We
are now ready to look at some applications of Newton's laws. There are three types of forces that we will
deal with in this course. The first is
applied forces. The other two are
natural forces and contact forces.
Applied
forces are those that arise due to the interaction of the object with some
other object. Usually the two objects
are in contact for a brief period. An
example of this type of force is the force exerted by a person pulling a block
or the tension in a line. These forces
always depend on the situation being examined.
Natural
forces are those that arise from fundamental interactions in nature. In reality, both direct action forces and
contact forces are applications of natural forces. There are four known natural forces. From our knowledge of particle physics we know that all of the
natural forces can be characterized as a force field times a charge. Each charge is associated with a particle.
The
first of the natural forces to be studied was the gravitational force. This
is the force that attracts to bodies together.
It creates weight and keeps the planets in orbit around the sun. It has an infinite range. The equivalent charge is the mass of the
body, and the associated particle is called the graviton. It is the only one we will look at this
semester.
The
second force that was known classically is the electromagnetic force. This
is more commonly recognized as electric forces and magnetic forces. As with gravity, it has infinite range. The charge is the electric charge, and the
associated particle is the photon. It
is studied in the second semester of this course.
The
third force is the strong nuclear force. This is the force that holds an atomic
nucleus together. It has a limited
range (roughly the diameter of a nucleus).
The charge is the strong nuclear charge, or color charge, and the
associated particles are called gluons.
The
last known natural force this the weak
nuclear force. This is another
short ranged nuclear force. It is
responsible for the decays of certain particles. The charge is the weak nuclear charge, or hypercharge, and the
associated particles are called the gauge bosons.
Since
the early 1970's there has been considerable effort to unify these four forces
into a single theory that describes the forces as a single force. To date, only the electromagnetic and weak
forces have been unified, although there are various theories, called grand unified theories, that unite
electromagnetic, strong and weak forces.
Attempts to unify these with gravity are found in the various superstring theories, or the theories of everything, that are being
postulated.
The
last class of forces we will deal with are the contact forces. These forces arose due to the contact of one
object with another. They can be
decomposed into two components. The normal force, which we encountered earlier, acts perpendicular to the
contact surfaces. The parallel
component is the friction force and
acts to either hold the objects at rest or to resist their motion. In general, the true nature of the friction
force is complex and difficult to describe.
However, there are certain general properties which can be described and
serve as an easy means to describe the force.
The main property is that the magnitude of the friction force is
proportional to the magnitude of the normal force. The constant of proportionality depends on whether the objects
are moving relative to each other or not.
If
the objects are moving relative to each other, then the magnitude of the
friction force is given by

f
= m_{k}N 
(6.1) 
where m_{k} is called the coefficient of kinetic friction. The coefficient
of kinetic friction is a property of the object and must be determined based on
its composition. Don't forget that the
friction force acts parallel to the contact surfaces and in the direction
opposite the motion.
If
the objects are at rest, then the friction force is approximately proportional
to the normal force. Its magnitude can
range from zero up to a maximum of m_{s}N. Thus
the friction force is given by

_{} 
(6.2) 
where m_{s} is the coefficient
of static friction. The equality holds only when the object is
about to begin motion. In most cases, when
sliding begins, the friction force decreases.
This means that, in general, the coefficient of kinetic friction is less
than the coefficient of static friction.
Example:
A
toboggan loaded with vacationing students slides down a long, snowcovered
slope having a coefficient of kinetic friction m_{k}. The
slope has just the right angle to make the toboggan slide with constant
speed. Find the angle of the slope.
Since
the toboggan is sliding at constant speed, the acceleration is zero. We chose a frame of reference that is
parallel to the slope. In this frame of
reference, the force equations are

component: mg
sinq  m_{k}N = 0
^
component: N
 mg cosq = 0
Look at the  equation
first
_{}
Substituting this into
the ^ equation and solving for q yields
_{}
Thus, a constant
velocity is achieved only if the tangent of the angle is equal to the
coefficient of kinetic friction, regardless of the number of people on the
toboggan.
Example:
A 75 kg man hangs from the middle of a tightly stretched
rope so that the angle between the rope and the horizontal direction is
5°. What is the tension in the rope?
Since
the two sections of the rope are symmetrical with respect to the man, the
tensions must have the same magnitude.
Thus, summing the forces in the y direction yields
_{}
How
do we go about solving force problems with acceleration? As always, we follow the steps outlined in
the first chapter, taking special attention to the magnitudes and directions of
the various forces involved. We then
set up the vector equations using F
= ma, and solve for the
acceleration and subsequent motion.
Example:
An
elevator and its load have a total mass of 800 kg. Find the tension T in the supporting cable when the
elevator, originally moving downward at 10 m/s, is brought to rest with
constant acceleration in a distance of 25 m.
We
first need to find the acceleration.
This can be found from the equation
_{}
All of the forces are in
one dimension, so we get
_{}
Notice that the tension
is greater than the weight. This is the
cause of the upward acceleration. Also
note that the relative signs on the acceleration and gravity were accounted for
in setting up the equation, so we use the positive values when substituting.
Example:
A
block of mass m_{1} moves on a level surface with a coefficient
of kinetic friction m. It is
connected by a light (massless), flexible cord passing over a small,
frictionless pulley to a second hanging block of mass m_{2}. What is the acceleration of the system, and
what is the tension in the cord connecting the two blocks?
We
treat each block as a separate system.
Starting with the block on the table we get
x
component: _{}
y
component: _{}
Solving for the
acceleration we get
_{}
The forces acting on the
second block are
x
component: 0
y
component: T – m_{2}g
= m_{2}a
Substituting the
previous result for a, we get
_{}
The acceleration can be
written in terms of the masses by substituting this back into the acceleration
equation.
As
continued examples of the applications of Newton's laws, we now turn to
circular motion. Recall that we saw
that if an object was undergoing circular motion at a constant speed, it
experienced centripetal acceleration which had a magnitude of
a_{c} = v^{2}/r
By Newton's laws, this
acceleration is related to a force.
This "force" is called the centripetal force and is given by
F
= mv^{2}/r
pointing towards the
center of the circle. Notice that the centripetal force is not a physical
force. The actual force used on the
left hand side of Newton's law must be supplied by some external force.
Example:
A
2 kg rock is spun at the end of a 3 m rope in a horizontal circle. If it has a constant speed of 5 m/s, what is
the tension in the rope?
The
only true force acting on the rock in the horizontal direction is the tension
in the rope. This must be the source of
the centripetal acceleration, so
_{}
Example:
A
car of mass m is rounding a flat curve with a coefficient of static
friction m. If the
curve has a radius of r, what is the maximum speed that the car can go
before it begins to slip?
If
there were no friction on the road, the car would continue to move in a
straight line. The friction between the
tires and the road cause the car to turn.
Hence, we see that the centripetal force is caused by the friction
force. Since the road is flat, the
normal force is just equal to the weight of the car, as so
_{}
Example:
At
what angle should the road be banked so that friction is not needed to keep the
car on the road?
Splitting
the normal force into components, we get
x
component: N
sinq = mv^{2}/r
y
component: N
cosq = mg
dividing the first
equation by the second one, we get
tanq = v^{2}/gr
If the curve has a
radius of 100 m and is part of the interstate (with a legal speed of 29.0563
m/s), then the banking angle should be
_{}
For
horizontal circular motion it is possible to have the speed be a constant. However, if the circle is vertical, we must
also include the tangential acceleration due to gravity.
Example:
What
is the minimum speed that a roller coaster can go in order to stay on the
tracks at the top of a loop if the loop is 20 m tall?
The
forces acting on the car at the top of the loop are its weight mg, and
the normal force of the tracks against the car N. These must equal the centripetal force.
mg
+ N = mv^{2}/r
If the velocity is to be
the minimum, then the normal force would be zero, so this becomes
_{}
Example:
A
rock is whirled at the end of a rope in a vertical circle. Find a general expression for the tension T. What is the magnitude of the total
acceleration?
Let
the rock have a mass m, the rope have a length r, and measure the
angle from the bottom of the circle.
Then for an arbitrary angle q, the forces are
tangential: _{}
normal: _{}
We can find the tension
from the normal component,
T
= m(v^{2}/r + g cosq)
The magnitude of the
total acceleration is
_{}
Notice that the tension is greatest at the bottom of the loop, and is smallest at the top.