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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/ 
Associated Topics:
High School Physics/Chemistry

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