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A 2000kg SUV can safely turn in a road curve with a radius of curvature of 63m and a banking of 17degrees. If the coefficient of friction between tires and road is ...

Question

A 2000kg SUV can safely turn in a road curve with a radius of curvature of 63m and a banking of 17degrees. If the coefficient of friction between tires and road is 0.72,determine the normal pressure of its tire exert on the road in terms of Newton.

A 2000kg SUV can safely turn in a road curve with a radius of curvature of 63m and a banking of 17degrees. If the coefficient of friction between tires and road is 0.72,determine the normal pressure of its tire exert on the road in terms of Newton.



Answers

Friction is needed for a car rounding a curve. But, if the road is banked, friction may not be required at all. What, then, supplies the needed centripetal force? (Hint: Consider vector components of the normal force on the car.)

Right this problem because it's the concept that the circular motion and we are going to use this equation to solve are open. So the next readily word falls is four times the maximum static frictional force between the tire and the road that is acting along the radial and more direction. And that is equivalent to the mass of the trip, into the square of the speed upon the radius R. And from this, the radius of the cough can witness uh M. V. Square upon four and the maximum static frictional force and substitute the value so the radius of the corvis, The mass of the truck is 1.5. And to industry kilogram into uh 20 medals, four seconds squared upon four times The maximum static fictional force that is eight into the industry look, are the radius of taco. The minimum radius Equals 18.8 m.

In this problem. First time changing the unit of pressure. The pressure is given to 40 K. P. A. Changing the unit, I just multiplied by 10 to depart three P by K. P. A. Which is equal to 24 multiplication and 10 to the power for P. Now we all know that weight is equal to MG, which is equal to 38 00 kg multiplication, 9.8 m per second squared, Which is equal to 37,240 Newton. Now we also know that the value of their physical to W by 40 years, So it is equal to 37,240 Newton by four, which is equal to 9310 Newton. Now we all know that the surface city of each tire is given by. The formula is equal to F by B, which is equal to 931 Little Newton by 2. 40 more duplicates and tend to the par three p. A. Which is equal to 0.0 0.3 88 m squared. As our final answer.

Put it This problem? You have a car traveling on highway in the section off the higher where highway we're moons. Radius are. Yeah. I strongly on the section of the highway. We're going Saw scheme. What is the safest velocity? That car can travel since the highway with the section of the high ones. No bank this unbanked so we can go our green for me. Find a groom. Well, wait. Okay. My works. Two continents, cities. They have more course in their opposite damage. He has Mitchell course pointing to us the religions of the center. Thank you. Government on the highway we can do of the submission. Of course. Next sequel office would have on one force in these fours. Sure. I mean, is given my That's a professional stop. The correction times the number, of course. And these in time, the sequel Reagan Assimilation times the mass of metal. So have army. No. You know that this information it's by Union Square. This something submission next, then? Because in this square, huh? You also have summation since and of course in life, which is violence. If we combine these two questions, we have Yeah, that's a quote. Um, And because actually, we can just stick to the value off and in the FX equation, so we have can rest. And this will end you and her last. The square. Oh, those end course. Well, then so for velocity, uh, I mean, equal to events. Jeanne, Raiders Square. And this is Missus. This is the velocity. The safest lots of earned it. I took the car. Car trouble for highway, that is, um, bent. The second part of the problem is asking what happened? If the velocity who's higher, it's higher. Then there's this quote safe for these were just going to happen. Is static. Were you? Put your force when it's not enough. If the velocity increases, there is not gonna be on all. The only parameters are dreaming now, Constant. Then there is no enough force to keep the car in the circular. But secular Battisti, the hungry, they're part of the kind. They escaped the worst outside of the court. And this is Dick. You think you increased velocity about this Save money. This is a solution of the crop

Hi there. So, for this problem, we have a race track Sure, that has a radius of 120 m. And we are also given that this curve is bank at an angle of 18° and at the coefficient of static friction between the tires and the wrong. And the runway is 0.3. And there is a race card with mass uh huh. 900 kg. Um that rounds this curve with the minimum speed needed to know slide down the banking. So the first question is, um, what is the normal force exerted on it by the road? Of detergent on the um on the car bided road? So, um, the race card in this case is accelerated toward the center of the circle of the curve and the card goes at the proper speed for the banking angle there will be no friction force. Um In this case it is going is lower than the proper speed. So it will tend to slide down the road so the friction force will be up the road. And with this information we just can apply Newton's second law first, we're going to draw the forces acting on it. And with diagram And from there we're going to apply needles, like a lot to obtain the equations. So I'm going to draw first the assets and put in here the white upsets and the assets gonna set. And this is the positive X access and this is the white assets. So we will have here, we have our car at this point. So the forces that are at in this car are the following. We have to wait, connect your idea scene, read the weight of the card. We also have um, the normal force that is perpendicular to the surface. So it will be here. So in that case this I'm going here corresponds to to that 18°. Given for these, given us any information for this problem. And we also know that since and the card is and since the car will be sliding down the road, the friction force will oppose this motion. So it will be too the right. And this is the friction force. A static friction force at this in here corresponds to the angle tita. Now the system is moving as I said, to the center with a rapidly acceleration. No. Um the procedure is that we need to apply into the second law. So the first thing that are going to do is to some the forces that are acting on the Y component. So that will be that the normal force or the normal force um cousin of tita is equal to the weight. And oh no. We also have a company of the for the friction, as you can see from the picture that is going to be a positive description. But this is given a satisfaction but this is given by the sign of tita because the angle is with respect to the um the assets and these minus the weight is equal to zero because the system is not moving in the why and direction just in the it's the eruption. So this is equal to zero. And for the summer forces in the X company we will have that this is in normal force sign of Teton positive because we said this part positive minus because we set these negative that's will be the friction force consigned of tita because that angle is well respected ads at six and this is equal to mass times they're radiantly or this centripetal acceleration. Again in this case we want the normal force from the beginning weekends. And we know that the definition of the friction or the static friction force has a maximum value when it is equal to the aesthetic coefficient times the normal force. So we're gonna substitute that in the first equation to see what we can obtain from there. So we will have the normal forest assign of data blast, the friction force which is normal sign of tita and minus the weight. We know that the way it is defined as the mass times acceleration to do gravity and this is equal to zero. So from here we can solve for end. We can um we can first fact arise the end and The N. four. So we will have to follow him so we can pass this to the other side. So we will have this is able to positive weight. And the normal force is just the weight over this sum of angles. So we substitute all of these values because we know we know all of them. We know that the mass The car is nine 100 kg and the acceleration due to gravity is 9.8 m/s square. The consign of the angle the angle is getting in from the beginning is 1818°, so that we put that in here Plus 0.3 signs of 18°. And this when we plug this into the calculator, whipped in a value off 8450 noodles. So this is the normal force that it's been asserted um, in the car by the road. Um, Now, for part B of this problem, we are asked to determine what is the car's speed. So to determine the car's speed, we can use the fact that the centripetal acceleration is defined as the velocity squared over the radios. So from here will seem that the velocity is equal to and the square root of the centripetal acceleration times the radius. We don't know the acceleration, but we can obtain it from this situation right here. Um We can solve for the acceleration and we can substitute the values for the normal force. So I'm going to use that equation. I'm going to put in here again. That is the normal force times this sign of detail minus the minus friction force. We know that that is defined as the normal force time study description the provision of aesthetic corruption times the cosine of theta. And this is equal to and sometime in the acceleration. So we can just so for the acceleration and just um divide this by M. So if we put all of these values, we know mhm. That the normal force is given, that it is this value that we just have obtained. We put that in here, we can't um as you can see we can characterize that value. So we'll have sign of 18 degrees -0.3, a sign of 18 degrees. And this over the mass of the car. So the mass of the car is nine 100 kilograms. When we plot this into the calculator, whipped into the acceleration is let me put this into the calculators. So yeah, 0.22 meters per second square. So to find the velocity, we substitute this value into the equation that we sitting here and we use the radios for this case. So the radios um as 120 m. So we blocked that value in here and this is equal to 5.16 or mm 17m/s of this is the velocity of the car. So this is it for this problem. Thank you.


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