Recall
Topics from Physics 217, especially simple one dimensional problems.
Newton's Laws
Newton expressed his three laws in words. These are repeated in the
text, for brevity, I will put them in a more mathematical form. In
mathematical expressions, boldface symbols or symbols with an arrow above
will represent vectors.
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If F = 0 then p = constant.
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F = dp/dt.
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To every action there is an equal and opposite reaction.
What are Fowles and Cassidy saying in their historical perspective?
The idea that bodies tend to remain in motion unless acted upon by an external
force is counter to our earth-bound experience, but can be arrived at by
some clever observations: i.e. Galileo's realization that,
although a mass sliding down a plane will come to rest, this is due to
friction. By reducing the friction, the mass will slide further and
further, and in the limit that the friction becomes zero, the mass will
slide indefinitely. Then, one can see that, since it does not take
a force to maintain constant motion, but rather to alter motion, F
= ma is the correct relationship, not F = mv.
The great success of Newton's Laws is to put the explanation of all
motion into one set of rules. As noted in the book, earlier attempts
required different rules for the orbits of planets and moons in space,
and the motion of rocks and apples on Earth.
Inertial Reference Frames
The correct application of Newton's Laws requires an inertial frame of
reference. A frame a reference means simply, the coordinate system
that is used for measuring positions, velocities, etc. in a problem.
An inertial frame of reference is one that is not accelerating -- we shall
see later that a non-inertial frame can be used, if understood, but will
produce "fictional" forces.
The concept of inertial reference frames is a bit tricky. Basically,
an inertial reference frame is an idealization, like Galileo's frictionless
surface. For many problems we will study, the surface of the Earth
is a reasonably good approximation to an inertial reference frame.
Later in the course we will work some problems where effects due to the
Earth's rotation become important, at which point we will (implicitly)
use the center of the Earth, with directions fixed in space, as the reference
frame.
Momentum, Mass and Force
The "quantity of motion" Newton refers to is what we now call linear momentum,
p.
Expressing Newton's second Law as F = dp/dt is more
general, and correct, than F = ma. The form
using ma implicitly assumes that the mass of a moving object
is constant, but there are many problems where the mass is not constant
(rockets, or sand filling a cart), and then the form involving the derivative
of the momentum is better. (Also, this form is correct for speeds
near the speed of light, where special relativity applies.)
If two bodies interact, with no other outside influence, then by Newton's
third Law, F1 = -F2, or by the
second law, dp1/dt = -dp2/dt.
This is equivalent to d/dt(p1+p2) = 0, or p1
+ p2 = constant. This is the law of conservation of linear
momentum. Conservation of linear momentum is one of the most fundamental
laws of Physics, and has thus far proven to apply in all situations!
For the time being, we will be dealing with "point-like" (non-relativistic)
objects, so we will use the second law expressed in the form
Fnet = ma ,
where the force is referred to as Fnet to remind
us that, to correctly use the second law, we must consider the total force
acting on an object. The basic problem we will address in this course
is, given the forces acting on an object and some knowledge of its initial
state (initial conditions), determine its motion. Last week we saw
an example of this when we determined the motion of an object subject to
a constant force (gravity) in one dimension.
Rectilinear Motion
The above equation is a vector equation. Initially we will consider
problems where motion occurs in only one dimension, so we will use the
scalar form Fnet = ma , where it is understood
that we are only interested in the component of the force in the direction
of allowed motion. This is called the "equation of motion".
Example 1 -- Frictional Force
As an example, let us consider again the problem of the block on an inclined
plane, only this time we will include a frictional force. [Include
diagram of block on an inclined plane, with force vectors, and components
perpendicular to and parallel to the plane, like Fig. 2.2.1 in the text.]
We let the frictional force be f = µN . (I will omit
the supscript k, since this is the only type of friction we will consider,
and therefore we can't be confused with the coefficient of static friction.)
The normal force of the plane on the block is N = mgcos[theta]
, therefore the frictional force is f = µmg cos[theta]
.
There are 2 cases to consider:
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The block is sliding down the plane. The frictional force is always
directed opposite the motion, in this case, opposite the component of the
weight. Fnet = mg sin[theta] - µmg cos[theta]
= mg ( sin[theta] - µ cos[theta]). The equation
of motion becomes [xddot] = F/m = g ( sin[theta] - µ
cos[theta]).
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The block is sliding up the plane. The frictional force is still
directed opposite the motion, but this time that is in the same direction
as the component of the weight. Fnet = mg sin[theta] +
µmg cos[theta] = mg ( sin[theta] + µ cos[theta]).
The equation of motion becomes [xddot] = F/m = g ( sin[theta]
+ µ cos[theta]).
There is no easy way to combine these two cases into one equation that
you can apply without first having to stop and consider in which direction
the block is moving. If a problem involves motion up and down the
plane, then you must divide the problem into two parts, get solutions to
each, and connect the solutions together as necessary to get a physically
meaningful result.
Example 2 -- Time Dependent Force
Suppose that we have a problem where the force changes with time.
We encounter such situations every day, in what is less affectionately
called "stop and go traffic". In rush hour traffic, cars and buses
accelerate and decelerate and accelerate again. For this example,
let us assume that a car is subject to a force given by F = F0sin(t)
where F0 is a known constant. Does this unfortunate
driver ever make her way home? In other words, what is the position
as a function of time, and does the position get arbitrarily large, or
is it bounded, like the force function?
[Include a simple diagram.]
Since it is not stated to the contrary, I am assuming that the car travels
over a flat surface in the x direction. Initially the car
is at rest at x=0. The only force of any consequence is the
oscillating force given in the problem. Thus, the equation of motion
is:
F0sin(t) = ma = m[vdot] = m[xddot]
At first, this may seem a bit daunting, but recall how we integrated
the time dependent velocity to determine the position in the constant force
problem. This time, we simply start with a time dependent force,
and integrate first to get the velocity, then again to get the position.
v = F0/m [Int from 0 to t] sin(t)dt + v0 = F0/m (1-cos(t)) + v0 =
F0/m (1-cos(t))
Since the car is at rest at t=0, it must be that v0=0
(check by setting t=0 in above equation). Now, what is the
position as a function of time? The position is the integral of the
velocity:
x = [Int from 0 to t] v(t)dt + x0 = F0/m [Int from 0 to t] (1-cos(t))dt
= F0/m ( t - sin(t))
This is a solution, but what does it mean and does it make sense?
Let's make a plot of the position versus time.
[Plot of position versus time.]
This function is a straight line with a slope of F0/m with a
sine function superimposed on top. Clearly this function does eventually
reach any arbitrarily large value of x. Therefore, this driver
would eventually reach her destination, despite the constant accelerations
and decelerations. Interestingly, the car is always moving forward,
that is, the velocity is always positive (or zero). I leave this
as an exercise for you to plot the velocity and verify this statement.