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Note 05
Applying the Laws of Motion
Sections Covered in the Text: Chapter 5
In this note we apply Newton’s Laws to solve some
specific problems. We describe an object in uniform
circular motion and consider the physics of an object
moving through a viscous medium. We begin with
the concept of friction.
The existence of the force of friction eluded the
natural philosophers such as Aristotle and others well
into modern times. Indeed, it could be said that the
understanding of the force of friction paved the way
for the laws of motion.
Forces of Friction
Whenever two objects in contact are moved one
against the other, some resistance to the movement is
observed. This resistance is attributed to a force of
friction over the area of the surfaces in contact. Two
forces have been identified, the force of static friction
and the force of kinetic friction. In both cases, the force
is directed opposite to the direction of motion or to
the direction of impending motion (in the case of
objects not actually moving). Both types of friction are
commonly quantified in terms of a coefficient. The
motion of an object through a viscous medium such as
air or a fluid is also resisted by a force of friction.
Static Friction
We all learn in childhood that in order to make an
object like a block move across a floor we need to
apply a force to the block. Careful study reveals that if
the force applied is not great enough, then the block
does not move. If the force is increased steadily, then
a condition is eventually reached at which the block
“breaks free” from the supporting surface and moves.
This effect is interpreted to be the result of a force of
static friction between the block and the supporting
surface.
The origin of the force is not difficult to identify. A
critical observation is that the same object can be
moved more easily over a smooth surface than a
rough one. This means that the underside of the block
and the supporting surface are not infinitely smooth.
Microscopic jagged points in both surfaces must catch
one another thereby resisting movement. Before the
onset of movement, the force of static friction equals
the applied force (as required by Newton’s third law)
and is directed opposite to the applied force (or
opposite to the direction of impending motion). This
force is found to be proportional to the normal force
the supporting surface exerts on the object (which is
the same as the weight of the object if the surface is
horizontal).
Kinetic Friction
We soon learn as well that once the block is moving,
less force is required to keep the block moving at
constant speed than is required to get the block
moving in the first place. This implies the existence of
a force of kinetic (moving) friction. If a force is applied
to keep the block moving at constant speed, then the
force of friction equals the applied force (as required
by the first law). This force, too, is found to be proportional to the normal force. All things being equal the
force of kinetic friction is less than the force of static
friction.
A wealth of experimental evidence reveals that the
forces of static and kinetic friction can be written
and
f s ≤ µs n
…[5-1a]
f k = µk n ,
…[5-1b]
where µ s and µ k are the coefficients of static and kinetic
friction, respectively, and n is the normal force. Eq[51a] gives an upper bound; the maximum force of
static friction is
f s.max = µsn ,
which exists just before the block moves. Figure 5-1
summarizes these and other observations about static
and kinetic friction.
Figure 5-1. How the forces of static and kinetic friction vary
with the force applied.
05-1
Note 05
Unhappily, the forces of static and kinetic friction are
“messy” forces in the sense that the corresponding
coefficients are not true physical constants; they
depend on the types of materials in contact, on the
smoothness of the surfaces and on environmental
factors such as humidity and the existence of
lubrication. Let us consider an example.
(a) Before the block moves, the three forces put the
block in a state of translational equilibrium. Applying
Newton’s second law in component form to the block
gives the equations
and
Example Problem 5-1
Determining µ s and µk Experimentally
For any two dry materials in contact, the coefficients
of static and kinetic friction can be calculated with the
help of the simple apparatus sketched in Figure 5-2. A
block made of the one material is placed on the surface of an inclined plane made of the other material.
The angle θ of the plane is increased slowly until the
block starts to move. (a) How is the coefficient of static
friction related to the critical angle θc at which the
block starts to move? (b) Describe how the coefficient
of kinetic friction might be subsequently found.
∑F
x
= mg sin θ – f s = 0 ,
…[5-2a]
∑F
= n – mg cosθ = 0 .
…[5-2b]
y
These equations apply for any angle θ of inclination.
But at the critical angle θc at which the block is on the
verge of slipping, the force of static friction has its
maximum magnitude, µsn. Thus eq[5-2a] at that point
gives
mg sin θ c = µsn .
…[5-3a]
Now rearranging eq[5-2b] we have
mg cos θc = n .
…[5-3b]
Dividing eq[5-3a] by eq[5-3b] gives
µs = tanθ c .
…[5-4a]
Thus the coefficient of static friction equals the tangent of the critical angle. Thus a measurement of θc
enables us to calculate µ s .
Figure 5-2. Free-body diagram of a block on an inclined
plane.
(b) The coefficient of kinetic friction can be found as
follows. Ordinarily, if the angle of inclination were
held constant after the block starts to move, then the
block would actually accelerate down the incline. But
if immediately after the block starts to move, the angle
were reduced to, say, θ’ so that the block moves down
the incline at constant speed then the first law
requires that the sum of the x-components of force is
zero, i.e.,
Fx = 0 .
∑
Thus from eq[5-2a]
Solution:
For convenience we choose a coordinate system in
which the x-axis points downwards parallel to the
incline and the y-axis points upwards perpendicular
to the incline. Three forces act on the block whether it
is moving or not: the component of the force of
gravity acting parallel to the incline downwards, the
normal force acting perpendicular to the incline and
the force of friction (static or kinetic depending on
whether the block is stationary or moving) acting
upwards parallel to the incline.
05-2
f k = mg sinθ ,
and
f k = µk n = µk mgcosθ '= mgsin θ ' .
Thus
µk = tanθ ' .
…[5-4b]
€ Thus the coefficient of kinetic friction equals the tangent of the angle θ’. Since θ’ is observed to be less than
θc it follows
€ that µk < µs .
Note 05
Example Problem 5-2
Calculating Coefficients of Friction
the corresponding position vectors. In the elapsed
time ∆t = tf – ti the position vector sweeps out an angle
∆θ.
A wooden block of mass 1.00 kg is placed on an inclined plane made of steel. The angle of inclination of
the plane is slowly increased, until at 32.0˚ the block
starts to move. Subsequently, when the angle is
reduced to 26.0˚, the block moves at a constant speed
down the incline. (a) Calculate the coefficients of static
and kinetic friction. (b) Calculate the maximum force
of static friction and the force of kinetic friction.
vi
f
Δr
i
ri
Δθ
vf
rf
vi
Δθ
Solution:
(a) From eqs[5-4] we have:
Δv
vf
µs = tan 32.0o = 0.625
and
µk = tan 26.0 o = 0.488 .
(b) The maximum force of static friction is
f s = mg sin θc
= (1.00kg)(9.80m.s –2 )sin32.0 o = 5.19 N.
and the force of kinetic friction is
f k = mg sinθ ' ,
= (1.00kg)(9.80m.s –2 )sin26.0 o = 4.30 N.
Figure 5-3. A particle moving clockwise with uniform circular motion.
We wish (1) to show that the particle is undergoing an
acceleration, (2) to calculate the magnitude of that
acceleration and (3) to show that the acceleration
vector is directed towards the center of the circle.
The first thing to appreciate about this kind of
motion is that although the particle is moving with
constant speed and the magnitude of the velocity
vector does not change, the direction of the velocity
vector does change and changes continuously. A
diagram illustrating the relationship between the
velocity vectors is shown to the right in the figure.
From this diagram it can be seen that
Thus the force required to start the block moving is
greater than the force required to keep the block
moving at constant speed once it is moving.
Newton’s Second Law Applied to a Particle
in Uniform Circular Motion
We have all seen objects being spun in a circle. A ball
on the end of a string is one example. With the tools
we have developed we can describe this motion.
Consider a ball modelled as a particle moving clockwise in a circle at constant speed v (Figure 5-3). Shown
are two positions i and f on the particle’s path. The
vectors locating i and f relative to the center of the
circle are also shown, along with the instantaneous
velocity vectors of the object at the two positions. The
instantaneous velocity vectors point along tangents to
the particle’s path and are therefore perpendicular to
Δv = v f – v i .
Since the direction of the particle’s velocity vector is
changing, the particle is, by definition, undergoing an
acceleration. According to its definition laid down in
Note 03, the average acceleration is the change in
velocity divided by the elapsed time:
a=
v f – v i Δv
=
.
t f – ti
Δt
…[5-5]
You should be able to see that the triangles formed by
the position and velocity vectors in the figure are
isosceles triangles and are similar having the same
inner angle ∆θ. Thus the following relationship
holds:1
1
These ratios define the inner angle in radians. There will be
05-3
Note 05
| Δv | | Δr |
=
.
v
r
Subsituting for |∆ v | from eq[5-5] we can obtain the
magnitude of the average acceleration:
| a | Δt | Δr |
=
v
r
v | Δr |
.
r Δt
| a |=
yields
We have already seen that since the velocity of the
ball is continuously changing, the ball is undergoing a
centripetal acceleration. Its magnitude is given by
eq[5-6] and its direction is towards the center of the
circle. This centripetal acceleration is caused by the
force exerted on the ball; the magnitude of this force is
the tension T in the string. We can therefore write in
accordance with the second law:
The magnitude a of the instantaneous acceleration
vector is obtained by taking the limit of the magnitude
of the average acceleration as ∆t → 0:
a = lim | a |=
Δ t→ 0
v
| Δr | v 2
lim
= .
r Δt → 0 Δt
r
Thus we have the magnitude of the acceleration. To
get the direction of the acceleration vector notice that
as ∆t → 0, so also does ∆θ → 0 and Δv becomes
perpendicular to v i or antiparallel to r i . Since it is
antiparallel to r i it is automatically pointing toward
the center of the circle. This acceleration is called a
centripetal acceleration (centripetal, meaning centerseeking) and is commonly denoted ac :
ac =
v2
.
r
…[5-6]
The time required for the particle to execute one complete revolution is called the period, and is commonly
denoted T. The distance travelled by the particle in
one revolution is the circumference of the circle, 2πr. T
is the distance travelled divided by the speed:
T=
2πr
.
v
We can also describe uniform motion in a circle in
terms of force. Figure 5-4 shows a more real-world
representation of Figure 5-3. A ball on the end of a
string is being moved at uniform speed v in a circle of
radius r. To avoid the question of how the ball stays
“up” we can assume that the ball is moving on a
frictionless horizontal table (not shown) and that we
are looking down from directly above the table.
more on the radian in Note 10.
05-4
Figure 5-4. A body in uniform circular motion. The hand
exerts a force on the ball whose magnitude is the tension T
in the string.
v2
∑ F = mac = m r .
The magnitude of the centripetal force required to
move the ball in a circle is the tension T in the string.
Example Problem 5-3
An Object in Uniform Circular Motion
A ball of mass 0.500 kg is moved in a horizontal circle
at a constant speed of 2.00 m.s–1 on a frictionless table
(as depicted in Figure 5-4). Calculate the centripetal
acceleration and the tension in the string.
Solution:
The magnitude of the centripetal acceleration is given
by eq[5-6]:
ac =
Thus
v 2 (2.00m.s –1) 2
=
= 8.00 m.s–2 .
r
(0.500m)
T = mac = (0.500kg)(8.00m.s –2 ) = 4.00 N.
Note 05
An object in the real world has to move through a
medium like air or water. We have already described
a number of examples of objects moving through air
and we have chosen to simply ignore the possible
effect of the air on the object’s motion. In the next section we examine this special kind of frictional force.
directed. Thus it has the form:
Motion in the Presence of VelocityDependent Resistive Forces
If the medium through which an object moves has
sufficient density and the object moves with sufficient
velocity, then the object will experience an appreciable
frictional force that depends in a non-obvious way on
its velocity relative to the medium. If the velocity is
low then the frictional force is observed to be directly
proportional to the velocity. If the velocity is high
then the force is observed to be proportional to the
square of the velocity. We consider here only the
former type.
Resistive Force Proportional to Object Velocity
We consider this physical situation illustrated in
Figure 5-5 because it is at the heart of the experiment
“Simple Measurements” that you will be doing in the
lab soon. The experiment is a simple one. A marble is
released at the top of a cylinder of shampoo and
allowed to fall slowly to the bottom. Two marks 0.500
m apart are etched on the cylinder to enable you to
measure the marble’s average velocity (which is also
the average speed in this case). (Average speed equals
distance travelled divided by elapsed time.) Then
from various measurements you can calculate the
viscosity of the fluid. 2
The expression you will use to calculate the
viscosity is derived on the assumption that the marble
moves downwards at a constant velocity called the
terminal velocity. Therefore a fundamental question
asked is “does the marble move downwards at a
velocity that is truly constant?” This question can be
answered at least in principle by calculating how long
the marble takes to reach a constant, maximum
velocity, and if this time elapses before the timing of
the marble’s fall is begun when the marble passes the
top mark on the cylinder.
We make the assumption that the resistive force R
exerted on the marble by the fluid is directly
proportional to the object’s velocity and is oppositely
2
The viscosity of a fluid is a measure of what could be called the
fluid’s “stickiness”. This topic belongs to the study of fluids (Notes
12 & 13). In this first experiment in the lab the major objective is to
master various kinds of measurements and not just to obtain the
viscosity.
Figure 5-5. (a) The free-body diagram of a marble falling
through a fluid, and (b) the velocity of the marble as a function of elapsed time. What is meant by the time constant is
shown.
R = –bv .
where b is a constant, dependent on factors such as
the viscosity of the fluid and the radius of the marble.3
The object is moving downwards so we can apply the
second law:
Fy = may ,
∑
or
Fg + R = ma ,
which gives
mg – bv = m
dv
.
dt
Rearranging we have
dv
b
= g– v.
dt
m
…[5-7]
This is a first-order differential equation in v. You should
3
For a falling marble the constant b has the form 6πη(d/2) where
d is the diameter of the marble and η is the viscosity.
05-5
Note 05
be able to show that a solution of the equation is: 4
–bt

mg 
v(t) =
1 – e m  ,
b 

…[5-8]
−t


= vT  1– e τ 


where we write vT = mg/b and τ = m/b. The factor vT
has units of velocity and is called the terminal velocity.
This is the velocity the object approaches after a
sufficiently long time, i.e., as t → ∞. The quantity τ has
units of time and is called the time constant (it goes by
other names in other areas of physics where the
variable has a similar form as eq[5-8]). 5 It is the time
required for the velocity of the object to approach to
within 1/e of the terminal velocity. In the fluid used
in the experiment “Simple Measurements” this time is
seen to be of the order of 10 milliseconds. The
conclusion to be drawn is that the marble does indeed
reach terminal velocity before timing actually begins.
The Fundamental Forces of Nature
As far as is known to physicists today, there are four
fundamental forces in nature. They are the gravitational force, the electromagnetic force, the nuclear force
and the weak force. We can discuss these only briefly.
The Gravitational Force
Isaac Newton was the first to describe the gravitational force in mathematical terms in 1686. According
to his description any two masses attract each other
with forces that are directly proportional to the
product of their masses and inversely proportional to
the square of the distance between their centers. In
other words, for any two masses m1 and m 2 a distance
r apart the magnitude of the force is
Fg = G
m1m2
.
r2
G is a constant that today is called the universal
gravitational constant. From numerous experiments its
accepted value is:6
G = 6.67x10 –11 N.m2.kg –2,
to 3 significant digits. We shall continue the study of
this force in later notes.
The Electromagnetic Force
We have seen that mass is a fundamental property of
matter. Charge is another. Charges attract or repel each
other with a force called the electromagnetic force . It has
a form similar to the gravitational force (with charge q
instead of mass m and the electric constant k instead of
the gravitational constant G). We shall discuss this
force in detail in Note 20.
The Nuclear and Weak Forces
The nuclear force is the force responsible for keeping
protons and neutrons bound together in the atomic
nucleus. If this force did not exist then mutual repulsion between protons in the nucleus would cause the
nucleus to fly apart. This force is a very short-range
force and drops to virtually zero much beyond the
nucleus itself. The weak force figures in radioactive
decay. Its description belongs in a higher-level course
in physics and is beyond the scope of these notes.
The Gravitational Field
The fundamental forces of nature are all non-contact
forces. When Newton proposed that the force of
gravity was, in essence, a non-contact force the idea
was resisted by the natural philosophers of the day, in
spite of the fact that the predictions based on it were
so accurate. Scientists trained on ropes and pulleys
found it hard to believe that two masses, like the
Earth and the Moon, could exert forces on one another
even though separated by many kilometers of empty
space.
The issue of contact was essentially side-stepped by
the brilliant English experimentalist, Michael Faraday,
in his proposal of the concept of the gravitational field.
Faraday conjectured that any object by virtue of its
mass sets up an entity in its surroundings called a
gravitational field. The field is continuous in the sense
of having a strength (a magnitude) and a direction at
every point. The field is therefore a vector field. A
second mass in the region of this field is, of necessity,
in contact with the field at that point. The field in
contact with the mass then gives rise to the gravitational force on the object.7 A field is associated with
4
To show that eq[5-8] is a solution of eq[5-7] it is sufficient to
show that when it and its first derivative are substituted into eq[5-7]
the LHS=RHS.
5
Time constants appear in other areas of physics, in particular
in the experiment “The Capacitor” in the PHYA21 lab.
6
Do not confuse the universal gravitational constant G with the
05-6
acceleration due to gravity g.
7
Throughout their lifetimes, neither Faraday nor Newton had
any idea as to how this might happen. Suffice us to say that the
existence of the field, either gravitational, electric, nuclear or weak,
has now been proven beyond doubt.
Note 05
each force of nature, and in each case the field is
regarded as the source of the force (and the source of
the energy transferred).
The gravitational field at the position of a test mass
mt is defined as the vector
g≡
Fg
,
mt
the field vector always points toward the center of the
Earth.
…[5-9]
where F g is the gravitational force on mt. In words,
the gravitational field at some point in space is the
gravitational force per unit mass at that point. In the
case of gravity the gravitational field vector is what
we know more familiarly as the acceleration due to
gravity vector g .
The definition, eq[5-9], provides a way of imagining
how a gravitational field might be mapped, or represented graphically. We imagine that a test mass mt can
be moved to various selected (arbitrary) positions in a
region of space and the force per unit mass measured
(by means of some instrument) at the chosen
positions. A vector can then be drawn at every chosen
position, with a magnitude equal to the magnitude of
the gravitational force per unit mass, and with the
direction of the gravitational force vector. The collection of vectors is the field map.
For example, this effort carried out in the neighborhood of a spherically symmetric source mass might
resemble Figure 5-6a in 2D space. The gravitational
field is seen to have a radial geometry. The field
always points toward the source mass and the
magnitude of the field decreases as r increases (with
an inverse square dependency). Of course, this picture
is only a representation.
If this procedure were carried out near the surface
of the Earth the result might resemble Figure 5-6b in
2D space. This field is seen to be uniform. The
magnitude of the field strength at every point is g and
Figure 5-6. 2D representations of the gravitational field near
a spherically symmetric source mass (a) and near the
surface of the Earth (b)
Having studied a number of applications of Newton’s
Laws we are ready to consider the concept of energy.
This we will do in Notes 08 and 09.
To Be Mastered
•
•
•
•
Definitions: force of static friction, force of kinetic friction, coefficient of static friction, coefficient of kinetic friction
Definitions: centripetal acceleration, uniform circular motion
Definitions: viscous force, time constant
Definitions: field, gravitational field vector
05-7
Note 05
Typical Quiz/Test/Exam Questions
1.
A block of mass 1.0 kg is in contact with a horizontal surface (see the figure). The coefficient of static friction
between the surfaces is 0.7 and the coefficient of kinetic friction is 0.6. Three cases are shown: (1) when the net
external force is zero, (2) when a force of 1.0 N is applied at an angle of 60˚ to the horizontal, and (3) when a
force of 2.0 N is applied at an angle of 60˚ to the horizontal.
1.0 N
block
(1)
2.0 N
60˚
block
block
(2)
(3)
60˚
Draw a free body diagram for the block in each case.
2.
A ball of mass M is tied to the end of a string of length R and spun in a vertical circle at a constant speed v (see
the figure).
A
T
D
B
C
(a) Draw free body diagrams for the ball at positions A, B, C and D.
(b) If M = 1.0 kg, R = 1.0 m and v = 4.00 m.s–1, calculate the tension T in the string, in N, at the four positions.
For convenience, take g = 10.0 m.s–2.
05-8