Download Chapter 4 Forces and Newton`s Law

Chapter 4
Forces and Newton's Law
GOALS
When you have mastered the concepts of this chapter, you will be able to achieve the
following goals.
Definitions
Define each of the following terms, and use each term in an operational definition:
force
coefficient of friction
weight
centripetal force
frictional force
weightlessness
Newton's Laws
State Newton's laws of motion and gravitation.
Resolution of Forces
Given the force acting on a system, draw a force diagram and/or resolve forces into
their components and/or solve for an unknown force.
Newton's Second Law Problems
Given any two of the three variables in Newton's second law, solve for the third.
Centripetal Force Problems
Given any three of the four variables in the centripetal force equation, calculate the
value of the fourth variable.
PREREQUISITES
Before you begin this chapter you should have achieved the goals of Chapter 3,
Kinematics, including uniformly accelerated motion and uniform circular motion.
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Chapter 4
Forces and Newton's Law
OVERVIEW - The statements which we call Newton's Laws of Motion are considered to
be a central feature of the study of the motion of objects. These laws generalize
much of our common experiences with forces and their influence on the state of
motion of objects. As you read and consider the Laws of Motion presented in this
chapter, make a point to relate the meaning of the laws to real experiences you
have had with bicycles, automobiles, electric trains, etc.
SUGGESTED STUDY PROCEDURE - As you begin your study of this module, begin by
carefully reading each of the Chapter Goals: Definitions, Newton's Laws,
Resolution of Forces, Newton's Second Law Problems, and Centripetal Force
Problems. Remember that on the next page of this Study Guide you will find a
paragraph concerning each of the terms listed under Definitions. Next, read text
sections 4.1-4.7. Remember that answers to questions asked in the text are given
in the second section of this Study Guide. As you read, be sure to consider the
examples given. If you have difficulties with the mathematical treatment of
forces and their components, refer to the initial chapter of this Study Guide
(Chapter 0).
At the end of the chapter, read the Chapter Summary and complete
Summary Exercises 1-17. Now do Algorithmic Problems 1, 2, 5, 6, 7, 8, 9, 10, and
11. Check your answers to each problem carefully. Now complete Problems and
Exercises 1, 2, 5, 8, 10, 12, 14, 26, 27, and 28.
Now turn to the Examples portion of this Study Guide and consider the
additional problems and questions presented. After you have completed this
task, you should be prepared to attempt the Practice Test on Forces and
Newton's Laws. This study procedure is outlined below.
--------------------------------------------------------------------------------------------------------------------Chapter Goals
Suggested
Summary
Algorithmic
Exercises
Text Readings
Exercises
Problems
& Problems
--------------------------------------------------------------------------------------------------------------------Definitions
4.1, 4.2, 4.3,
1-6
4.4,4.5,4.7
Newton's Laws
4.2, 4.3
7-11
2, 9, 10
8
Resolution of
Forces
4.2, 4.4
12, 13
2, 10, 11
1, 2, 5, 10,
12, 26
Newton's Second
Law Problems
4.2, 4.3, 4.4,
14-17
1,5,6,7
8,28
4.6
Centripetal
Force Problems
4.6
8
14,27
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DEFINITIONS
FORCE - A measure of the strength of an interaction; a push or pull; the effect of a force
is to alter the state of motion of a body. A vector quantity with a magnitude
measured in newtons (N).
The strength of the earth's gravitational field is measured by the weight of an
object. Weight is a force acting vertically downward.
WEIGHT - The product of the mass and the acceleration due to gravity.
Another way to think of g is to treat it as the strength of the force of gravity near
the surface of the earth. Then the weight is a downward, vertical force.
FRICTIONAL FORCE - Force that opposes the motion of an object.
In all of the cases of motion you commonly encounter there is a force acting to
oppose the motion. When you are coasting down a slope on your bicycle, what
are the forces that tend to slow you down? All of them can be classified as
frictional forces.
COEFFICIENT OF FRICTION - Ratio of the maximum frictional force to the net force
pressing the surface together.
Try rubbing two sheets of paper together, then try rubbing two sheets of
sandpaper together. Which requires the greater force to slide them along? This
illustrates the larger coefficient of friction of sandpaper. If you wet the
sandpaper, what happens to the force required to slide the paper along?
CENTRIPETAL FORCE - The unbalanced force acting on the body towards the center
of rotation in circular motion.
The radial acceleration of an object going around in a circle is proportional to
centripetal force.
WEIGHTLESSNESS - The gravitational attraction of the earth for an object in motion
around the earth is precisely equal to the centrifugal force of the body in its orbit.
When you are in such an orbiting spacecraft floating objects and persons are the
rule rather than the exception. Ask your physics instructor to show you some of
the interesting films made from NASA photographs.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
SECTION 4.4 Force of Friction
Suppose you think of oil as composed of many smooth planes loosely packed
together. Then the oil could slide easily along the planes and would greatly reduce
friction in those directions. This is not a very good model for oil. Can you think of a
better one? At least this smooth planes model for oil offers an explanation for the
reduction of friction in certain directions.
Many frictional forces are essential. They enable us to place ourselves in a fixed
position and remain there even when subjected to small external forces. If your chair
were resting on a frictionless floor, the slightest breeze would blow you away from the
table where you are studying physics. That would be too bad, wouldn't it?
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Indeed, it would be difficult to live as you presently do in a frictionless world. If
we got started moving, how would we stop?
GENERAL COMMENTS (valuable insights included in the , free of
charge)
NEWTON'S LAWS OF MOTION as given in Latin in his famous book PRINCIPIA.
Lex. I - Corpus omneperseverare in statu suo quiesendi vel movendi uniformiter in directum,
nisi quatenus illuda viriblus impressis cogitur.
Lex. II - Mutationem motus, proportionalem esse vi motrici impressae, et fieri secujdum lina
lineam rectam qua vis illa imprimitur.
Lex. III - Actioni contrarium semper et aequalem esse reactionem; sive corporum duorum
actiones in se mutuo semper esse aequales et in partes contrarias dirigi.
COMMENTS ON NEWTON'S FIRST LAW
The inertia of a body may be defined as that property of a body that tends to resist a
change in its state of rest or motion. Mass is defined as a quantitative measure of
inertia.
COMMENTS ON NEWTON'S SECOND LAW
Another way of stating the classical version of Newton's second law is: when a body
is acted upon by a constant force, its
resulting acceleration is proportional to the force and inversely proportional to its
mass; a = F/m
COMMENTS ON NEWTON'S THIRD LAW
It is important to note that the action force and the reaction force in Newton's third
law of motion act on different bodies.
EXAMPLES
RESOLUTION OF FORCES
1. A beanbag chair is at rest on a driveway inclined at 22ø. The chair has a weight
of 68.0 newtons. (a) Draw the diagram of the chair and label all of the forces
acting on the chair. (b) Calculate the values for all forces, etc., for this situation.
What Data Are Given?
The weight of the chair (68.0 N) and the angle of incline of the plane (22ø) are
given. The velocity of the chair is zero throughout.
What Data Are Implied?
The weight is a force acting vertically down in a region where the strength of the
earth's gravitational force produces an acceleration of 9.80 m/s2. The chair is not
moving because of the presence of friction, so a frictional force is implied in this
problem. Likewise, the plane is supporting the chair so a normal force is acting
on the chair because the plane is implicit in this problem.
What Physics Principles Are Involved?
Both the second and third laws of Newton are necessary to understand this
problem as well as knowledge of the vector nature of force. Before we go on to
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the algebraic solution of this problem we are ready to use a force diagram to
answer part (a) and get us ready to solve part (b).
(a) Force Diagram
What Equations Are to be Used?
We can make use of Newton's Second Law. We know the velocity remains
constant (at zero) so the acceleration is zero; thus, the net force on the chair must
be zero. That means the vector sum of the forces on the chair must be zero;
W+N+f=0
Algebraic Solution
The vector equation can be written as two separate scalar equations, a vertical
component equation and a horizontal component equation.
Vertical forces = 0 = W = N cos θ - f sin θ where positive is downward.
(1)
Horizontal forces = 0 = N sin q - f cos θ where positive is to the right.
(2)
These equations can be combined to solve for the magnitudes of the two
unknown forces f and N.
Rearrange Equation (2) and divide by cos θ:
f = N sin θ / cos θ) = N tan θ
(3)
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Rearrange equation (1), divide by cos θ and substitute the friction for f,
N = (W/cos θ) - (f sin θ/cos θ = (W/cos θ) - N tan2 θ
(4)
Now combine the terms containing the normal force N, so
N (1 + tan2 θ) = W/cos θ or
N = W / [cos θ (1 + tan2 θ)]
(5)
Substitute this expression into Equation (3) to obtain
f = W tan θ / [cos θ (1 + tan2θ]
(6)
Numerical Solution
W = 68.0 N; θ = 22ø, so cos θ = 0.927; tan θ = 0.404.
N = 68.0 N / [0.927(1 + (.404)2)] = 63.1 N perpendicular to the driveway.
f = N tan θ = (63.1 N) (.404) = 25.5 N, parallel and up the driveway.
Thinking About the Answer
Notice that both answers come out in the correct units, newtons. The normal
force is much larger than the friction force. This seems reasonable for a low angle
driveway. Notice as the angle decreases in Equations 5 and 6 the normal force
approaches a value of W and the friction force goes to zero. Is that what you
expect?
NEWTON'S SECOND LAW PROBLEMS
2. Assume the beanbag chair given in the previous problem is lying at rest on an
icy (friction ÷ zero) driveway. What happens? Give quantitative results.
What Data Are Given
Same as in the previous problem except now the friction force f is equal to zero.
What Data Are Implied?
Same as in the previous problem, except the frictional force is zero. Motion along
the driveway can be deduced.
What Physics Principles Are Involved?
As before, Newton's 2nd and 3rd laws. Now Newton's 2nd law shows that the
beanbag chair will undergo some acceleration down the driveway since the net
force cannot be zero. This is clearly seen by removing the frictional force from the
previous force diagram. Then you are left with
It is not possible for the weight and the normal force to add to zero. Hence, the
chair must experience some net force and some acceleration.
What Equations Are to be Used?
The vector equation for the 2nd law in this case is given by
ma = W + N
(7)
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However, we know the chair can only move along the driveway since there is no
force acting to lift the chair. Hence we can transform Equation (7) into a scalar
equation by considering only motion and forces that act parallel to the driveway.
The normal force has no components parallel to the driveway. The weight has
only one component, W sin q, acting down the driveway, so we can write
ma = W sin θ {motion down the driveway}
(8)
Algebraic Solution
Remember, only the weight W, g, and q are given, so we need to replace the
mass on the left side of Equation (8) by W/g, thus
(W/g)a = W sin q or
a = g sin θ
(9)
Numerical Solution
g = 9.80 m/s2, sin 22ø = 0.375
a = (9.80)(0.375) = 3.68 m/s2, down the driveway.
Thinking About the Answer
Note the units are correct, and that if the slope of the driveway is decreased to
zero degrees the acceleration goes to zero; i.e., the chair remains at rest, does that
seem reasonable to you?
CENTRIPETAL FORCE PROBLEMS
3. An 80.0 kg earth satellite circles the earth once every 94.2 minutes at a mean
altitude of 522 km. Use 6.38 x 106 m as the radius of the earth. (a) What is the
gravitational force acting on the satellite while in orbit? (b) What fraction of its
force at the surface of the earth is it? (c) If you were riding inside the satellite
what forces would you experience?
What Data Are Given?
The mass of the satellite is 80.0 kg. Data which can be used to calculate its
velocity and the radius of its circular path are given.
What Data Are Implied?
The problem intends for you to assume the path is a circle with the center of the
earth as the center of the path, hence the radius of the path is 6.38 x 106 m + 522
km = 6.38 x 106 + 0.52 x 106 = 6.90 x 106 m.
What Physics Principles Are Involved?
You need to recognize that the gravitational force must have the same magnitude
as the centripetal force. Furthermore, for an observer inside the satellite the
gravitational force will appear to be canceled by the centrifugal force and the
condition of weightlessness will be experienced.
What Equations Are to be Used?
The equation for centripetal force is needed.
F = mv2/r
(4.6)
In order to calculate the velocity v we need to divide the circumference of the
orbit by the time required for one pass around it.
v = 2π(radius)/time for one orbit
(10)
Algebraic Solutions
(a) In orbit Fo = mv2/r but v = 2πr/t
Fo = m4π2r2) / rt2 = 4π2mr /t2
(11)
(b) On the surface Fe = mg
Ratio of orbit force to surface force is given by F0/Fe = 4π2r/gt2
(12)
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Numerical Solution
r = 6.90 x 106 m
t = 94.2 minutes = 5.65 x 103 seconds
m = 80.0 kg; g = 9.80 m/s2
(a) Putting these values in Equation (11)
Fo = (80.0 kg)4π2(6.90 x106) / (5.65 x 103)2 = Fo = 6.82 x 102 kg/s2
Fo = 6.82 x 102 N
(b) Then Fo/mg = 0.87 which is 87%.
In summary, the force on the satellite in orbit is 682 N. Since its weight on earth
is 784 N, the orbital force is 87% of its weight on the earth. As an observer inside
the satellite you would observe the effects of the gravitational force being
balanced by a centrifugal force and you would experience weightlessness.
Thinking About the Answers
Notice again that the units come out correctly, 1 newton is a kilogram meter per
second squared. According to Equation (12) the ratio of the orbit force to the
surface of the earth force is independent of the mass of the object since the
symbol (m) does not appear in Equation (12). Does that seem reasonable to you?
PRACTICE TEST
1. Newton's Universal Law of Gravitation applies to all massive objects,
including planets, automobiles, and people. However, many people
misrepresent the gravitational law. Use your knowledge of Newton's Law
and write a short correction for the following statements.
a. Because of its larger mass, the sun attracts the earth with a greater force than
the earth attracts the sun.
b. The earth's rotation produces our gravitation. If the rotation were to stop, we
would drift into space.
c. A body taken to an altitude of about 100 miles has no weight because the
gravitational force does not extend out into space beyond the atmosphere.
2. The Fig. shown below shows a modified Russell traction apparatus for femoral
fixation.
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a. Construct a force diagram isolating point B showing all the forces which are at
that point.
b. Find the total horizontal force applied to the leg at B by the apparatus when a
4 kilogram mass is suspended at A.
3. A 500 kilogram subcompact car is parked on top of a long straight incline
whose inclination is 30ø from the horizontal.
a. Draw a free body diagram isolating the car and showing all the forces holding
the car in equilibrium.
b. Assume that the brakes are released and the car rolls (without friction) down
the plane. Using Newton's second law, predict the car's acceleration.
4. Astronaut Owen Garriott (mass = 65 kilograms) is placed aboard a spinning
device used to determine his ability to withstand high "g" forces.
a. If the apparatus is revolving at 1
revolution per second, what force does
Garriott experience due to this rotation
motion?
b. How many times his weight is this
force? (how many "g" 's does he
experience?)
ANSWERS:
1. a) Newton's Universal Law states that these forces must be exactly equal in
magnitude, opposite in direction.
b) Rotation of the earth has little effect on our weight. Actually, if rotation were to
cease, our apparent weight (as measured by a spring balance on the surface of
the earth) would increase slightly.
c) The gravitational force extends indefinitely into space. In a space ship which is
circling the earth, we say an astronaut is weightless because in the condition of
"falling" around the earth, he exerts no force on a spring balance in his ship.
Actually, his actual weight (attraction to the earth) is 95% of the force at the
earth's surface.
2. 69.6 N
3. 5 m/s2
4. 25,600 N, 39 g's
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