Newton’s Laws
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What is Force?
• A force is a push or pull on an object.
• Unbalanced forces cause an object to
accelerate…
– To speed up
– To slow down
– To change direction
• Force is a vector!
Types of Forces
• Contact forces involve contact between
bodies.
– Normal, Friction
• Field forces act without necessity of
contact.
– Gravity, Electromagnetic, Strong, Weak
• Question: Is there really any such thing
as a contact force?
Forces and Equilibrium
• If the net force (SF) on a body is zero, it is
in equilibrium.
• An object in equilibrium may be moving
relative to us (dynamic equilibrium).
• An object in equilibrium may appear to be
at rest (static equilibrium).
Galileo’s Thought Experiment
© The Physics Classroom, Tom Henderson 1996-2007
Galileo’s Thought Experiment
This thought experiment lead to
Newton’s First Law.
© The Physics Classroom, Tom Henderson 1996-2007
Newton’s First Law
• Newton’s 1st Law is also called the Law of
Inertia.
• A body in motion stays in motion in a
straight line unless acted upon by an
external force.
• This law is commonly applied to the
horizontal component of velocity, which is
assumed not to change during the flight of
a projectile.
Newton’s Second Law
• Newton’s 2nd Law says that a body acted
upon by a net external force will
accelerate.
• The acceleration is proportional to the net
force and inversely proportional to the
mass. It is in the direction of the net force.
• SF = ma
Newton’s Third Law
• Newton’s 3rd Law is commonly stated “for
every action there exists an equal and
opposite reaction”.
• If A exerts a force F on B, then B exerts a
force of -F on A.
Commonly Confused Terms
• Inertia: or the resistance of an object to
being accelerated
• Mass: the same thing as inertia (to a
physicist).
• Weight: gravitational attraction
inertia = mass  weight
Sample Problem: Three forces act upon a 3.0 kg body
moving at constant velocity. F1 = (4i – 6j + k) N and F2 = (i – 2j
- 8k) N. Find F3.
• Sample Problem: Two forces, F1 = (4i – 6j + k) N and F2
= (i – 2j - 8k) N, act upon a body of mass 3.0 kg. No other
forces act upon the body at this time. What do you know
must be true?
• Sample problem: A tug-of-war team ties a rope to a tree and
pulls hard horizontally to create a tension of 30,000 N in the
rope. Suppose the team pulls equally hard when, instead of a
tree, the other end of the rope is being pulled by another tug-ofwar team such that no movement occurs. What is the tension in
the rope in the second case?
A systematic approach for
1st or 2nd Law Problems
1. Identify the system to be analyzed. This may
be only a part of a more complicated system.
This is really the most important step.
2. Select a reference frame or coordinate system,
stationary or moving, but not accelerating,
within which to analyze your system.
3. Identify all forces acting on the system. Ignore
the others. This is best done by drawing!
4. Set up ΣF = ma equations for each dimension.
Solve for relevant unknowns.
5. Use kinematics or calculus where necessary to
calculate more about the motion.
• Sample Problem: A 5.00-g bullet leaves the muzzle of a rifle
with a speed of 320 m/s. The bullet is accelerated by
expanding gases while it travels down the 0.820 m long
barrel. Assuming constant acceleration and negligible
friction, what is the force on the bullet?
• Sample Problem: A 3.00 kg mass undergoes an acceleration
given by a = (2.50i + 4.10j) m/s2. Find the resultant force F and
its magnitude.
Monday, October 7, 2013
The Normal Force
Tension
Normal force
• The force that keeps one object from
invading another object is called the
normal force.
• “Normal” means “perpendicular”.
• Always determine the normal force by
considering all forces that have
components perpendicular to a surface.
Problem: determine the normal force acting on a 5.0 kg
box sitting on a flat table.
Problem: Now determine the normal force acting on a
5.0 kg box sitting on a flat table.
F=16 N
40°
Problem: Now determine the normal force acting on a
this 5.0 kg box sitting on a ramp at angle q=30o.
Problem: Assume that the ramp is frictionless. What is
the acceleration of the block down the ramp?
F=20 N
20°
30o
Tension
• A pulling force.
• Generally exists in a rope, string, or
cable.
• Arises at the molecular level, when a
rope, string, or cable resists being
pulled apart.
Tension (static 2D)
The sum of the horizontal and vertical
components of the tension are equal to
zero if the system is not accelerating.
30o
45o
1
2
3
15 kg
Problem: Determine the tension in all three ropes.
30o
45o
1
2
3
15 kg
Problem: What is the tension is the cable attached to a 5,000
kg elevator that starts on the ground floor at rest and
accelerates upward, reaching a speed of 3.0 m/s in 2 seconds?
M
Tuesday, October 8th, 2013
Magic Pulleys
Pulley problems
• Pulley’s simply bend the coordinate
system
m1
m2
Sample problem: derive a formula for acceleration, assuming the
table is frictionless.
m1
m2
Sample problem: derive a formula for acceleration,
assuming the table is frictionless.
q
m2
Sample problem: derive a formula for the tension T in the
string.
q
m2
Atwood machine
m2
m1
• A device for measuring g.
• If m1 and m2 are nearly the
same, slows down freefall such
that acceleration can be
measured.
• Then, g can be measured.
Problem: For the Atwood machine shown, derive
an equation which can be used to find g, the
gravitational acceleration, from a measured value of
acceleration.
m2
m1
Atwood Machine Pre-lab
• Purpose: to determine gravitational
acceleration using the Atwood machine.
• Hypothesis:?
• Theoretical Background: See previous
slide.
• Procedure: ?
• Analysis Plan: ?
• Data: You are collecting data to determine
acceleration. Mass is important, too, You’ll
need data tables ready to record data into.
Atwood Machine Lab
Thursday, October 10th, 2013
Atwood Machine lab
• Write a full lab report in your lab notebook. The
pre-lab you did yesterday has many of the
relevant sections. Today, collect your data and
analyze your results. Tabulate your data neatly.
• Calculate g from your mass and acceleration
values. Do a standard deviation analysis on g,
and do a percent error calculation on g as well.
• Report results and write a conclusion.
Friction
Tuesday, October 15, 2013
Problem: How high up the frictionless ramp will the block slide?
v = 12.0 m/s
5.0 kg
20o
Problem: Calculate acceleration of the 5 kg block. Table and pulley are
magic and frictionless. Calculate Tension in the string.
1.0 kg
20o
Friction
• Friction opposes a sliding motion.
• Static friction exists before sliding
occurs
– (fs  sN).
• Kinetic friction exists after sliding
occurs
–fk = kN
y
y
x
Draw a free body diagram for
a braking car.
x
Draw a free body diagram for
a car accelerating from rest.
Question:
• Why is it disadvantageous for cars to skid
to a stop?
• Draw free body diagrams that include friction for a body
which is
q
Sliding down a ramp
q
Sliding up a ramp
Sample problem: A 1.00 kg book is held against a wall by
pressing it against the wall with a force of 50.00 N. What
must be the minimum coefficient of friction between the book
and the wall, such that the book does not slide down the wall?
F
Problem: Assume a coefficient of static friction of 1.0 between
tires and road. What is the minimum length of time it would take
to accelerate a car from 0 to 60 mph?
Problem: Assume a coefficient of static friction of 1.0 between tires and road
and a coefficient of kinetic friction of 0.80 between tires and road. How far
would a car travel down a 15o incline after the driver applies the brakes if it
skids to a stop? Assume the speed before brakes are applied is 26 m/s.
Review Time!
• Homework Problems --?
• Clicker Quiz – Newton’s Laws
Centripetal Force
Thursday, October 17th, 2013
Centripetal Force
• Inwardly directed force which causes a
body to turn; perpendicular to velocity.
• Centripetal force always arises from
other forces, and is not a unique kind of
force.
– Sources include gravity, friction, tension,
electromagnetic, normal.
• ΣF = ma
•
a = v2/r
• ΣF = m v2/r
Highway Curves
z
R
Friction turns the vehicle
r
Normal force turns the vehicle
Problem: Find the minimum safe turning radius for a car
traveling at 60 mph on a flat roadway, assuming a
coefficient of static friction of 0.70.
Problem: Derive the expression for the best banking angle
of a roadway given the radius of curvature and the likely
speed of the vehicles.
Centripetal Force Lab
Conical Pendulum
z
L
What provides
the centripetal
force for a
conical
pendulum?
T
r
mg
Problem: Derive the expression for the period of a conical
pendulum with respect to the string length and radius of
rotation.
L cosq
T  2
g
Centripetal Force Lab
• This lab report will include only three parts:
– Theoretical development section.
– Data table from your class.
– A graph the conical pendulum data from the
data such that the acceleration due to gravity
can be obtained from the slope of a line.
Example class
Conical Pendulum Data
# osc.
Total time
(t, s)
Period
(T, s)
Radius
(R, m)
Length
(L,m)
10
14.9
1.49
0.07
0.70
5
14.96
2.95
0.50
2.21
5
5.97
1.19
0.09
0.57
2.70
0.31
1.86
3
1.31
10
11.02
0.06
0.39
3rd Period Conical Pendulum Data
# osc.
Total time
(t, s)
Period
(T, s)
Radius
(R, m)
Length
(L,m)
Non-uniform circular motion
Monday, October 21, 2013
Non-uniform Circular Motion
• Consider circular motion in which either speed
of the rotating object is changing, or the forces
on the rotating object are changing.
• If the speed changes, there is a tangential as
well as a centripetal component to the force.
• In some cases, the magnitude of the
centripetal force changes as the circular
motion occurs.
Problem: You swing a 0.25-kg rock in a vertical circle on a 0.80 m
long rope at 2.0 Hz. What is the tension in the rope a) at the top
and b) at the bottom of your swing?
Problem: A 40.0 kg child sits in a swing supported by 3.00 m long
chains. If the tension in each chain at the lowest point is 350 N,
find a) the child’s speed at the lowest point and b) the force
exerted by the seat on the child at the lowest point.
Problem: A 900-kg automobile is traveling along a hilly road. If it
is to remain with its wheels on the road, what is the maximum
speed it can have as it tops a hill with a radius of curvature of
20.0 m?
Non-Constant Forces
(Forces that are functions of time)
Tuesday, October 22, 2013
Non-constant Forces
• Up until this time, we have mainly dealt
with forces that are constant. These
produce a uniform, constant acceleration.
Kinematic equations can be used with
these forces.
• However, not all forces are constant.
• Forces can vary with time.
• Forces can vary with velocity.
• Forces can vary with position.
Calculus Concepts for
Forces That Vary With Time
• The Derivative
• The derivative yields tangent lines and slopes.
• Use the derivative to go from position -> velocity ->
acceleration
• The Integral
• The integral yields the area under a curve.
• Use the integral to go from acceleration -> velocity ->
position
• In these two approaches, F = ma, so methods
for determining or using acceleration are
extendable to force.
Tutorial on the Integral
• Mathematically, the integral is used to calculate a sum
composed of many, many tiny parts.
• In physics, the integral is used to find a measurable change
resulting from very small incremental changes. It’s useful to
think of it as a fancy addition process or a multiplication
process, depending on the situation.
• Here’s a specific example. Velocity times time gives
displacement. If the velocity is changing with time, but a
very tiny time change is used to calculate a very tiny
displacement, we can nonetheless assume the velocity was
constant during that tiny time change.
• If we calculate tiny displacements this way (recalculating our
velocity for each time increment), then add the tiny
displacements up to get a larger displacement, we have
done “integration”.
Velocity as a Function of Time
Displacement
• Velocity can be represented as follows:
dx
v
dt
• Rearrangement of this expression yields:
dx  vdt
• What this means is that we can calculate a tiny
displacement dx from the velocity v at a given
time times a tiny time increment dt.
Summing the Displacements
• When we sum up these tiny displacements, we
use the following notation:

xf
xo
tf
dx   vdt
to
• This notation indicates that we are summing up
all the little displacements dx starting at position
xo at time to until we reach a final position and
time, xf and tf. The velocity v may be a function of
time, and may be slightly different for one time
increment dt and the next time increment.
Evaluating Integrals
• We will evaluate polynomial integrals by reversing
the process we used in taking a polynomial
derivative. This process is sometimes called “antidifferentiation”, or “doing an anti-derivative”.
• The general method for doing an anti-derivative is:
n 1
dx
At
if
 At n then x 
C
dt
n 1
• Can you see how this is the reverse of taking a
derivative? (Note: This “indefinite integral” requires
us to add a constant C to compensate for constants
that may have been lost during differentiation…but
we will do “definite integrals” that do not require C.)
Evaluating Definite
Integrals
• When a “definite integral”
with “limits of integration”
is to be evaluated, first
do the “anti-derivative”.
Then evaluate the
resulting functions using
the top limits, evaluate
them again using the
bottom limits, then
subtract the two values
to get the answer. (Note:
do each side of the
equation separately.)
dx
 v where v  3t 2  2t  5
dt
Starting time: 1 s; ending time: 2 s.
Starting position:4 m

xf
4
2
dx   (3t  2t  5)dt
2
1
xf
x4
2
 3t
2t
5t 


 
2
1 1
 3
3
2
x f  4   t  t  5t 
3
2
2.0
1.0
x f  4  22  7  15
x f  19
Evaluating Definite
Integrals
• When a “definite integral”
with “limits of integration”
is to be evaluated, first
do the “anti-derivative”.
Then evaluate the
resulting functions using
the top limits, evaluate
them again using the
bottom limits, then
subtract the two values
to get the answer. (Note:
do each side of the
equation separately.)
dx
 v where v  3t 2  2t  5
dt
Starting time: 1 s; ending time: 2 s.
Starting position:4 m

xf
4
2
dx   (3t  2t  5)dt
2
1
xf
x4
2
 3t
2t
5t 


 
2
1 1
 3
3
2
x f  4   t  t  5t 
3
2
2.0
1.0
x f  4  22  7  15
x f  19
Evaluating Definite
Integrals
• When a “definite integral”
with “limits of integration”
is to be evaluated, first
do the “anti-derivative”.
Then evaluate the
resulting functions using
the top limits, evaluate
them again using the
bottom limits, then
subtract the two values
to get the answer. (Note:
do each side of the
equation separately.)
dx
 v where v  3t 2  2t  5
dt
Starting time: 1 s; ending time: 2 s.
Starting position:4 m

xf
4
2
dx   (3t  2t  5)dt
2
1
xf
x4
2
 3t
2t
5t 


 
2
1 1
 3
3
2
x f  4   t  t  5t 
3
2
2.0
1.0
x f  4  22  7  15
x f  19
Evaluating Definite
Integrals
• When a “definite integral”
with “limits of integration”
is to be evaluated, first
do the “anti-derivative”.
Then evaluate the
resulting functions using
the top limits, evaluate
them again using the
bottom limits, then
subtract the two values
to get the answer. (Note:
do each side of the
equation separately.)
dx
2
 v where v  3t  2t  5
dt
Starting time: 1 s; ending time: 2 s.
Starting position:4 m

xf
4
2
dx   (3t 2  2t  5)dt
1
xf
x4
2
 3t
2t
5t 


 
2
1 1
 3
3
2
x f  4   t  t  5t 
3
2
2.0
1.0
x f  4  22  7  15
x f  19
Evaluating Definite
Integrals
• When a “definite integral”
with “limits of integration”
is to be evaluated, first
do the “anti-derivative”.
Then evaluate the
resulting functions using
the top limits, evaluate
them again using the
bottom limits, then
subtract the two values
to get the answer. (Note:
do each side of the
equation separately.)
dx
2
 v where v  3t  2t  5
dt
Starting time: 1 s; ending time: 2 s.
Starting position:4 m

xf
4
2
dx   (3t 2  2t  5)dt
1
xf
x4
2
 3t
2t
5t 


 
2
1 1
 3
3
2
x f  4   t  t  5t 
3
2
2
1
x f  4  22  7  15
x f  19
• Sample problem: Consider a force that is a function of time:
F(t) = (3.0 t – 0.5 t2)N
• If this force acts upon a 0.2 kg particle at rest for 3.0 seconds, what is the
resulting velocity and position of the particle?
• Sample problem: Consider a force that is a function of time:
F(t) = (16 t2 – 8 t + 4)N
• If this force acts upon a 4 kg particle at rest for 1.0 seconds, what is the
resulting change in velocity of the particle?
Review: Graphical Integration
• When we are given a graph, we can
perform graphical integration.
• This is equivalent of “multiplying the axes”
or “determining the area under the curve”.
• Let’s take a look at an example.
v (m/s)
Displacement
1.0
t (s)
-1.0
1.0
2.0
3.0
4.0
Consider the graph of velocity versus time shown above. Now
determine, first from an equation and then directly from the
graph, the displacement during the first second (from 0 to 1.0 s).
Displacement
v (m/s)
v  vo  at
v  0.60t
dx
 0.60t
dt
dx  0.60tdt
1.0

xf
0
1.0
dx   0.6tdt
0
t (s)
-1.0
1.0
2.0
3.0
4.0
The velocity follows the function shown above between zero and 1.0
s. An integral is performed on that function between zero and 1.0 s
to determine the displacement during that time interval.
Displacement
v (m/s)

xf
0
dx   0.6tdt
0
xf
1.0
1.0
xx
0
0.60t

2
2 1.0
0
x  0.30m
t (s)
-1.0
1.0
2.0
3.0
4.0
Evaluation of the integral yields a displacement of 0.30 m. This is
exactly the same displacement you would get by taking the area of
the colored triangle.
Non-Constant Forces
(Forces that are functions of
velocity)
Drag Forces
• Drag forces are functions of velocity,
rather than functions of time. They
– slow an object down as it passes through a
fluid.
– act in opposite direction to velocity.
– vary with velocity or velocity squared.
– depend upon the fluid and the projectile
characteristics.
– will impose a terminal velocity if a falling
object reaches a high enough speed.
Drag as a Function of Velocity
• In general, a drag force acts like a
velocity-dependent kinetic friction.
The general expression is
– fD = bv + cv2
• b and c depend upon
– shape and size of object
– properties of fluid
• b is more important at low velocity
• c is more important at high velocity
Drag Force in Free Fall
fD
fD
fD
mg
mg
mg
mg
when fD
equals
mg,
terminal
velocity
has been
reached
Drag Force in Free Fall
FD = bv + c v2
for fast moving objects
FD
FD = c v2
for slow moving objects
FD = b v
mg
c = 1/2 D r A
Where
D = drag coefficient
r = density of fluid
A = cross-sectional area
of projectile
Problem: For slow-moving objects, show that vT = mg/b,
where vT is terminal velocity.
Problem: For slow-moving objects, show that prior to attaining
terminal velocity, the velocity of an object subjected to the drag
force varies as:
bt
 
mg 
m
v(t ) 
1  e 
b 

Problem: For fast-moving objects, show that terminal velocity is
2mg
vT 
Dr A
Problem: Derive an expression for the velocity of a falling fastmoving object as a function of time. The expression you derive
should be valid prior to attainment of terminal velocity, as well as
after terminal velocity is attained.