Download Chapter 4: Newton`s Three Laws of Motion First

Lecture 6 : Introduction to Newton’s Laws of Classical Mechanics
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Chapter 4: Newton’s Three Laws of Motion
First Law: The Law of Inertia
An object at rest will remain at rest unless and until acted upon by an external
force.
An object moving at constant velocity will continue to move at constant velocity
unless and until acted upon by an external force.
These two statements are the same when one considers that whether or not an
object has velocity depends upon one’s frame of reference.
Second Law: Relation between Force and Acceleration
An object will experience an acceleration a in direct proportion to the force F
exerted on the object. The constant of the proportionality is the object’s mass
m.
~ = m~a
F
(4.2)
If there is no force then there must be no acceleration. If there is no acceleration
then there must be no force. If there is force then there must be acceleration. If
there is acceleration then there must be a force.
Third Law: Action and Reaction Forces
Every (action) force which exists has an equal an opposite (reaction) force, but
the action-reaction forces NEVER act on the same body.
A force F~AB is exerted on body A by body B. By Newton’s Third Law there
must also be a force F~BA which is exerted on body B by body A.
F~AB = −F~BA
(4.6)
“It takes two to tango” could be just another way of stating Newton’s Third
Law. In other words, all forces are the result of two bodies (particles) acting on
one another.
Lecture 6 : Introduction to Newton’s Laws of Classical Mechanics
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Insight into Newton’s First Law of Motion
First Law: The Law of Inertia
An object at rest will remain at rest unless and until acted upon by an external
force.
An object moving at constant velocity will continue to move at constant velocity
unless and until acted upon by an external force.
This may be the most difficult law of all to grasp. In fact, for at least 2,000
years, until the time of Galileo and Newton, the law was simply not recognized
by humankind.
The View of Aristotle
The view which pervaded human thought until the 1600s was that objects were
“naturally” at rest on the surface of the Earth. In order to keep an object
moving, some force was necessary, although the word “force” would not have
been used. Once the force was taken away, then the object would slow down
and return to its “natural” state of rest.
Of course, there were certain objects such as the Sun, the Moon, the stars and
the planets, which appeared to be in a perpetual state of motion, unaided by
external force. Since this motion was not “natural”, then the objects had to be
“supernatural” or gods.
The View of Newton (Galileo and others of the time)
The present scientific view of motion is called Newtonian Mechanics, after Sir
Issac Newton. It was Newton who discerned the connection between motion and
force, and he was able to dispel the notion of a “natural” state of rest. His First
Law was truly a revolution, and we can best appreciate it through the air track
demonstration shown in class.
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Astronomy and Newton’s First Law of Motion
We conclude the study of Newton’s First Law by returning to the observations
of the seemingly perpetual motion of astronomical objects such as the Sun, the
Moon, and the planets.
Here again is the First Law:
An object at rest will remain at rest unless and until acted upon by an external
force.
An object moving at constant velocity will continue to move at constant velocity
unless and until acted upon by an external force.
Now we ask these questions relative to the First Law:
1) Is the motion of the planets consistent with the First Law?
2) Specifically, are all of these objects moving with constant velocity, and how
does one define constant velocity?
3) If a given astronomical object such as the Moon is not moving with constant
velocity, what inference can we derive from the First Law
With the answer to this last question, we (as was Newton) are drawn into the
discussion of the Second Law.
Lecture 6 : Introduction to Newton’s Laws of Classical Mechanics
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CHAPTER 5: Newton’s Second Law of Motion
The fundamental equation of mechanics is Newton’s Second Law of Motion:
F = ma
(4.2)
A FORCE acting on an object with mass m will produce an acceleration a. Note
that this is a vector equation, so it actually represents three separate equations
for the X, Y , and Z components of the force and the acceleration.
More than one force acting on an object
If there is more than one force acting on an object, the left hand side of the
Newton’s second law is simply replaced by the vector sum of the forces acting
on the body:
ΣN
(4.3a)
i=1 Fi = ma
=⇒ ΣN
i=1 Fix = max
ΣN
i=1 Fiy = may
ΣN
i=1 Fiz = maz
(4.3b)
It is very important that you realize that this equation applies to the forces
acting on the object. The most difficult thing for most students is to recognize
all, and only all, the forces which act on a particular object. Sometimes students
will leave out certain forces, and other times students will include forces which
are not acting on the object in question. The secret is to focus your attention
on the object in question.
ACCELERATED MOTION
Accelerated motion occurs when the left hand side of Eq. 4.2 is not zero. Then
the object must have an acceleration. Many of the problems you will have to
work out will involved computing the acceleration of an object when the object
experiences certain forces. Typical cases will include the force of gravity (weight),
the normal forces exerted by supporting walls and floors, and the static or kinetic
frictional forces.
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Example: Newton’s Second Law of Motion (page 111)
A hockey puck with a mass of 0.3 kg slides on the horizontal frictionless surface
of an ice rink. Two forces act on the puck. The force F1 has a magnitude of 5
N and acts at an angle of −20o , and the force F2 has a magnitude of 8 N and
acts at an angle of +60o with respect to the x axis. Determine the acceleration
of the puck. (Fig. 4.4 page 113) Solution This is a straightforward application
of Newton’s second law
F
(4.2)
F = ma =⇒ a =
m
We have to compute the net or resultant force acting on the mass of the puck.
The acceleration vector a will be in the same direction as F, and with a magnitude equal to the magnitude F divided by m.
Force
F1
F2
Fnet
Magnitude
(N)
5.0
8.0
Fnet =
Angle
(deg.)
−20.0
+60.0
θnet =
X Comp. Y Comp.
(N)
(N)
+4.70
−1.71
+4.00
+6.93
+8.70
5.22
So, by the usual methods of analytic vector addition, we find
q
Fnet = (8.70)2 + (5.22)2 = 10.15 N
θnet = tan−1
Fy
5.22
= tan−1
= 31.0o
Fx
8.70
Now to compute the acceleration
a=
10.15
Fnet
=
= 33.8 m/s2
m
0.3
The direction of a is exactly the same as that of Fnet :
θa = 31.0o .
Lecture 6 : Introduction to Newton’s Laws of Classical Mechanics
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Chapter 4: Newton’s Second Law F~ = m~a
First Law: The Law of Inertia
An object at rest will remain at rest unless and until acted upon by an external
force.
An object moving at constant velocity will continue to move at constant velocity
unless and until acted upon by an external force.
These two statements are the same when one considers that whether or not an
object has velocity depends upon one’s frame of reference.
Second Law: Relation between Force and Acceleration
An object will experience an acceleration a in direct proportion to the force F
exerted on the object. The constant of the proportionality is the object’s mass
m.
~ = m~a
F
(4.2)
If there is no force then there must be no acceleration. If there is no acceleration
then there must be no force. If there is force then there must be acceleration. If
there is acceleration then there must be a force.
Third Law: Action and Reaction Forces
Every (action) force which exists has an equal an opposite (reaction) force, but
the action-reaction forces NEVER act on the same body.
A force F~AB is exerted on body A by body B. By Newton’s Third Law there
must also be a force F~BA which is exerted on body B by body A.
F~AB = −F~BA
(4.6)
“It takes two to tango” could be just another way of stating Newton’s Third
Law. In other words, all forces are the result of two bodies (particles) acting on
one another.
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CHAPTER 4: Newton’s Second Law of Motion
The fundamental equation of mechanics is Newton’s Second Law of Motion:
F = ma
(4.2)
A FORCE acting on an object with mass m will produce an acceleration a. Note
that this is a vector equation, so it actually represents three separate equations
for the X, Y , and Z components of the force and the acceleration.
More than one force acting on an object
If there is more than one force acting on an object, the left hand side of the
Newton’s second law is simply replaced by the vector sum of the forces acting
on the body:
ΣN
(4.3a)
i=1 Fi = ma
=⇒ ΣN
i=1 Fix = max
ΣN
i=1 Fiy = may
ΣN
i=1 Fiz = maz
(4.3b)
It is very important that you realize that this equation applies to the forces
acting on the object. The most difficult thing for most students is to recognize
all, and only all, the forces which act on a particular object. Sometimes students
will leave out certain forces, and other times students will include forces which
are not acting on the object in question. The secret is to focus your attention
on the object in question.
ACCELERATED MOTION
Accelerated motion occurs when the left hand side of Eq. 4.2 is not zero. Then
the object must have an acceleration. Many of the problems you will have to
work out will involved computing the acceleration of an object when the object
experiences certain forces. Typical cases will include the force of gravity (weight),
the normal forces exerted by supporting walls and floors, and the static or kinetic
frictional forces.
Lecture 6 : Introduction to Newton’s Laws of Classical Mechanics
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Newton’s Third Law of Motion
When two objects interact, the force of the first object on the second object
(≡ F12 ) is equal in magnitude and opposite in direction to the force of the
second object on the first object (≡ F21 ) i.e. F12 = −F21
The force from the first object is sometimes called the action force, and the
opposing force from the second object is then called the reaction force. This
leads to another formulation of Newton’s third law
To every action force there is an equal and opposite reaction force
In a sense, forces always come in pairs. However, it is important that you realize
that the action and the reaction forces NEVER act on the same object. If they
did, then the whole theory of mechanics would make no sense since there could
never be a net force acting on an object.
Examples of action–reaction forces are the Sun exerts a gravitational force on
the Earth keeping the Earth in orbit. By Newton’s third law, the Earth must
also be exerting an equal and opposite force on the Sun. More familiar examples
are, you swing a bat at a pitched ball. You hit the ball and it goes flying away
in the opposite direction. You felt a force on your hands from the bat which has
been transferred from the ball. The same thing happens if you kick a ball. You
feel a force on your foot, and the ball felt a force from your foot.
The last example is a cantaloupe resting on a table. The earth pulls down on
the cantaloupe with a force FCE which is the cantaloupe’s weight. The table
exerts a force FCT which cancels out the cantaloupe’s weight, so the cantaloupe
does not move. The forces FCE and FCT are NOT an action-reaction pair of
forces because the act on the same body. However, there is a force FEC which is
the gravity force of the cantaloupe on the Earth, and a force FT C which is the
force of the cantaloupe against the table.
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Applications of Newton’s Laws of Motion
While the three Newton’s Laws of Motion look extremely simple at first sight
(and really are deep down) , there are a wide variety of problems to which they
can be applied. And, at first glance, the solutions to those problems will likely
be difficult for you.
There are two types of problems: Static Equilibrium and Accelerated Motion. Both of these problems are dealt with in Chapter 4.
Static Equilibrium
This is the case of Newton’s second law where there is no net force on an object.
Typically the object is at rest in the Earth’s frame of reference.
ΣFi = 0
=⇒ ΣFix = 0
ΣFiy = 0
=⇒ the sum of the Forces acting Right = the sum of the Forces acting Left
=⇒ the sum of the Forces acting Up = the sum of the forces acting Down
Steps in Solving Statics Problems
The “Free Body” Approach
Generally you will be confronted with one or more objects on which are exerted
certain forces. Usually there will be the weight forces, and the objects will exert
forces on one another. You will be asked to solve for one or more unknown forces.
You should then follow a procedure call the Free Body Approach, explained
on the next page
Lecture 6 : Introduction to Newton’s Laws of Classical Mechanics
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The Free Body Approach to Mechanics Problems
1) Isolate the object(s) of interest.
2) Identify and draw all the forces acting on the object(s). This is called a free
body diagram.
3) If there is more than one object, make sure you understand which forces
are acting on which objects.
4) For each object, resolve the forces into their rectangular components.
5) Apply the Newton’s Second Law equations
ΣFix = 0
and
ΣFiy = 0
to the separate object(s).
6) Solve for the unknown force(s).
Lecture 6 : Introduction to Newton’s Laws of Classical Mechanics
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Worked Example in STATIC EQUILIBRIUM
A traffic light weighing 100 N hangs from a cable which in turn is tied to two
other cables fastened to an overhead support as shown in the figure on page 122.
The upper two cables make angles of 37o and 53o with the horizontal. Find the
tensions in the three cables.
Solution First construct the diagram of the physical set–up, and then isolate
the objects of interest. First there is the traffic light, and second there is the
knot where the three cables are connected. (These are knotty problems; there
will often be a knot which is at rest while subject to three or more tensions.)
From the diagram (see Fig. 4.10 on page 122) we immediately deduce the magnitude of T3 = 100 N. Then we can proceed to the free body diagram of the
knot. We know that the vector sum T1 + T2 + T3 = 0, so we can write down
the usual table for the analytic addition of vectors:
Force
T1
T2
T3
Magnitude Angle X Comp.
(N)
(deg.)
(N)
T1
143
T1 cos 143o
T2
53
T2 cos 53o
100
270
0.00
Y Comp.
(N)
T1 sin 143o
T2 sin 53o
−100.0
ΣFx = 0 =⇒ T1 cos 143o + T2 cos 53o = 0
ΣFy = 0 =⇒ T1 sin 143o + T2 sin 53o − 100 = 0
We have two equations in two unknowns (T1 and T2 ). From the first:
− cos 143o T2 = T1
= 1.33T1
cos 53o
Substitute this in the second equation to obtain
T1 sin 143o + 1.33T1 sin 53o − 100 = 0 =⇒ T1 = 60.0 N, and T2 = 79.8 N