Download lecture notes on Newton`s laws

PHYS1110H, 2011 Fall.
Shijie Zhong
1. Newton’s Laws
Newton’s 1st law: A body in uniform motion remains in uniform motion,
and a body at rest remains at rest, unless acted on by a nonzero net force.
A body with zero net force is also called an isolated body. Therefore, the 1st
law can also be stated as “An isolated body in uniform motion remains in
uniform motion and at rest remains at rest.”
Motion is always measured with respect to a reference frame. A reference
frame in which the Newton’s 1st law is valid is called as inertial reference
frame. We may say that Newton’s 1st law defines inertial reference frame.
Newton’s 2nd law: Net force on a body is equal to the product of its mass
multiplying its acceleration, or


F = ma .
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(1.1)
Remarks:
1) Newton’s 2nd law is valid only in an inertial reference frame (i.e., where
the acceleration is measured).
2) Net force means that it could be a vector sum of multiple forces acting on
the body.
3) Newton’s 2nd law is consistent with Newton’s 1st law. From the 2nd law,
for net force of zero (i.e., the isolated body), the acceleration is zero, or
constant motion. However, Newton’s 2nd law cannot substitute the first law,
because the 2nd law is only valid for inertial systems which are defined by
the 1st law.
4) Another definition of Newton’s 2nd law is: The rate at which a body’s
momentum changes is equal to the net force acting on the body.

dp 
= F,
dt
(1.2)
where the moment of a body is defined as
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

p = mv .
Note
2
(1.3)
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More on the momentum is given in later sections.
Easy to prove that (1.1) and (1.2) are the same, for mass m is constant.




dp d(mv )
dv  dm
dv
 
=
=m +v
=m
= ma = F .
dt
dt
dt
dt
dt
(1.4)
Mass
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What’s mass? Newton’s 2nd law provides an operational definition of mass.
Suppose that one applies the same amount of force to pull two bodies of
different mass: m1 and m2, where m1 could be one unit of mass. We can then
measure accelerations of these two bodies (using ruler and watch) under the
same force, a1 and a2. Then we can determine m2, according to the Newton’s
2nd law,
m2 =
a1
m1 .
a2
(1.5)
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Larger mass has smaller acceleration, for a given force.
Of course, in practice, we cannot use this operational definition to find mass
for all the objects. For example, we cannot determine the mass for a
mountain using this method. However, we can use other experiments and
theories derived from the physics to determine the mass.
Mass’ unit is kilogram (kg).
Force
In the last paragraph, we said that we applied the same amount of force to
the two bodies of different mass. How do we know we apply the same
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amount of force? We can use a spring to pull the body and use the stretch of
the spring as a measure of force magnitude (i.e., the same amount of stretch
means the same amount of force).
Newton’s 2nd law can also be used to provide an operational definition of
force, now that we define mass. One unit force (i.e., 1 Newton) is the force
that causes one unit acceleration (i.e., 1 m/s2) to a body of one unit mass
(i.e., 1 kg).
Gravitational force: for abody of mass m in a gravitational field (e.g., on


the surface of Earth), W = mg , where g is gravitational acceleration (for
Earth or Mars). The gravitational force is also called weight in physics,
although in daily life, we often call mass m as weight.
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Other forces include
electromagnetic
forces, friction force, and so on.
Newton’s
third law: Forces always appear as pairs. If body b exerts force

Fb−a on body a, then there must be a force Fa−b on body b due to body a,


such that Fa−b = −Fb−a (i.e., these two forces are equal in magnitude but
opposite in direction).
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There is no lone force!
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You stand on the surface of the Earth. Earth exerts a gravitational force on
you (i.e., pull you down). The Earth’s surface exerts a normal force to hold
you up.
Inertial systems.
Example. Suppose that you, an astronaut driving a spaceship A, and your
colleague driving another spaceship B, are chasing an enemy spaceship C.
Suppose that the three ships move in the same direction. You want to figure
out the force that the enemy ship C applies to itself. You and your colleague
have devices on your ships to measure the distances of spaceship C to your
ships at different times.
So, you turn off engine and measure the distance between your ship and ship
C at different time as XA(t). And you asked your colleague to do the same,
and your colleague measured the distance between ships B and C as XB(t).
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From the distance XA(t), you determine velocity VA(t)=dXA(t)/dt, and
acceleration aA(t)=dVA(t)/dt for ship C. You found that VA(t) increases
linearly with time and aA(t) is a constant at 1000 m/s2. Therefore, you
concluded that the force that ship C’s engine exerts is
FA=MaA(t)=1000M Newton,
where M is the mass of the ship C (you may not know, but it may not
matter).
Your colleague, after following the same procedures, found that aB(t) is a
constant at 950 m/s2. Therefore, your colleague concluded that the force for
ship C is 950M Newton, different from yours.
It turned out that your colleague did not completely shut down the engine
and ship B actually accelerates with respect to you or an inertial system at an
acceleration of 50 m/s2. That is, ship B is not an inertial system, and
Newton’s second law does not really work for ship B (Your colleague can
use a pencil to determine whether s/he is in an inertial system. How?).
One may introduce a fictitious force to a non-inertial system. We may
discuss this in the future.
Limitations of Newton’s laws.
1) Newtonian mechanics fails for a particle motion that is comparable to the
speed of light and that is at atomic length scale. Under those conditions,
relativity and quantum mechanics are needed.
2) Newton’s laws in the forms given apply to particles and are inconvenient
to describe a continuous system such as a fluid. However, the same
principles can be extended to develop Newtonian mechanics for continuum.
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2. Applications of Newton’s Laws.
There are some simple rules that may help us use Newton’s laws to solve
problems.
1) Divide the system into smaller systems, each of which can be treated
as a point mass.
2) Draw a force or free-body diagram for each mass.
3) Introduce a coordinate system (i.e., positive x and y axis).
4) Apply Newton’s third law.
5) Apply constraints to different bodies.
6) Set up Newton’s 2nd law for each mass, and solve the equations.
We will use some example problems to demonstrate these procedures.
Example 1: The astronauts’ tug-of-war.
Two astronauts, A and B with mass Ma and Mb, initially at rest in free
space, pull on either end of a rope that has mass Mp (but negligibly small).
Astronaut A is stronger than Astronaut B. Find their motion, if each pulls the
rope as hard as they can.
The diagram shows that the system is divided into three smaller systems,
with three point masses: Ma, Mb, and Mp, and also how the forces acting on
each point mass.
For the coordinate system, 1-D is sufficient, and only x axis is introduced in
the diagram. In this coordinate system, some forces are positive (along x
axis), but others are negative (opposite to the x axis).


According to the Newton’s third law, we may state that Fa = −Fa ' , and


Fb = −Fb '.
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
For the rope, assuming that its acceleration is a p in positive x direction, the
nd
Newton’s

 2 law:

Fb + Fa = M p a p ,
(2.1)
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in vector form. Considering the directions of these vectors, the scalar form:
Fb − Fa = M p a p .
(2.2)
Since
the

 mass of the rope, Mp, is negligibly small, this leads to Fb=Fa, and
Fa '= − Fb ' . We do not need to consider the rope, and we can reduce the
number of point masses from 3 to 2. (Note that in many problems, the mass
of rope (or pulley) can be ignored, and we do not need to consider it in
analyses. However, if the mass of the rope or pulley is given and you are not
told that they can be ignored, then you will need to include them in the
analysis).
Now
Newton’s
2nd law for masses Ma and Mb, in vector form:
 consider



Fa '= M a a a ,
Fb '= M b ab .
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Or in scalar form: Fa '= M a a a , −Fa '= M b ab , where we consider the


directions of accelerations and Fa '= − Fb ' from the Newton’s third law.
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a a = Fa '/ M a , and ab = −Fa '/ M b . The negative sign for ab means that the
acceleration
€ of mass B has
€ an opposite direction from mass A.
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Note
astronaut A is stronger than astronaut B, we still have
 that
 although
from the Newton’s third law, i.e., the force exerting on B by A is
Fa '= − Fb ' €
the same as the force on A by B.
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Example 2: Freight train.
Three freight cars, each with mass M, are pulled by a force F. Ignore the
friction. Determine the force on each car and their acceleration.
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The system is divided into 3 smaller systems with three point masses, and
force diagram is given in the figure for each point mass. Also, positive x and
y axes are defined as in the figure.
Since the cars remain on the ground all the time, there cannot be any motion
in y direction at any time, and consequently ay=0. Therefore, normal force
on a car cancels/balances the weight of the car. There is no need to consider
the vertical direction.




According to the Newton’s third law, we have F1 = −F1 ' , and F2 = −F2 '.




The cars move together, therefore a1 = a 2 = a 3 = a .
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We
 cannow consider Newton’s second law for each car.
(2.3)
F1 = Ma ,
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


(2.4)
F2 + F1′ = Ma ,
 

(2.5)
F + F2′ = Ma .
In their scalar form:
(2.6)
F1 = Ma ,
(2.7)
F2 − F1 = Ma ,
(2.8)
F − F2 = Ma .
We can solve these three equations (2.6), (2.7), and (2.8), for three
unknowns F1, F2 and a.
a=F/(3M), F1=F/3, and F2=2F/3.
Constraints: For the last problem, the relation between point mass’
accelerations is straightforward – they are the same because they are
constrained to move together. However, this is not always true. Consider the
following problem.
Example 3. Two masses are connected by a string that passes over a pulley.
The two masses are different and are therefore not balanced. The pulley
itself is pulled upward with some non-zero acceleration. How are the
accelerations of masses 1 and 2, and pulley related to each other?
Set up a coordinate system with y axis pointing upward. The length of the
string is fixed at L. Based on the diagram, we have:
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L = πR + (y p − y1 ) + (y p − y2 ) ,
(2.9)
where R is the radius of the pulley.
Differentiate equation (2.9) with respect to
time t twice (i.e., second order derivative),
and lead to
d 2 yp
d 2 y1 d 2 y2
0=2 2 − 2 − 2 ,
dt
dt
dt
2a p = a1 + a 2 .
(2.11)
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(2.10)
When the pulley is fixed or moved at a
constant velocity (i.e., ap=0), the accelerations
of the two masses are equal in magnitude and
opposite in directions.
Example 4. Two pulleys are connected by
a string as in the diagram. The lower
pulley with a hanging block 1 can move
vertically, while the other pulley is fixed to
the ceiling and is connected to another
block 2. How are the accelerations for the
two blocks related?
Use the coordinate system. Block 1 has the
same y acceleration as the lower pulley.
Similar to the last problem, the length of
the string is constant.
L = πR + (y2 − h) + (y1 − h) + πR + y1, or
L = 2 πR + y2 + 2y1 − 2h ,
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(2.12)
where R is the radius of the pulleys and h is the distance of pulley 2 to the
ceiling, both constant. Differentiate equation (2.12) with respect to time t
twice (i.e., second order derivative), and lead to
(2.13)
a 2 = −2a1.
The acceleration of the second block is twice of that of the first block, but
they have opposite directions.
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