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Problem 50 Hard Difficulty

Multiple-Concept Example 17 reviews the basic concepts involved in this problem. Air rushing over the wings of high-performance race cars generates unwanted horizontal air resistance but also causes
vertical downforce, which helps the cars hug the track more securely. The coefficient of static friction between the track and the tires of a $690-\mathrm{kg}$ race car is $0.87 .$ What is the magnitude of the maximum acceleration at which the car can speed up without its tires slipping when a $4060-\mathrm{N}$ downforce and an $1190-\mathrm{N}$ horizontal-air-resistance force act on it?


12 $\mathrm{m} / \mathrm{s}^{2}$


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Video Transcript

in this question. There are five forces acting on the car. The weight forced. The normal force there is exerted by the ground because the car is on the ground. Then we have that downward force due to the wind and that, uh, backwards force that he's outside due to Tween. Then there is the fictional force. Well, the frictional force is rather counterintuitive because usually we would think that if the car is going in this direction, then the frictional force would be in that direction. But it is not the case here. And why does he serve well? It is very similar to how can we walk on the ground? So what happens when we are walking is that here is the floor and then we push the floor and then the floor pushes us, allowing us to walk. So when you are walking in this direction, the floor is pushing you in that direction. So it's the floor that expression you. Then the frictional force pushes you in the single election you are moving. It's a very safe you singler thing that happens with a car. So the car is moving in this direction. Then it's tires are rotating like that. So when it's rotating, if there waas no frictional force, the car will be in place on the tires would be rotating and sliding on the floor. But thanks to the fictional forces, what happens when the tire tries to rotate is that there is a relative movement between the tires and the ground. And then the frictional force tries to seize that relative movement, and as a result, we can see what happens. This part off the tire tries to move in this direction. Then the frictional force opposed itself to that movement, pushing the tire in that direction to thanks to that Ah, way off work off the frictional force, the car is able to move. So what the car is doing all the way is pushing the wrong backwards, and the ground is pushing the car from towards. So this is why we have a frictional force that is pointing in the same direction as the velocity off the car. Now we can proceed to the question. So the question is, what is the highest possible acceleration such that the tires don't sleep on the ground? To answer that question, we have to apply Newton's second law in this direction in the vertical direction which I will call the Y direction on the horizontal direction, which I will call the X Direction and knows that I had the red chose in my reference frame everything that is pointing to the right. It's positive, everything that is pointing up its positive. Everything that is pointing to the left is negative and everything that is pointing downwards negative, too. Then applying Newton's second law on the horizontal direction, we get the following. So, for the horizontal direction, Newton's second law tells us that the net force on that direction is equals to the mass off the car times next generation, off the car in the direction. Then what are the components? What forces are composing this net force? So in the horizontal direction, we have to forces the fictional forests and these resistant force. So the frictional force minus that resistant force is equal to the mass off the car times acceleration off that car. Okay, then we can calculate acceleration off the car as follows acceleration off the car in the X direction is it goes to the fictional force minors that 1190 force divided by the mass off the car, which is 690. Then remember that the frictional force is given by the following expression. A fictional force is equal to the static frictional coefficient times the normal force. Again, the car is moving, but we use the static frictional coefficient. Why is that? This is because this point off the tire that is touching the ground isn't moving with respect to the ground. If it were moving, then the car will be sleeping for the ground, and this is not what's happening. Therefore, we have to use the static fictional force because this point off the tire, the points that is in contact with the ground is at rest with respect to the ground, then the fiction there or result of acceleration. Zico's too. That's static. Frictional coefficient 0.87 times the normal force, minus 11 90 divided by a 690. Now we have to complete what is the value off the normal force? For that, we apply Newton's second law on the Y direction. It tells us that the net force in that direction is it close to the mask off the car times explanation in the vertical direction notes that the car is at rest in that direction. So this is a question zero. Therefore the net force in the Y direction is it close to zero? The force of that composed these net force the normal force, the weight force and that downward force provoked by the wind. Then we have the following the normal force minus these downwards force minus the weight force is because 20 Then the normal force is there close to that downward force plus the weak force. No, I have the following The don't wards forced is it goes to 40 60 Newtons. Then we have the weight force. The weight force is given by the following expression the mass off the car times acceleration of gravity. The mass off the car is 690. And remember, the acceleration of gravity is approximately 9.8 meters per second square near the surface off the earth. So we have this, then the normal force is it close to 10,820 to deal terms. So the acceleration off the car in the X direction is equals to 0.87 times, 10,820 to minus 11. 90 divided by 690. And this gives us an acceleration off approximately 12 meters per second squared and this is the maximum possible acceleration. Any acceleration above this value will make the tires off. The car is not on the ground. And we'll note that these expression for the static frictional force is useful when we are copulating the maximum possible frictional force. So this means that we are using the maximum possible frictional force to calculate acceleration.