Transcript Lecture04

Physics 111 Lecture 04
Force and Motion I: The Laws of Motion
SJ 8th Ed.: Ch. 5.1 – 5.7
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5.1
5.2
5.3
5.4
5.5
5.6
5.7
Dynamics - Some history
Force Causes Acceleration
Newton’s First Law: zero net force
Mass
Newton’s Second Law
Free Body Diagrams
Gravitation
Newton’s Third Law
Application to Sample Problems
The Concept of a Force
Newton’s First Law and
Inertial Frames
Mass
Newton’s Second Law
Gravitational Force and Weight
Newton’s Third Law
Using Newton’s Second Law
Copyright R. Janow – Spring 2012
Dynamics - Newton’s Laws of Motion
Kinematics described motion only – no real Physics.
Why does a particle have a certain acceleration?
New concepts (in 17th century):
• Forces - pushes or pulls - cause acceleration
• Inertia (mass) measures how much matter is being accelerated –
resistance to acceleration
Newton’s 3 Laws of Motion:
• Codified kinematics work by Galileo and other early experimenters
• Introduced mathematics (calculus) as the language of Physics
• Allowed detailed, quantitative prediction and control (engineering).
• Ushered in the “Enlightenment” & “Clockwork Universe”.
• Are accurate enough (with gravity) to predict all common
motions plus those of celestial bodies.
• Fail only for v ~ c and quantum scale (very small).
Sir Isaac Newton 1642 – 1727
• Formulated basic laws of mechanics
• Invented calculus in parallel with Liebnitz
• Discovered Law of Universal Gravitation
• Many discoveries dealing with light and optics
• Ran the Royal Mint for many years during old age
• Many rivalries & conflicts, few friends,
no spouse or children, prototype “geek”
Copyright R. Janow – Spring 2012
Force: causes any change in the velocities of
particles
Newton’s
A force is that which
definition:
causes an acceleration
Contact forces involve “physical contact” between objects
F = - kx
Field forces act through empty space without physical contact
Action at a distance through intervening space? How?
Given atomic physics, Is there any such thing as a real contact force?
The four fundamental forces of nature are:
Gravitation, Electromagnetic, Nuclear,
and Weak force
Copyright R. Janow – Spring 2012
Force: what causes any change in the velocities
of particles
Units: 1 Newton = force that causes 1 kg to accelerate at 1 m/s2
1 Pound = force that causes 1 slug to accelerate at 1 ft/s2
Forces are VECTORS
Operate with vector rules

F
Replace a force acting at a point by its
components at the same point.

Fy ĵ
Fx î

F1
Superposition: A set of forces at a point
have the same effect that their vector
resultant force would

F3

F2

Fnet

Notation:

Fnet 

 Fi
i
Definition: A body is in EQUILIBRIUM
if the net force applied
to it
equals
zero
Copyright
R. Janow
– Spring 2012
Relative motion - Inertial Reference Frames in 1 Dimension
A frame of reference amounts to selecting a coordinate system.
• Describe point P in both frames using x1, v1, a1 and x2, v2, a2.
• Origins coincide at t = 0
• Constant v12 = relative velocity of origin of 1 in frame 2
• x12 is the distance between origins at some time t. 2
y1
y
Transform the coordinates:
x2 (t)  x1(t)  x12(t)
v12
Transform the velocities:
dx 2
dt

dx1
dt

dx12
dt
v2  v1  v12
a2 
dt

dv1
dt
Then a12 = 0 and

dv12
dt
a1  a2
x2
P
x2
Find the accelerations:
dv2
x1
x12
dv1
dt
 a1
dv12
dt
 a12  0
The acceleration of a moving object is the same for a pair of inertial frames.
Example: An accelerating car viewed from a train and the ground, with the train
itself moving at constant velocity
Inertial frames can not be rotating or accelerating relative to one another or to the
Copyright R. Janow – Spring 2012
fixed stars. Non-inertial frames  fictitious forces.
x1
Newton’s First Law (1686)
Don’t moving objects come to a stop if you stop pushing?
• Stopping implies negative acceleration, due to friction
or other forces opposing motion.
• What effect does inertia have on a curve on an icy road?
“The Law of Inertia”: A body’s velocity is constant (i.e., a = 0) if
the net force acting on it equals zero
Alternate statement: A body remains in uniform motion along a
straight line at constant speed (or remains at rest) unless it is
acted on by a net external force.
Above assume an “inertial reference frame”:
• Equations of physics look simplest in inertial systems.
• Non-inertial frames (e.g., rotating) require fictitious forces &
accelerations in physics equations (e.g., centrifugal and Coriolis forces).
First Law (our text): An object that does not interact with other objects
(no net force, isolated) can be put into a reference frame in which the
object has zero acceleration (i.e., it’s motion can be transformed to an
inertial frame if it is not in one already).
Copyright R. Janow – Spring 2012
Mass: the Measure of Inertia
Apply the same force to different objects.
Different accelerations result. Why?
Example: Apply same force to baseball, bowling ball, automobile, RR train
Mass measures inertia:
• the amount of “matter” in a body (i.e., how many atoms of each type)
• resistance to changes in velocity (i.e., acceleration) when a force acts
For a given force applied to m1 and m2:
m1
m2

a2
a1
Mass is a scalar:
A mass and resulting acceleration on it
are inversely proportional
Attach m to m
1
2
The result behaveswith m  m  m
1
2
Mass is intrinsic to an object:
• it doesn’t depend on the environment or state of motion (for v << c), or time.
Don’t confuse mass with weight (a force):
W  mg
If m  2 kg then Wearth  20 N. but Wmoon  3.3 N.
The same “inertial mass” value also measures “gravitational mass” –
- a particle’s effect in producing gravitational pull
on other
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R. Janowmasses
– Spring 2012
Newton’s Second Law
SIMPLE PROPORTIONALITY WHEN IN AN INERTIAL FRAME

Fnet 
Summarizing:


 Fi  ma
i
Vector sum of forces
acting ON a particle
Inertia of
particle
Acceleration resulting
from Fnet
DIRECTION OF ACCELERATION AND NET FORCE ARE THE SAME
Cartesian component equations:
Fnet,x   Fix  ma x
Fnet,y 
i
Other ways to
Write 2nd Law:

Fnet 
Units for Force:
[F]
SYSTEM
SI
CGS
British
1 dyne = 10-5 N

Fiy 


d2r
 Fi  m dt2
i
[m] [a]
FORCE
Newton (N)
Dyne
Pound (lb)
1 gm = 10-3 kg
ma y
Fnet,z 
 Fiz 
ma z
i
i




dp
Fnet 
where p  mv
dt
 M L / T2
MASS
Kg
gm
slug
ACCELERATION
m/s2
cm/s2
ft/s2
1 lb = 4.45 kg
1 kg
WEIGHS
2.2 lb
1 slug = 14. 59 kg
Copyright R. Janow – Spring 2012
Tug of war
4-1: Three students can all pull on the ring (see sketch) with identical
forces of magnitude F, but in different directions with respect to the +x
axis. One of them pulls along the +x axis with force F1 as shown.
What should the other two angles be to minimize the magnitude of the
ring’s acceleration?
a) q2 = 0, q3 = 0
b) q2 = 180, q3 = -180
c) q2 = 60, q3 = -60
d) q2 = 120, q3 = -120
e) q2 = 150, q3 = -150

F2

F3
q3
q2

F1
x
4-2: What should the other two angles be to maximize the magnitude of the ring’s
acceleration?
a) q2 = 0, q3 = 0
b) q2 = 180, q3 = -180
c) q2 = 60, q3 = -60
d) q2 = 120, q3 = -120
e) q2 = 150, q3 = -150
Copyright R. Janow – Spring 2012
Example:
A hockey puck whose mass is 0.30 kg is
sliding on a frictionless ice surface. Two
forces act horizontally on it as shown in the
sketch. Find the magnitude and direction of the
puck’s acceleration.
m = 0.3 kg
Apply 2nd Law to x and y directions
Fnet,x  F1x  F2x  ma x  F1 cos(20)  F2 cos(60)
Fnet,y  F1y  F2y  ma y  F1 sin(20)  F2 sin(60)
Evaluate:
Fnet,x  8.7 N
Fnet,y  5.2 N
ax 
ay 
Fnet,x
m
Fnet,y
m

8.7
 29 m / s2
0.3

5.2
 17 m / s2
0.3
FBD
Convert to polar coordinates:
a  [a2x  a2y ]1 / 2  34 m/s2
q  tan-1 (ay / ax )  31o
The net force and acceleration vectors have the same direction
A unit vector in that direction is:

ay
ax
a
29
17
â 

î 
ĵ 
î 
ĵ  0.85 î  0.50 ĵ
|a| |a|
|a|
34
34
Copyright R. Janow – Spring 2012
Measuring Force and Mass
On frictionless surface (e.g., air track): Apply enough horizontal
force F0 to give the standard mass m0 the standard acceleration a0
m0 = 1 kg
a0 = 1 m/s2
The (standard) force unit thereby defined = 1N.
F0  m0a0
Note: x-components only above, F & a are in same direction
Measuring another mass:
• Apply standard force, record resulting acceleration
a  F0 / m
m  F0 / a
What about forces in the y – direction, left out above?
Copyright R. Janow – Spring 2012
Gravitational Force, Weight, “Normal” Force
Mass in free fall on Earth’s surface accelerates at g:
FBD
m
g has the same direction (toward center of Earth)
and magnitude for all masses (in a volume large
compared to a human).

Fg   mg ĵ
g = 9.8 m/s2 = 32.2 ft/s2 at the Earth’s surface

Fg  " the weight"
• “Action at a distance”
• The weight is independent of how a mass is
moving (perhaps other forces also act).
Mass in contact with a “horizontal” surface (table, air track, …)
may be in equilibrium for y: " Equilibrium" for y   Fy  0  a y
FBD
ay = 0  N = Fg
(m does not accelerate)
N  " normal force"  perpendicular to surface
m

Fg   mg ĵ
N
N is a contact force that adjusts to Fg
Fg pushes on the surface –
- the surface pushes back with N
If ay not = 0, N does not equal Fg
Copyright R. Janow – Spring 2012
Newton’s Third Law
Bodies interact by pushing or pulling on each other
3rd Law (antique version): Each action has an equal and opposite reaction
More modern version: When two bodies interact the forces that each
exerts on the other are always equal in magnitude and opposite in direction
Example: gravity acting at a distance

F12  force on object 1 due to object 2

F21  force on object 2 due to object 1


F12   F21
Example: box on a level surface
m

Fg   mg ĵ
N
Fs
Fe
•
•
•
•
•
•
•
Fg is the pull of Earth on the box (weight)
The 3rd law reaction is Fe - the box’s pull on the Earth
N is the surface’s push on the box
Fs is the box’s push on the surface
ONLY forces on the box (Fg & N) affect it’s motion
Fg & N are NOT a 3rd law pair (Fsb & N are a pair)
Why then does Fg = N ???
If you ever find a force w/o the 3rd law reaction,
Janow – Spring 2012
you can build a perpetualCopyright
motionR.machine
Free Body Diagrams (FBDs)
Drawing the FBDs is the most important step in analyzing motion.
• DRAW FBDs FIRST – before you start writing down equations.
• Pictorial sketches are not the same as FBDs.
• Model bodies in your system as point particles. There may be
several. Sometimes you can treat the system as one object.
• Choose coordinates.
• Include in FBDs only forces that act ON your system.
• Exclude forces exerted BY bodies in your system on other bodies.
• Neglect internal forces. When you break up a system for analysis,
INCLUDE formerly internal forces that become external.
• Don’t forget action-at-a-distance forces (fields) such a gravity
M
N
m

Fg
N
m

Fg
N

Fg   Mg ĵ
Is this a FBD?
m
'
Fg   mgĵ
Copyright R. Janow – Spring 2012
FBDs: show only forces on bodies
Friction
Frictionless
m
N
F
Fg
No friction parallel to surface
ax  F / m
Sliding or static friction f (later)
m
f
N
F
Fg
Friction is a resistive force parallel to surface
ax  (F  f ) / m
f N
Contact force - always opposed to motion
Cords
Tension only, no compression
Pulling creates tension T – the force transmitted by the cord
T is the same everywhere in a zero-mass, unstretchable cord
Support
FBD for body FBD for cord
FBD for support
T’ = T
T
T
F
3rd law pair
m
rd law pair
3
m
T’
Fg
T
Fg  T  T'  F
Equilibrium
Copyright R. Janow – Spring 2012
Newton’s Laws - Summary
Newton’s First Law

A body’s velocity is constant ( a  0 ) if the net external force on it is zero
- Motion is along a straight line
- Find and use an inertial frame of reference
Newton’s Second Law

Fnet 


F

net
force
vector

m
a
i
i
In Cartesian components:
Fx
 ma x
Fy
 ma y
Fz
 maz
Newton’s Third Law
If body A exerts a foce on body B, then body B exerts and equal and
Opposite force on body A.
Copyright R. Janow – Spring 2012
Method for solving Newton’s Second Law problems
• Draw or sketch system. Adopt coordinates. Name the variables,
• Draw free body diagrams. Show forces acting on particles. Include
gravity (weights), contact forces, normal forces, friction.
• Apply Second Law to each part



Fnet   Fi  ma
• Make sure there are enough (N) equations; Extra conditions connecting
unknowns (constraint equations) may be applicable
• Simplify and solve the set of (simultaneous) equations.
• Interpret resulting formulas. Do they make intuitive sense? Are the
units correct? Refer back to the sketches and original problem
• Calculate numerical results, and sanity check anwers (e.g., right order of
magnitude?)
Systems with several components may have several unknowns….
…and need an equal number of independent equations
Copyright R. Janow – Spring 2012
Example: Traffic Light in Equilibrium
Conceptualize:
cables are massless and don’t break
no motion
Categorize:
equilibrium problem – accelerations = 0
Model as particles in equilibrium
2 FBDs
FBD of Knot
FBD of Light
T3
Fg = 122 N
FBD of light yields T3 = Fg ( alight = 0)
FBD of knot yields:
Fx 
Fy 
0  T2 cos(53o )  T1 cos(37o )
0  T2 sin(53o )  T1 sin(37o )  T3
Solve upper for T2:
Fg
cos( 37o )
T2 
T1  1.33T1
o
cos( 53 )
T1 [sin( 37o )  1.33 sin(53o )]  1.22 N
T1  73.4 N
T  97.4 N
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Copyright
Example: Particle Motion Under a Net Force
y
Mass m is moving on a frictionless horizontal
surface, acted on by an external force of
magnitude F, making an angle q with the x-axis.
F
m
Find expressions for the acceleration
along the horizontal and vertical directions
Along x:
Fx  F cos(q)
 ma x
ax 
N
q
x
F cos(q)
m
W
W = weight
Along y: Is ay zero, or does particle accelerate upward?
Fy  F sin(q)  N  W  may
Set ay = 0. Crossover to ay > 0 when N also = 0, i.e, when
0  F sin(q)  N  W 
Fsin(q)  W
When Fy is > the weight, the
particle accelerates upward
ay 
When Fy is < the weight, the
particle does not accelerate
ay  0
F sin(q)  W
m
N0
N  W  F sin(q)
Copyright R. Janow – Spring 2012
Bootstraps
4-3: The man and the platform weigh a total of 500 N. He pulls upward
on the rope with force F. What force would he need to exert in order to
accelerate upward with a = 0.1 g? Is this possible?
a) F = 50 N
b) F = 1000 N
c) F = 550 N
d) F = 500 N
e) He cannot lift himself by his own bootstraps at all.
Copyright R. Janow – Spring 2012
Example: Sliding and Hanging Blocks
Block S, mass M is sliding on a frictionless
horizontal surface. Block H, mass m hangs from a
massless, unstretchable cord wrapped over a
massless pulley.
Find expressions for the accelerations of the
blocks and the tension in the cord.
N
T
No friction,
No mass
W  Mg
T’
Apply 2nd Law to blocks S & H separately for x & y
FBD for Block S
FBD for Block H
N
T’
T
a’
a
choose
positive down
W = Mg
Fxs  T  Ma
Fys  N  W 
W’ = mg
Ma ys  0
Fxh  maxh  0
Fyh  W'  T'  ma'
Constraints: T’ = T, a’ = a
Find formula for T
Mm
T  Ma 
g
mM
W’ = mg
Eliminate T,T’,a’
W'  Ma  ma
mg  ma  Ma
could also use
m
a

g
system approach
m

M
To find this Copyright R. Janow – Spring 2012
Example: Block Sliding on a Ramp [“Inclined Plane”]
Mass m is accelerating along a frictionless
inclined surface as shown, making an angle q
with the horizontal.
m
N
a
W = mg
q
Find expressions for the acceleration
and the normal force
Forces acting ON the block are N and W
• N is normal to the surface
• W is vertical as usual
Choose x-y axes aligned to ramp, for which:
Assume
a y  0,
a x  a (positive down & right)
Wy  W cos(q) Wx  Wsin(q)
Apply 2nd Law to x and y
Fy  N  Wy  N  mgcos(q)  0
N  mgcos(q)
Fx 
Wx  ma x  mg sin(q)
ax  a  gsin(q)
Does not depend on m
Why does the block accelerate?
Do you expect N to equal W?
y
FBD
N
Wx
90q
Wy
Check:
What happens as q  90o
as q  0o
q
q
W
Copyright R. Janow – Spring 2012
x
Example: “Atwood’s Machine” with Massless Pulley
No forces or
motion along x
Both masses have the same acceleration a (constraint). The
tension T in the unstretchable cord is the same on both sides
of the massless pulley (another constraint).
Find expressions for the acceleration and the tension
In the cord. Find the force Wtot supporting the pulley
FBD for m1
a
T
FBD for m2
Fy1  T  m1g  m1a
m1g
T
a
Fy2  m2g  T  m2a
m2g
Add the equations
T  m1g  m2g  T  m2a  m1a
a = 0 for m1 = m2
Wtot
T
T
a
a
m2  m1
m2  m1
a
g
a is clockwise for m2 > m1 a = g for m1 = 0 or a = - g for m2 = 0
Subtract the equations
T  m1g  m2g  T   m2a  m1a
FBD
for
pulley
Wtot
Wtot  2T  0
T
2m1m2
m2  m1
Wtot 
T = 0 for m1 or m2 equals zero.
T = m1g for m2 = m1
g
4m1m2
m2  m1
g
Wtot = 2m1g if m2 = m1
Otherwise not so
Copyright R. Janow – Spring 2012
Example: Your Weight in an Elevator
FBD for
A passenger whose mass m = 72.2 kg is standing on a
passenger
platform scale in an elevator.
What weight does the scale read for him as the elevator (and
+
himself) accelerates up and down?
• Fg = mg is the real weight, which doesn’t change
• The building is an inertial frame, as is the elevator
when traveling at constant speed.
• When the elevator and passenger are accelerating
their non-inertial frame  fictitious forces
N is the scale reading = apparent weight
• Apply Second Law in the building’s reference frame
Fy  N  mg 
ma
No forces or
motion along x
N
Fg  mg
N  m (a  g)
Interpretations:
N
N
N
N
=
>
<
=
mg for a
mg for a
mg for a
0 for a =
= 0
> 0
< 0
-g
Normal weight for elevator at rest or moving with constant v
Increased apparent weight for elevator accelerating up
Decreased apparent weight for elevator accelerating down
Free fall - weightless
mg = force exerted by gravity
velocity has no effect on N – apparent weight
Fictitious force appears in non-inertial frame
Copyright R. Janow – Spring 2012