Copyright R. Janow – Spring 2012
Physics 111 Lecture 04Force and Motion I: The Laws of MotionSJ 8th Ed.: Ch. 5.1 – 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
5.1 The Concept of a Force5.2 Newton’s First Law and Inertial Frames5.3 Mass5.4 Newton’s Second Law5.5 Gravitational Force and Weight5.6 Newton’s Third Law5.7 Using Newton’s Second Law
Copyright R. Janow – Spring 2012
Dynamics - Newton’s Laws of MotionKinematics 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
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”
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).
Copyright R. Janow – Spring 2012
Force: causes any change in the velocities of particles A force is that
which causes an acceleration
Newton’s definition: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
Replace a force acting at a point by its components at the same point.
iFx
jFy F
Superposition: A set of forces at a point have the same effect that their vector resultant force would
F1
3F
2F
i
inet F F
Notation:
netF
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.
Transform the coordinates:)t(x)t(x)t(x 1212
Transform the velocities:
dt
dx
dt
dx
dt
dx 1212 1212 v v v
Then a12 = 0 and
Find the accelerations:
012122 a
dt
dv a
dt
dv
dt
dv
dt
dv
dt
dva 12
121
1
a a 21
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 velocityInertial frames can not be rotating or accelerating relative to one another or to the fixed stars. Non-inertial frames fictitious forces.
x2
y2
x1
y1
x12
x2
x1
P
v12
Copyright R. Janow – Spring 2012
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 InertiaApply 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:
1
2
2
1a
a
m
m A mass and resulting acceleration on it
are inversely proportional
The same “inertial mass” value also measures “gravitational mass” –- a particle’s effect in producing gravitational pull on other masses
Mass is a scalar: 2121 m mm m to m withbehaves result The Attach
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):
mgW N. 3.3 Wbut N. 20 Wthen kg m If moonearth 2
Copyright R. Janow – Spring 2012
Newton’s Second Law
am F F i
inet
Summarizing:
Vector sum of forces acting ON a particle
Inertia of particle
Acceleration resulting from Fnet
SIMPLE PROPORTIONALITY WHEN IN AN INERTIAL FRAME
Cartesian component equations: ma F F x
iixx,net ma F F y
iiyy,net ma F F z
iizz,net
Other ways toWrite 2nd Law:
dt
rdm F F
2
iinet 2
vmp
dt
pd F wherenet
DIRECTION OF ACCELERATION AND NET FORCE ARE THE SAME
Units for Force: 2T/L M ]a[ ]m[ ]F[
SYSTEM FORCE MASS ACCELERATION SI Newton (N) Kg m/s2
CGS Dyne gm cm/s2
British Pound (lb) slug ft/s2
1 dyne = 10-5 N 1 gm = 10-3 kg 1 lb = 4.45 kg 1 slug = 14. 59 kg
1 kgWEIGHS
2.2 lb
Copyright R. Janow – Spring 2012
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) = 0, 3 = 0b) = 180, 3 = -180 c) = 60, 3 = -60 d) = 120, 3 = -120 e) = 150, 3 = -150
Tug of war
x1F
2F
3F 3
2
4-2: What should the other two angles be to maximize the magnitude of the ring’s acceleration?
a) = 0, 3 = 0b) = 180, 3 = -180 c) = 60, 3 = -60 d) = 120, 3 = -120 e) = 150, 3 = -150
Copyright R. Janow – Spring 2012
Example: A hockey puck whose mass is 0.30 kg issliding on a frictionless ice surface. Twoforces act horizontally on it as shown in thesketch. Find the magnitude and direction of the puck’s acceleration.Apply 2nd Law to x and y directions
)cos(F )cos(F ma F F F 1x2x1xx,net 6020 2
)sin(F )sin(F ma F F F 1y2y1yy,net 6020 2
N 8.7 F x,net
N 5.2 F y,net
229 s/m 0.3
8.7
m
F a xnet,
x
2s/m 17 0.3
5.2
m
F a ynet,
y
Evaluate:
FBD
m = 0.3 kg
Convert to polar coordinates:o
xy-12/2
y2x )a/a( tan m/s 34 ]a [a a 3121
The net force and acceleration vectors have the same directionA unit vector in that direction is:
j 0.50 i 0.85 j34
17 i
34
29 j
|a|
a i
|a|
a
|a|
a a yx
Copyright R. Janow – Spring 2012
Measuring Force and Mass
On frictionless surface (e.g., air track): Apply enough horizontalforce F0 to give the standard mass m0 the standard acceleration a0
m0 = 1 kg a0 = 1 m/s2
00 am F 0The (standard) force unit thereby defined = 1N.
Note: x-components only above, F & a are in same direction
What about forces in the y – direction, left out above?
Measuring another mass:• Apply standard force, record resulting acceleration
m/ F a 0 a/ F m 0
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).
jmg Fg g = 9.8 m/s2 = 32.2 ft/s2 at the Earth’s surface
• “Action at a distance”• The weight is independent of how a mass is moving (perhaps other forces also act).
weight"the" F g
Mass in contact with a “horizontal” surface (table, air track, …) may be in equilibrium for y: a 0 F yfor m"Equilibriu" yy
FBD
m
jmg Fg
N
ay = 0 N = Fg (m does not accelerate) surface to larperpendicu force" normal" N
N is a contact force that adjusts to Fg
Fg pushes on the surface – - the surface pushes back with NIf ay not = 0, N does not equal Fg
Copyright R. Janow – Spring 2012
Newton’s Third LawBodies interact by pushing or pulling on each other3rd Law (antique version): Each action has an equal and opposite reaction
More modern version: When two bodies interact the forces that eachexerts on the other are always equal in magnitude and opposite in direction
F F 2112
Example: gravity acting at a distance
2 object to due 1 object on force F 12
1 object to due 2 object on force F 21
If you ever find a force w/o the 3rd law reaction, you can build a perpetual motion
machine
Example: box on a level surface
• 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 ???
m
jmg Fg
N
Fe
Fs
Copyright R. Janow – Spring 2012
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
Fg
N
Fg
m
NM
jMg Fg
N
m
jmg F'g
Is this a FBD?
Copyright R. Janow – Spring 2012
FrictionFrictionless
m
gF
NF
Sliding or static friction f (later)
m
gF
NF
f
No friction parallel to surface Friction is a resistive force parallel to surface
m/Fax m/)fF(ax Nf
Contact force - always opposed to motion
CordsTension only, no compression Pulling creates tension T – the force transmitted by the cordT is the same everywhere in a zero-mass, unstretchable cord
3rd law pairT
m
Support FBD for body
mT
Fg
FBD for cord
T
T’ = T
3rd law pair
FBD for support
T’
F
Equilibrium F'TTFg
FBDs: show only forces on bodies
Copyright R. Janow – Spring 2012
Newton’s Laws - Summary
Newton’s First Law
A body’s velocity is constant ( ) if the net external force on it is zero0a
- Motion is along a straight line- Find and use an inertial frame of reference
Newton’s Second Law
am vector force net F Fi
inet
In Cartesian components:
zzyyxx ma F ma F ma F
Newton’s Third Law
If body A exerts a foce on body B, then body B exerts and equal andOpposite force on body A.
Copyright R. Janow – Spring 2012
Method for solving Newton’s Second Law problems
Systems with several components may have several unknowns….
…and need an equal number of independent equations
• 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
• 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?)
amFF inet
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Example: Traffic Light in Equilibrium
Conceptualize:cables are massless and don’t breakno motion
Categorize: equilibrium problem – accelerations = 0Model as particles in equilibrium
2 FBDs
T3
Fg
FBD of LightFBD of Knot
Fg = 122 N
Solve upper for T2:
FBD of light yields T3 = Fg ( alight = 0) FBD of knot yields:
)cos(T)cos(T0 F oo2x 3753 1
31 3753 T)sin(T)sin(T0 F oo2y
1153
37 331 T.TT)ocos(
)ocos(2
N . )]sin( .)[sin( T oo 22153331371
N . T 4731 N . T 4972
Copyright R. Janow – Spring 2012
Example: Particle Motion Under a Net Force
mN
F
W
x
yMass m is moving on a frictionless horizontal surface, acted on by an external force of magnitude F, making an angle with the x-axis.
Find expressions for the accelerationalong the horizontal and vertical directions
W = weightAlong x: xx ma )cos(F F m
)cos(F a x
Set ay = 0. Crossover to ay > 0 when N also = 0, i.e, when
Along y: Is ay zero, or does particle accelerate upward?
yy ma W N )sin(F F
W N )sin(F 0 W )Fsin(
When Fy is > the weight, the particle accelerates upward
m
W )sin(F a y
N 0
When Fy is < the weight, the particle does not accelerate
a y 0 )sin(F W N
Copyright R. Janow – Spring 2012
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 Nb) F = 1000 Nc) F = 550 N d) F = 500 N e) He cannot lift himself by his own bootstraps at all.
Bootstraps
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.
W’ = mg
MgW
NT
T’
No friction,No mass
W = MgW = Mg
N T
FBD for Block S
a
Ma T F xs 0 ysys Ma W N F
0 xhxh ma F
'ma T' W'F yh
W’ = mgW’ = mg
T’
FBD for Block H
a’
choosepositive down
Apply 2nd Law to blocks S & H separately for x & y
Constraints: T’ = T, a’ = aEliminate T,T’,a’ma Ma W'
Ma ma mg
g Mm
m a
Find formula for T
gMm
Mm Ma T
could also usesystem approachTo 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 with the horizontal.
Find expressions for the accelerationand the normal force
m Na
W = mg
Why does the block accelerate? Do you expect N to equal W?Forces acting ON the block are N and W
• N is normal to the surface• W is vertical as usualChoose x-y axes aligned to ramp, for which:
y
x
N
W
Wy
Wx
right) & down (positive xyAssume a a ,a 0FBD
Apply 2nd Law to x and y
) Wsin( W)cos(WW xy
)sin(mgma W F xxx
0 )mgcos(N WN F yy )mgcos( N
)gsin( a a x
Does not depend on m
Check:What happens as 90o
as 0o
Copyright R. Janow – Spring 2012
Example: “Atwood’s Machine” with Massless Pulley
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 tensionIn the cord. Find the force Wtot supporting the pulley
a
a
Wtot
No forces or motion along x
am gmT F 1y 11
T
m1g
a
FBD for m1
T
m2g
a
FBD for m2
am T gm F 2y 22
Add the equationsam am T gm gm T 21 12 g
m m
mm a
2
2
1
1
a = 0 for m1 = m2 a is clockwise for m2 > m1 a = g for m1 = 0 or a = - g for m2 = 0
Subtract the equationsam am T gm gm T 21 12 g
m m
m2m T
2
1
1
2
T = 0 for m1 or m2 equals zero. T = m1g for m2 = m1
Wtot
T T
0 2T Wtot FBD for pulley
g m m
m4m W
2
1tot
1
2
Wtot = 2m1g if m2 = m1
Otherwise not so
Copyright R. Janow – Spring 2012
Example: Your Weight in an Elevator
A passenger whose mass m = 72.2 kg is standing on a platform scale in an elevator.What weight does the scale read for him as the elevator (and himself) accelerates up and down?
No forces or motion along x
N
mg F g
FBD for passenger
+
• 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
am mgN F y )ga( m N
Interpretations:N = mg for a = 0 Normal weight for elevator at rest or moving with constant v
N > mg for a > 0 Increased apparent weight for elevator accelerating up
N < mg for a < 0 Decreased apparent weight for elevator accelerating down
mg = force exerted by gravity
N = 0 for a = -g Free fall - weightless
Fictitious force appears in non-inertial frame
velocity has no effect on N – apparent weight