Download Chapter 4 Newton`s Laws of Motion

Chapter 4
Newton’s Laws of Motion
4.1
Forces and Interactions
Fundamental forces. There are four types of fundamental forces: electromagnetic, weak, strong and gravitational. The first two had been successfully
unified into electroweak theory and there are ongoing attempts to unify its
with strong force. The task proved to be very difficult, but there are good
reasons to believe that it is at least possible (for example, the so-called GUT
theories). The situation is much worse with regards to gravitational force
which is manifests itself not through exchange of participles like other forces,
but through curvature of space-time. The string theory (perhaps the most
promising candidate of unifying all forces) does describe a way of how to
think about gravity (perturbatively), but it is too naive to expect that we
will know the final answer any time soon. With regards to our everyday experience the (microscopic) forces and interactions manifest themselves trough
(macroscopic) forces and interactions. For example, tension force from a
rope, normal or friction forces from a surface or weight due to gravitational
attraction. To these types of macroscopic forces we will refer as forces which
are 3D vectors with direction and magnitude (representing its strength).
Superposition of forces. There might be a number of different forces
acting on a given object, and the total force (or net force) is given by
!
⃗ net =
⃗i = F
⃗1 +F
⃗2 +F
⃗ 3 + ...
F
F
(4.1)
i
where the usual vector addition is used, i.e.
"
#
!
!
!
(Fnetx , Fnety , Fnety ) =
Fix ,
Fiy ,
Fiz .
i
41
i
i
(4.2)
42
CHAPTER 4. NEWTON’S LAWS OF MOTION
Strength of the net force is then
Fnet =
$
2
2
2
Fnetx
+ Fnety
+ Fnetz
%"
#2 "
#2 "
#2
&
& !
!
!
Fix +
Fix +
Fix .
='
i
i
i
(4.3)
Example 4.1. Three professional wrestlers are fighting over a champion’s belt. The forces are on horizontal plane and have magnitudes and
directions:
F1 = 250 N and θ1 = 127◦
F2 = 50 N and θ2 = 0◦
F3 = 120 N and θ3 = 270◦ .
(4.4)
Find the components of the net force on the belt and it magnitude and direction.
Step 1: Coordinate system. The 2D coordinate system was already chosen
for us in the problem.
Step 2: What is given? We know that the three wrestlers apply the
following forces to the belt
F⃗1 = (cos(127◦) · 250 N, sin(127◦) · 250 N) = (−150 N, 200 N)
F⃗2 = (cos(0◦ ) · 50 N, sin(0◦ ) · 50 N) = (50 N, 0 N)
F⃗3 = (cos(270◦) · 120 N, sin(270◦) · 120 N) = (0 N, −120 N)
(4.5)
Step 3: What do we have to find? From Eq. (4.1) the net force is
⃗ net = (−100 N, 80 N)
F
(4.6)
CHAPTER 4. NEWTON’S LAWS OF MOTION
and its magnitude and directions are
$
(−100 N)2 + (80 N)2 = 128 N
Fnet =
80 N
θ = arctan
+ 180◦ = −39◦ + 180◦ = 141◦ .
−100 N
43
(4.7)
Note that θ had to be adjusted 180◦ due to the range where arctan function
is defined.
4.2
Newton’s First Law
First Law. A body acted on by no net force, i.e.
!
⃗i = 0
F
(4.8)
i
has a constant velocity (which may be zero) and zero acceleration. (This
tendency for a body to continue its motion is know as inertia and is extremely
important concept in theory of general relativity.)
Example 4.2. In the classic 1950 science-fiction film Rocketship X-M,
a space-ship is moving in the vacuum of the outerspace, far from any star or
planet, when it engine dies. As a result, the spaceship slows down and stops.
What does Newton’s first law say about this scene?
This is in a conflict with Newton’s first law. The spaceship should continue to move if there are no forces acting on it (no force from engine since
it died and no significant gravitational force since all stars/planets are far
away.)
Example 4.3. You are driving a Maserati GranTurismo S on a straight
testing track at a constant speed of 250 km/h. You pass a 1971 Volkswagen Beetle doing a constant speed 75 km/h. On which car is the net force
greater?
Since both cars move with constant velocities the net force on each car is
zero.
Inertial frames. It is important to note that the Newton’s first law is
not obeyed in all reference frames (e.g. inside of an accelerating train). Those
frames of references where it is obeyed is called inertial frame of reference
(e.g. inside a train moving with constant velocity) and for this reason it is
sometimes called the law of inertia. (Note that surface of Earth is not exactly
an inertial reference frame (why?) but it is pretty close to being inertial.)
For inertial reference frames one can easily go from one frame to another
using Eqs. (3.62) and (3.63) which makes such frames particularly useful.
CHAPTER 4. NEWTON’S LAWS OF MOTION
44
Examples. In which of the following situations is there zero net force
on the body?
An airplane flying due north as a steady 120 m/s and at a constant altitude?
A car driving straight up a hill with a 3o slope at a s constant 90 km/h
A hawk circling at a a constant 20 km/h at a constant height of 15 m above
an open field?
A box with slick, firctionless surface in the back truck as the truck accelerates
on a a level road at 5 m/s2 .
4.3
Newton’s Second Law
Second Law. If a net external force acts on a body, the body accelerates.
The direction of acceleration is the same as the direction of the net force.
The mass of the body times the acceleration vector of the body equals to the
net force vector, i.e.
!
⃗ i = m⃗a
F
(4.9)
i
or
( ⃗
Fi
⃗a = i .
(4.10)
m
Since Eq. (4.9) is a vector equation in 3D it is equivalent to three equations
!
Fix = max
i
!
Fiy = may
i
!
Fiz = maz .
(4.11)
i
Units. With the help of second law we can now relate the units of force
to units of mass and acceleration. Since each equation must have the same
units on both sides we see that
N = kg · m/s2 .
(4.12)
So if a 1 kg object moves with acceleration 1 m/s2 then there must be a net
fore of 1 N applied to it. In British system of units
1 lb = 4.448 N.
(4.13)
Example 4.4. A worker applies a constant horizontal force with magnitude 20 N to a box with mass m = 40 kg resting on a level floor with negligible
CHAPTER 4. NEWTON’S LAWS OF MOTION
45
friction. What is the acceleration of the box?
By using the second law (described by Eq. (4.10)) in 1D and converting
the units (described by Eq. (4.12)) we get
a=
4.4
20 N
= 0.5 m/s2 .
40 kg
(4.14)
Mass and Weight
Weight. Is a gravitational force acting on an object close to the surface of
the Earth (to be precisely at the see level)
⃗ = m⃗g .
w
(4.15)
It is a vector, but one often refers to the magnitude of the weight force as
weight
w = mg
(4.16)
(or even to mass itself just because one, i.e. m, is simply related to the other,
i.e. w.) One should however be careful when weight is measured above or
below the sea level (e.g. on the airplane), as the weight force can vary.
Example 4.7. A 2.49 × 104 N Rolls-Royce Phantom traveling in the +x
direction makes an emergency stop; the x-component of the net force acting
on it is −1.83 × 104 N. What is its acceleration?
The mass of Rolls-Royce Phantom is
m=
w
2.49 × 104 N
= 2540 kg
=
g
9.8 m/s2
(4.17)
CHAPTER 4. NEWTON’S LAWS OF MOTION
46
and the net force must only be along the direction of motion (call it x-axis)
Fx = −1.83 × 104 N.
(4.18)
(Note that there are two more forces action on the car in vertical direction
(weight and normal force), but they must balance each other or otherwise
the car would be moving in vertical direction. More on this coming in the
next section.) By plugging (4.17) and (4.18) into second law along x-axis
(4.11) we get
Fx
= −7.20 m/s2 .
(4.19)
a=
m
4.5
Newton’s Third Law
Third Law. If a body A exerts a force on body B (an “action”), then body
B exerts a force on body A (a “reaction”). These two forces have the same
magnitude, but are opposite in direction. These two forces act on different
bodies.
⃗ A on B = −F
⃗ B on A .
F
(4.20)
Example 4.8. After your sport car breaks down, you start to push it to
the nearest repair shop. While the car is starting to move, how does the force
exert on the car compare to the force the car exerts on you? How do these
forces compare when you are pushing the car along at a constant speed?
The magnitude of forces in every action-reaction pair is always the same
for any given setup. Thus the force exerted on the car has the same magnitude as the force car exerted on you. This is true when the car is moving
with or without acceleration. On the other hand the force that needs to be
applied to start moving is (usually) large than the force that needs to be
applied to continue motion with constant speed. We will come back to why
this is the case in next chapter in context of frictional forces.
Example 4.9. An apple sits at rest on a table, in equilibrium (i.e. static).
What forces act on the apple? What is the reaction force to each of the forces
acting on the apple? What are the action-reaction pairs?
CHAPTER 4. NEWTON’S LAWS OF MOTION
47
Apple experiences a gravitational attraction to the Earth (or weight)
⃗
⃗ table on apple . These
Fearth on apple as well as normal force from the table, F
forces must balance each other,
⃗ earth on apple + F
⃗ table on apple = 0 or F
⃗ earth on apple = −F
⃗ table on apple . (4.21)
F
but they do not form an action-reaction pair.
Since the Earth attracts apple, the apple must attract the Earth and this
is the first action-reaction pair
⃗ earth on apple = −F
⃗ apple on earth .
F
(4.22)
Since table repels the apple, the apple must also repel the earth and this the
second action-reaction pair
⃗ table on apple = −F
⃗ apple on table .
F
(4.23)
Example 4.10. A stonemason drags a marble block across a floor by
pulling on a (weightless) rope attached to the block. The block is not necessarily in equilibrium. How are the various forces related? What are the
action-reaction pairs?
CHAPTER 4. NEWTON’S LAWS OF MOTION
48
There are three objects of interest: the block, the rope and the mason.
The block is in contact with the rope and so it forms the first action-reaction
pair:
⃗ Block on Rope = −F
⃗ Rope on Block .
F
The rope is in contact with the mason and so it forms the second actionreaction pair:
⃗ Rope on Manson = −F
⃗ Manson on Rope .
F
Note that if the rope is weightless accelerating, then
FManson on Rope > FBlock on Rope .
⃗ Floor on Manson and F
⃗ Floor on Block
How do you think the friction forces F
Examples. You are driving a car on a country road when a mosquito
splatters on the windshield. Which has the greater magnitude: the force that
the car exerts on the mosquito or the force that the mosquito exerted on the
car? Or are the magnitudes the same? If they are the different, how can you
reconcile this fact with Newton’s third law? If they are equal, why is mosquito
splattered and damaged while the car is undamaged?
4.6
Free-body diagram
Free-body diagram. A useful tool in solving problems on Newton’s laws
is to draw free-body diagram, where a chosen body appears by itself without
of its surroundings, and with vectors drawn to show the magnitude and
directions of all the forces that act on the body. the tricky part here is to
include all of the forces that act on a chosen body, but not to include any
other forces that act on other bodies.