9.2 centimeter. And we wanna also convert that two m. So 9.2 times 10 to the power of negative two m. And then calculating this, we will then get the final pressure to be 5.16 times 10 to the power of five pascals. But the answer choices are all in P SI. So we will need to perform the conversion into P SI. So then we wanna multiply this with one P si divided by 6895 pascals. And then we can cross out the pascals and then the pressure will then come up to value of 74.8 P SI. So then that will be the final pressure at 74.8 P si off air in the tires, which will correspond to option D in our answer choices. So option D will be the answer to this particular practice problem and that'll be it for this video. If you guys still have any sort of confusion. Please make sure to check out our other lesson videos on similar topics and that'll be it for this one. Thank you.

Verified Solution

Key Concepts

Pressure

Force and Weight

Area of Contact