Copyright R. Janow – Spring 2012 Physics 111 Lecture 04 Force and Motion I: The Laws of Motion SJ 8th Ed.: Ch. 5.1 – 5.7 Dynamics - Some history Force

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


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).


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


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


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


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).


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


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


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


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


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


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


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


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?


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


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.


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


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


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


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


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


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


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


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

Documents

Copyright R. Janow – Spring 2012 Physics 111 Lecture 04 Force and Motion I: The Laws of Motion SJ 8th Ed.: Ch. 5.1 – 5.7 Dynamics - Some history Force