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You are driving with a friend who is sitting to your right on the passenger side of the front seat of your car. You would like to be closer to your friend, so you decide to use physics to achieve your romantic goal by making a quick turn. (a) Which way (to the left or the right) should you turn the car to get your friend to slide toward you? (b) If the coefficient of static friction between your friend and the car seat is 0.55 and you keep driving at a constant speed of $15 \mathrm{m} / \mathrm{s},$ what is the maximum radius you could make your turn and still have your friend slide your way?

$0.28,$ no

Physics 101 Mechanics

Chapter 6

Circular Motion and Gravitatio

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rutgers, The State University of New Jersey

Hope College

University of Winnipeg

Lectures

04:34

In physics, kinematics is the description of the motion of objects and systems in the frame of reference defined by the observer. An observer has to be specified, otherwise the term is meaningless.

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In mathematics, a proof is a sequence of statements given to explain how a conclusion is derived from premises known or assumed to be true. The proof attempts to demonstrate that the conclusion is a logical consequence of the premises, and is one of the most important goals of mathematics.

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so to solve this problem, we need Teo. No fewer questions. So first of all your friends lights toward you when the friction force on her is insufficient to make her turn as you turn. So we can set f s, which is a fiction force as, um us times. Ah, the normal force. And then we can apply thie force equals mass times acceleration equation on your friend and then we can on and we can set her exploration are to be particular set a radial acceleration as we squared over r for returning because if ah, you're turning then you have the radius. Are you have the acceleration that is pointed towards thie radius off the circular part and ah, that's fi squared over r right. So we know that we'll be using these equations toe solve the problem now in part A. Ah, it's just we need to figure out the direction so we can see that without the friction, your friend moves in a straight line. So if there's no friction, then you're just going in a straight line right on DH sensi have friction. That's why you have all this. Relax, oration turn or the Yeah, relax. Elation down that's coming into picture. So if you done right toward her, then she would be closer to you, right? So the direction ah, would be turning right. All right, Esso in part B First, let's draw the fever, the diagrams. So on your friend. So this right here is your friend and ah, if we look at the forces that's acting on your friend is so there's one friction force acting on the right and then, ah, we have the normal force and then the gravitational for that's compensating the normal force. So we can separate the forces on X and white and then count the are calculated, the individual extraction forces. And why direction forces. So for X for why Direction force, We have the normal force, which is balancing the gravitational force. So that's why we're sitting n equals m d you and is the normal force. And then for ah, the extraction force, we have, um, the friction force which is equal to mass times X and anarchist. It is a radial or the radial exploration that that's acting on her. So the friction forces mew est times and so we're using this equation. Right? Here s so friction force is equal to M times, eh? Ah, radial acceleration. And we can use this expression right here where this gives us the relation between the relax, elation and velocity and the radius off the circle. So we substitute Reese grade over Big are So it's actually write bigger this to be consistent. So yeah, we substitute the squared over r and here we get N V squared over r, and then we have Mu s, and then we're substituting this quantity here to get mg. So from there we see that m is common on both sides. So we get rid of em down there and if we solve for the radius, we see that our is basically V squared over new times us time. See where amuses thee coefficient of friction. So if we tell you know the numbers we see that the radius eyes 42 meters. Now there's one thing that ah, you might have noticed, Andi. That's if you if our coefficient of friction new s is going up, then our radius should go down because we see there's an inverse relation between radius and mu s or decryption friction. So if muses going up, then our radius are really soft. Turn will be smaller. Thank you

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