Newton's Laws Applied to an Accelerating Car
Date: 05/26/2005 at 01:09:58 From: Pun Subject: Car acceleration Newton laws When a car accelerates, what external force is acting on it? When a person sits in the car, why does he feel "backwards movements" on him? My teacher said it's something to do with inertia of the body and Newton's Law 1 and Newton's Law 3. Do you feel a forward or backwards movement first? Can you explain every single aspect when the car accelerates until the person is moving in the same state of motion as the car?
Date: 05/26/2005 at 08:30:30 From: Doctor Rick Subject: Re: Car acceleration Newton laws Hi, Pun. When a car accelerates, it is because the road is pushing it forward! Here's a little more detail. The car's engine turns the wheels. When the tire turns at a constant rate, the point at the bottom of the tire is stationary for a moment; the wheel acts like a lever with the fulcrum at the bottom, because friction of the tire against the road keeps it from slipping. The axle, at the center of the wheel, moves at the speed of the car; the point at the top of the wheel moves twice as fast. When the car is accelerating, the wheels undergo angular or rotational acceleration: they turn faster and faster. This is accomplished by angular force (called torque) exerted by the engine on the axle. At the bottom of the tire, this torque takes the form of a force acting backward, pushing against the road. But as long as the tire is not slipping (skidding), the point at the bottom of the tire continues to be stationary: it is not accelerating. By Newton's first law, it is "an object at rest and staying at rest" and this is only possible if the net force on it is zero. The axle is exerting a force on that point, pushing backward against the road. In order for the net force to be zero, there must be another force of equal magnitude, pushing forward on the point at the bottom of the tire. This force is exerted by the road; it is Newton's third law "equal and opposite" reaction force. When we turn our attention from the point at the bottom of the tire to the entire car, we find that this reaction force of the road against the tire, pushing forward, is the only unbalanced force acting on the car. (I'm assuming a flat road, so that the force of gravity pulling down on the car is balanced by the "normal force" of the road pushing up on the car, keeping the car from sinking into the road.) When an unbalanced force acts on an object (the car), the object accelerates in the direction of the force (Newton's second law). So it's the road pushing on the car that makes it accelerate forward. Of course the power comes from the car's engine, but in terms of external forces, that's the way it is! Now we can turn our attention to you, sitting there with your seat belt on. As the car moves, your body is subject to Newton's first law: it is minding its own business, moving along at constant speed (let's say 60 miles per hour) along with the car. But as the car accelerates, the next moment the seat back is moving slightly faster than you are, so it catches up with you. You and the seat back can't be in the same place at the same time, so the seatback pushes forward against you: it exerts a force directed forward. By Newton's third law, your body exerts a reaction force against the seat back, directed backward. Let's get some math into the picture. Let's say the road is exerting a forward-pointing force F against the car, the car (without you) has mass M and its acceleration is A. You have mass m, and the car is pushing against you with force f. Your body pushes back against the seat back with equal force f, and it accelerates with acceleration a. I can set up two free-body diagrams, one for the car and one for you: f<---- CAR ---->F YOU ---->f MA ma By Newton's second law, I derive the following equations from the free-body diagrams: F-f = MA f = ma I use the second equation to substitute into the first equation: F - ma = MA I'd need another equation in order to solve for both accelerations, a and A. This equation would have to do with how hard your body and the seat back push back when something gets too close. It won't be a simple equation, but I can assure you that eventually the seat back must stop getting closer to you: its acceleration and yours become equal. Once that has happened, we have F - ma = Ma F = Ma + ma F = (M+m)a a = F/(M+m) This is no surprise, because when we consider the car plus you to be a single free body, we get the diagram CAR+YOU ----> F (M+m)a All the time the car continues accelerating with constant acceleration a, you will feel that force of the seat pushing forward against you, and you will be pushing backward against the seat. If you put a bathroom scale between you and the seat back, it would register a weight. The scale normally measures the force with which the earth pushes up on you to keep you from falling through; in this case it is measuring the force with which the seat back pushes forward on you to get you going faster. That's why when the car accelerates, you feel as if you are pushed back into the seat. One more step: Let's say the car has reached 65 miles per hour, which is the speed limit, so it stops accelerating and goes at a steady 65 miles per hour. Now there is no more force between you and the seat. But when the car turns off the highway and decelerates to 45 miles per hour, the forces reverse direction. Now your body is moving faster than the car, so it moves forward until the seat belt stops it. In order to stop you from moving forward, the seat belt exerts a force on you, and you exert an equal and opposite forward force against the seat belt. When the car gets down to 45 miles per hour and stops decelerating, the forces go away again. How's that? Does it answer all your questions? Thanks for asking. If you'd like to discuss a point further, I'll be happy to do so. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.