Nicholas J. Giordano
www.cengage.com/physics/giordano
Forces and Motion in One Dimension
Applications of Newton’s Laws
• We will learn how Newton’s Laws apply in various
situations
• We will begin with motion in one dimension
• Idealized case – most real motion is three-dimensional
• Mathematics is easier
• Basic ideas and approaches can be applied to motion in
higher dimensions
Introduction
Motion of a Spacecraft
• Assume the spacecraft
travels along a straightline path from one galaxy
to another
• Look at the motion in
terms of position, velocity
and acceleration
• No forces acting on the
spacecraft so no
acceleration
Section 3.1
Insights: Problem Set Up
• Start with a picture
• Include the coordinate axes you have chosen for the
problem
• Determine a positive direction
• Be careful with units
• Be sure they are all in the same system of measurement
• Include the units with your calculations
• Check the units that they correctly match the quantity being
calculated
Section 3.1
Notation Simplification
• For cases involving one-dimensional motion the usual
vector notation can be simplified
• The direction is specified by the sign (+ or - ) of the
quantity
• This notation corresponds to specifying the components of
the quantity along the coordinate axis
Section 3.1
Motion Example – Constant Velocity = 0
• The velocity is zero
• On the graph, the line is at v = 0
• The position is constant, but generally not zero
• Not moving, so no change in position
• On the graph, a horizontal line
Section 3.1
Motion Example –
Constant Velocity ≠ 0
• The velocity is not zero
• On the graph, the line is at v = constant value
• The position is changing steadily
• The same Δ x occurs each second
• On the graph, an upward sloping straight line
• The slope of the position-time curve is equal to the value of the
Section 3.1
velocity
Motion Example – Constant Acceleration
• A constant force is applied
• The acceleration is a constant
• Value depends on the total force
exerted on (and mass of) the
object
• The value is equal to the slope of
the line on the v-t graph
• The velocity is changing
• On the graph, this is an upward
sloping straight line
• The position is changing
• Not the same change each second
• On the graph, this is a curved line
Section 3.1
Equations to Describe Motion with Constant
Acceleration
• v = vo + a t
• vo is the velocity at some initial time t = 0
• It depends on what happened prior to t = 0
• x = xo + vo t + ½ a t2
• xo is the position at some initial time t = 0
• v2 = vo² + 2 a (xf - xo)
• Eliminates t from the equation
• Which equation to use depends on what information you
are given in the problem and what you are asked to find
Section 3.1
Constant Acceleration Equations, Summary
Section 3.1
Note About Units
• Units can still be cancelled and combined as shown in
Chapter 1
• When dealing with units of force, it is simplest to express
Newtons in terms of fundamental units before combining
and/or canceling
• Remember: 1 N = 1 kg m / s2
• Always include units in your calculations and check that
they correctly match the quantity being calculated
Weight and Normal Forces
• There are many cases where gravity plays an important
role
• Weight is associated with gravity
• The normal force is another force often encountered in
problems
Section 3.2
Weight
• Weight is the force of gravity exerted by the Earth on an
object
• Denoted by
• If an object has a mass m, then
• The force of gravity is a consequence of Newton’s Law of
Universal Gravitation
• The value of g is approximately the same for all locations
near the surface of the Earth
• g ≈ 9.8 m/s²
• The weight will be different on another planet
• Since it is due to the gravitational attraction of that planet
Section 3.2
Weight, cont.
• The value of g is independent of the mass of the object
• The weight of an object is proportional to the mass of the
object
• g is commonly referred to as the “acceleration due to
gravity”
• Weight will be measured in Newtons
• It is a force
• Since weight acts vertically, it will be along the y-axis
• Since the weight acts downward, Fgrav = - m g
• It acts toward the center of the Earth
Section 3.2
Normal Force
• A normal force acts perpendicularly to the plane of contact
• Normal forces are common in nature
• They occur whenever the surfaces of two objects come in
contact
Section 3.2
Weight and Normal Force, Example
• The weight is directed
downward
• The normal force is directed
upward
• The person is at rest
• a=0
• ΣF = -m g + N = m a = 0
• N=mg
• In this case, the normal is
equal in magnitude to the
weight and opposite in
direction to the person’s
weight
Section 3.2
Free Body Diagram
• A free body diagram should be used for analysis using
Newton’s Second Law
• It is a simplified diagram showing all the forces acting on
each object involved in the problem
Section 3.2
Problem Solving with Newton’s Laws
• Recognize the objects of interest
• List all the forces acting on each object
• Sketch the problem
• Start with a drawing that shows all the objects of interest
• Include all the forces acting on the objects
• Make a separate sketch showing all the forces acting on each
using a dot for the object
•
This is the free body diagram
• Forces in the free body diagram should be represented by arrows
• The direction of the arrow will show the direction of the force
• Include only the forces acting on that particular object
Section 3.2
Problem Solving with
Newton’s Laws, cont.
• Identify the relationships
• Represent unknown quantities by an appropriate letter
•
Many unknowns have a particular symbol associated with them
• Generally, you will use Newton’s Laws to solve for the
unknowns
• Solve
• Use the information in the free body diagram to write Newton’s
Second Law
• You may need to include some algebraic steps to actually solve
for the unknown
• Check
• Consider what your answer means and check that it makes sense
Section 3.2
Apparent Weight
• The normal force is not
always equal to the weight
• One example is moving
upward in an accelerating
elevator
• Letting upward be positive:
• ΣF = m a = N – mg
• So N = m a + m g
• If the elevator moved
downward, N = m g – m a
• In this case, the normal force
is called the object’s
apparent weight
Section 3.2
Weight and Mass
•
•
•
•
•
Weight and mass are closely related
They are not the same
Mass is an intrinsic property of the object
Weight depends on the location of the object
Example:
• Assume mass on Earth is 10.0 kg
• The weight on the surface of the Earth would be 98.0 N
• Fgrav = m g
• The mass on the Moon would be 10.0 kg
• The weight on the Moon would be ~ 16.3 N
• gMoon ~ 1/6 gEarth
Section 3.2
Forces and Interactions
• Newton recognized that two objects could exert forces on
each other even when they are not in direct contact
• This is known as action-at-a-distance
• Gravitational force is an example
• Applies to many other forces
• Contact on a microscopic scale
• Contact forces are a result of electric forces between atoms
that are in very close proximity
• The atom-atom interaction is described by quantum
mechanics, not Newtonian Mechanics
Section 3.2
More About Mass
• The force of gravity depends on the mass of the object
• This is called gravitational mass
• The mass of an object also determines how the object will
move in response to forces
• This is called inertial mass
• Physicists believe that the inertial mass is precisely equal
to the gravitational mass
• Why will be addressed later
Section 3.2
Adding Friction
• Friction is another force commonly encountered in
problems
• Friction can be
• Kinetic
•
Related to moving
• Static
•
When objects are at rest
Section 3.3
An Object Coming to Rest
• Aristotle’s idea was that rest was the natural state of
terrestrial objects
• Newton’s view is that an object comes to rest because a
force acts on it
• This force is often due to a phenomena called friction
Section 3.3
Friction
• The force of friction opposes
the motion
• The magnitude of the
frictional force is related to
the magnitude of the normal
force
• Force of kinetic friction
• Ffriction = μk N
• μk is called the coefficient
of kinetic friction
•
A pure number with no units
Section 3.3
Motion with Friction
• Set up the problem as usual, including the force of friction
• For example, as seen in fig. 3.11:
• ΣF = Ffriction = -μk N = m a
• From the y-direction, Fgrav = N = m g
• Therefore, -μk m g = m a and a = - μk g
• Once you have found the acceleration, other quantities
involved with motion can also be found
• See Table 3.2 for values for coefficients of friction for
various materials
• The values depend on the smoothness of the surfaces
Section 3.3
Static Friction
• In many situations, the
relevant surfaces are not
slipping (moving) with
respect to each other
• This situation involves
static friction
• The amount of the push
can vary without the
object moving
Section 3.3
Static Friction, cont.
• |Ffriction | ≤ μs N
• μs is called the coefficient of static friction
•
Static indicates that the two surfaces are not moving relative to
each other
• If the push is increased, the force of static friction also
increases and again cancels the force of the push
• The magnitude of the static friction has an upper limit of
μs N
Section 3.3
Kinetic Friction Compared
to Static Friction
• Consider both the kinetic and static friction cases
• Use the different coefficients of friction
• The force of kinetic friction is just Ffriction = μk N
• The force of static friction varies by |Ffriction | ≤ μs N
• For a given combination of surfaces, generally μs > μk
• It is more difficult to start something moving than it is to
keep it moving once started
Section 3.3
Friction and Walking
• The person “pushes” off
during each step
• The bottoms of his shoes
exert a force on the ground
• This is
• If the shoes do not slip, the
force is due to static
friction
• The shoes do not move
relative to the ground
Section 3.3
Friction and Walking, cont.
• Newton’s Third Law tells
us there is a reaction force
•
• This force propels the
person as he moves
• If the surface was so
slippery that there was no
frictional force, the person
would slip
Section 3.3
Friction and Rolling
• The car’s tire does not slip
• There is a frictional force
between the tire and road
• There is a reaction force on
the tire
• This is the force that propels
the car forward
Section 3.3
Friction and Motion
• The force of friction makes walking and running
possible
• Friction plays a key role in rolling motion
Section 3.3
Free Fall
• A specific type of motion
• Only gravity acts on the
object when it is in free
fall
• There is actually some air
drag, but it is generally
considered negligible
• We will analyze the
motion in terms of
acceleration, velocity, and
position
Section 3.4
Free Fall – Acceleration
• Take t = 0 to be the instant after the ball leaves the thrower’s
hand
• Before then, there is an additional force from her hand acting on
the ball
•
This would not be an example of free fall since a force other than gravity
would be acting on the object
• Choose a coordinate system that measures position as the
height y above the ground
• Using Newton’s Second Law:
• The negative sign means gravity is directed downward
Section 3.4
Free Fall – Velocity and Position – Equations
• Find the velocity and position as functions of time:
• Note the acceleration is constant (and equal to g)
• The velocity and acceleration are not always in the
same direction
Section 3.4
Free Fall – Velocity and Position – Graphs
• The motion can be expressed
graphically as well
• The velocity can be positive,
negative, or zero
• The slope of the v-t graph is
constant and negative for the
entire motion
• The constant negative
acceleration is shown in
graph C
Section 3.4
Free Fall – Final Notes
• When the ball starts and stops at the same location
(assume the ground), the ball’s speed just before it hits the
ground is equal to its initial speed
• The velocities are in opposite directions
• The time spent on the way up is equal to the time spent
falling back down
• Shows the symmetry in the problem
Section 3.4
Transmitting Forces
• Strings exert a force on the objects they are connected to
• Also applies to cables, ropes, etc.
• The mass of the cable may have to be taken into account
• Pulleys can redirect forces
• Forces can be amplified
Section 3.5
Tension
• Strings exert a force on the
objects they are connected to
• Cables and ropes act the
same way
• The strings exert force due to
their tension
• The ends of the string both
exert a force of magnitude T
on the supports where they
are connected
• T is the tension in the string
Section 3.5
Tension Example – Elevator Cable
• Two forces are acting on
the compartment
• Gravity acting downward
• Tension in cable acting
upward, T
• Assume an acceleration
upward
• Applying Newton’s
Second Law gives
T = mg + ma
Section 3.5
Tension Example – Elevator, cont.
• Now consider the cable
• Assume the cable is massless
• Applying Newton’s Second Law gives: TC = T
• The tension is the same for all points along the cable
• This is true for all massless cables
• Tension has force units
Section 3.5
Cables with Mass
• Apply Newton’s Second
Law to the cable
• To support the cable, the
upper tension, T1 must be
larger than the tension from
the box, T2
• T1 = T2 + mcable g
• If no acceleration
• Can assume a massless cable
if the mass of the cable is
small compared to the other
tensions present
Section 3.5
Single Pulleys
• We often need to change
the direction of the force
• A simple pulley changes
the direction of the force,
but not the magnitude
• See diagram
• Assume the rope and pulley
are both massless
• Assume the cable does not
slip on the pulley
Section 3.5
Pulleys To Amplify Forces
• The person exerts a force of
T on the rope
• The rope exerts a force of 2T
on the pulley
• This force can be used to lift
an object
• More complex sets of
pulleys can amplify an
applied force by greater
factors
• The distance decreases to
compensate for the increase
in force
Section 3.5
Reasoning and Relationships –
Problem Notes
• We may need to identify important information that is
“missing” from the initial description of the problem
• We need to recognize that additional information is needed
• Then make reasonable estimates of the “missing” quantities
• An approximate mathematical solution and an
approximate numerical answer are generally sufficient
• The estimates of the “missing” values will vary from case to
case
Section 3.6
Reasoning and Relationships –
Problem Solving Strategy
• Recognize the principle
• Determine the key physics ideas central to the problem
• What principles connect the quantity you want to calculate with
the quantities you know
• Sketch the problem
• Show all the given information
• Draw a free body diagram, if needed
•
Include all the forces, velocities, etc.
• Identify the relationships
• Motion equations are an example of a set of relationships
• If some values are unknown, make estimates for these values
Section 3.6
Reasoning and Relationships –
Problem Solving Strategy, cont.
• Solve
• An exact mathematical solution typically is not needed
• Cast the problem into one that is easy to solve
mathematically
• Check
• Consider what your answer means
• Check to be sure the answer makes sense
Section 3.6
Parachutes, Air Drag, and Terminal Velocity
• Many constant acceleration problems are good
approximations to real-life examples
• Some interesting and important cases occur when
acceleration is not constant
• These cases include an object moving through a fluid
• The object is subject to a resistive drag force arising from
contact with the fluid molecules
Section 3.7
Air Drag
• There are situations where air drag cannot be neglected
• Air drag depends on speed, so at higher speeds it becomes more
of an effect
• Also depends on the area, so large areas cause a large air drag
• An estimate of air drag can be found by using
• Fdrag = ½ ρ A v2
• A more complete equation is
• Fdrag = ½ CD ρ A v2
• CD is the drag coefficient and depends on the aerodynamic shape
• CD is 1 for boxy shapes and less than 1 for many streamlined
shapes
Section 3.7
Skydiving Example
• Assume a skydiver opens
their chute at t = 0
• Initially, the skydiver can be
considered in free fall
• Velocity is small, so air drag
is negligible
• As the velocity increases, so
does the air drag
• Eventually, the skydiver will
reach a constant velocity
• Called the terminal
velocity
Section 3.7
Terminal Velocity
• Once the skydiver reaches terminal velocity, the net force
acting on her is zero
• The drag force is (nearly) equal in magnitude but opposite
in direction of the force of gravity
• The magnitude of the terminal velocity is
• Air drag is generally small for slowly moving objects, but
it is very important in some cases
Section 3.7
Bacterium Example
• Consider the motion of a small
object in a liquid
• Stokes showed that the drag
force on a spherical object
moving slowly through a fluid
is given by
• C is a constant that depends on
the properties of the fluid
• r is the radius of the object
• The negative sign indicates the
drag force is always directed
opposite of the velocity
Section 3.8