Lecture 07, Page 1 Physics 2211 Spring 2005 Dr. Bill Holm Physics 2211: Lecture 07 l A Gallery of Forces l Newton’s 2nd Law of Motion l Newton’s 1st Law

Lecture 07, Page 1 Physics 2211 Spring 2005Dr. Bill Holm

Physics 2211: Lecture 07Physics 2211: Lecture 07

A Gallery of Forces Newton’s 2nd Law of Motion Newton’s 1st Law of Motion


ForceForce

A force is a push or a pull.

A force is caused by an agent

and acts on an object. More

precisely, the object and agent

INTERACT.

A force is a vector.

1F��

agent

object


Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame.

.netF F ma

Law 2: For any object,

Review: Newton's LawsReview: Newton's Laws

Law 3: Forces occur in action-reaction pairs: where is the force due to object a acting on object b.

ab baF F

abF


WeightWeight

The Earth is the agent (on this planet)The Earth is the agent (on this planet)

long-range forcelong-range force

no contact needed!no contact needed!

W

The “weight force” pulls the box down (toward the center of the Earth)


GravityGravity

Typical student mass m = 55kgg = 9.8 m/s2.Fg = mg = (55 kg)x(9.8 m/s2 )

Fg = 539 N (weight)

What is the force of gravity exerted by the earth on a typical physics student?

ES gF F mg

SEF mg


ExampleExampleMass vs. WeightMass vs. Weight

An astronaut on Earth kicks a bowling ball and hurts his foot. A year later, the same astronaut kicks a bowling ball on the moon with the same force. His foot hurts...

(1) more (2) less (3) the same

Ouch!

The masses of both the bowling ball and the astronaut remain the same, so his foot will feel the same resistance and hurt the same as before.


Wow!

That’s light. However the weights of the

bowling ball and the astronaut are less:

W = mgMoon gMoon < gEarth

ExampleExampleMass vs. WeightMass vs. Weight

Thus it would be easier for the astronaut to pick up the bowling ball on the Moon than on the Earth.


The Spring ForceThe Spring Force

A compressed spring pushes on the object.

A stretched spring pullson the object.

The spring is the agentThe spring is the agent


SpringsSprings

Hooke’s Law: Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position (linear restoring force).

FX = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality.

relaxed position

FX = 0

xx = 0


SpringsSprings

relaxed position

FX = -kx > 0

xx 0




SpringsSprings

FX = - kx < 0

xx > 0

relaxed position




TensionTension

The rope is the agentThe rope is the agent

The rope force (tension) is a pull along the rope, away from the object


Tools: Ropes & StringsTools: Ropes & Strings

Ropes & strings can be used to pull from a distance. TensionTension (T) at a certain position in a rope is the magnitude of the

force acting across a cross-section of the rope at that position.The force you would feel if you cut the rope and grabbed the ends.An action-reaction pair.

cut

T

T

T



Consider a horizontal segment of rope having mass m:Draw a free-body diagram (ignore gravity).

T1 T2

m

a x x

Using Newton’s 2nd law (in xx direction): FNET = T2 - T1 = ma

So if m = 0 (i.e., the rope is light) then T1 =T2



An ideal (massless) rope has constant tension along the rope.

T = Tg

T = 0

T T

If a rope has mass, the tension can vary along the rope For example, a heavy rope

hanging from the ceiling...

We will deal mostly with ideal massless ropes.



The direction of the force provided by a rope is along the direction of the rope:

mg

T

m

Since ay = 0 (box not moving),

T = mg


Component perpendicularComponent perpendicular((normalnormal) to surface) to surface

NORMAL FORCE—the surfacepushes outward against the object.

n��

Component parallelComponent parallelto surfaceto surface

FRICTION FORCE pulls in a direction to (try to) prevent slipping of the object

f��

(one force — two components)(one force — two components)The Surface Contact ForceThe Surface Contact Force


Surface Contact ForceSurface Contact Force(microscopic view)(microscopic view)

The NORMAL force is caused by the molecular-scale repulsion between the object and the surface.

FRICTION is caused by the making and breaking of molecular bonds between the two surfaces.


There are two types of FrictionThere are two types of Friction

Static FrictionStatic Friction

FRICTION successfully prevents

slipping of the object along

surface

Kinetic FrictionKinetic Friction

FRICTION doesn’t prevent slipping of

object along surface.

sf��

kf��

v


Rolling frictionRolling friction

| | | |r rf N��

coefficient of rolling friction



21ˆ

4D Av v��

Unit vector that points in the direction of the velocity of the body

A force experienced by a body that moves through air.

Drag

Experiment shows that to a good approximation,

2Cross sectional of the body measured in marea

2 2Square of the body's measured in m /s .velocity


Drag ForceDrag Force

The fluid is the agent.The fluid is the agent.

Skin DragSkin Drag (like friction) (like friction)

Form DragForm Drag (rowboat) (rowboat)

DRAG occurs when an objectmoves in a gas or liquid. Likefriction, the force of dragalways points opposite to thedirection of motion

Drag is very small for most of the Drag is very small for most of the

objects we will discuss. We will alwaysobjects we will discuss. We will always

neglect dragneglect drag in our models unless we in our models unless we

explicitly state otherwise.explicitly state otherwise.


Throw a ball upward verticallyThrow a ball upward vertically

Drag adds to theweight as it rises

Drag decreases asthe ball slows down

Drag increasesas the ball speedsup slows

Drag opposes the weight as it falls


4T

mgv

A

Terminal velocityTerminal velocity

At the terminal velocity, the drag force balances the force of gravity.

0yF D w 214 0TAv mg


Slope approaches zero asv approaches terminal velocity

Without drag, v = - at

As the speed (and thus drag) increases,the slope decreases

Drop a particle from rest (v = 0)

Tv


The dimensions of a 1500 kg car, as seen from the front, are 1.6 m wide by 1.4m high.

At what speed does the drag force equal the force of rolling friction?


Linear Restoring Force:

Friction Force:

Fluid Force:

Empirical (Contact) ForcesEmpirical (Contact) Forces(examples)(examples)

F N

F kx

nF bv

Characterized by variable, experimentallydetermined “constants”: k, , b, etc.

These forces are all electromagnetic in origin.

; s s k kF N F N

( , (1 small); 2 large)n v n v


ThrustThrust

THRUST is a contact force exerted on rockets and jets (and leaky balloons) by exhaust gases (the agent).


Electric and Magnetic Forces Electric and Magnetic Forces (Non-Contact)(Non-Contact)

Magnetic Field

q

-q

E��

E��

E��

E��

E��

E��

E��

E��

Electric Field


Identifying Forces---The SkierIdentifying Forces---The Skier

1. Identify SYSTEM

2. Draw PICTURE with a closed curve around the SYSTEM. (Everything

outside the curve is theenvironment.

3. Find contact points between SYSTEM & ENVIRONMENT—Name and label the CONTACT FORCES

4. Name and labelLONG-RANGE FORCES

Tension T��

Weight W��

Normal force N

Friction force f��


Newton’s 2nd Law of Motion

Newton’s 1st Law of Motion


netnet

Fa F ma

m

Newton’s 2nd Law

the connection between force and motion

The acceleration of an object is proportional to the net

force that acts upon it.net

a

F

��

The constant of proportionality is called the "mass".

The unit of mass is the kilogram (kg).m

Mass is an intrinsic property of matter. 2The unit of force is [F] = [m][a] = kg-m s = N (Newton)


Newton’s First LawNewton’s First Law

An object moves with if and only if

the total force acting on the object i .

s net

v

F

constant velocity

zero

��

dyn am 0ics: neta F m a ��

0 0kinematics: v v at v v ��

0is just a special case wheret s 0 Re .v ��


Mechanical Equilibrium: ( 0) netF

Static Equilibrium(object at rest)

Dynamic Equilibrium(object moving with constant velocity)

An object moving in a straight line at constantvelocity is in dynamic equilibrium

N

W


Problem: AccelerometerProblem: Accelerometer

A weight of mass m is hung from the ceiling of a car with a massless string. The car travels on a horizontal road, and has an acceleration a in the x direction. The string makes an angle with respect to the vertical (y) axis. Solve for in terms of a and g.

i

a


AccelerometerAccelerometer

Draw a free body diagram (FBD) for the mass: What are all of the forces acting?

(gravitational force)mg

m

x

y

a

(string tension) T



Resolve forces into componentsResolve forces into components:

Eliminate T :

T sin = ma

T cos = mgtan

a

g=>

:x Fx = Tx = T sin = ma

:y Fy = Ty - mg = T cos - mg = 0

Sum forces in each dimension separatelySum forces in each dimension separately:

ˆ

yT j

ˆxT i

x

y

a

T

ˆ mg mg j



Say the car goes from 0 to 60 mph uniformly in 10 seconds: 60 mph = (60 x 0.45) m/s = 27 m/s.Acceleration a = Δv/Δt = 2.7 m/s2. So a/g = 2.7 / 9.8 = 0.28 .

= arctan (a/g) = 15.6 deg

a

tan a

g Let’s put in some numbers:


Problem: Inclined planeProblem: Inclined plane

A block of mass m slides down a frictionless ramp that makes angle with respect to the horizontal. What is its acceleration a ?

m a


Inclined planeInclined plane

Define convenient axes parallel and perpendicular to plane: Acceleration a is in x direction only.

a

x

ym


a

x

y

Resolve forces into components & sum forces in x and y directions separately:

Inclined planeInclined plane

N - mg cos = may= 0

:yF N = mg cos

a = g sin

Assume forcesare acting at “center of mass”of block.

Draw a FBD.

mg

N

mg sin =max = ma

:xF

mg sin i

- mg cos j


Angles of an Inclined planeAngles of an Inclined plane

Lines are perpendicular, so the angles are the same!

- mg cos jmg


VocabularyVocabulary

A Reference FrameReference Frame is the (x,y,z) coordinate system you choose for making measurements.

An Inertial Reference Frame is a frame where Newton’s Laws

are valid.

USE NEWTON’S LAWS TO TEST

FOR INERTIAL REFERENCE FRAMES


Test for Inertial Reference FrameTest for Inertial Reference Frame

The ball's motion doesn't change (no acObservatio celern : ation)

The plane is an inertial refereConclusi nce fron : ame.

EXAMPLE: Airplane parked on runway A ball is placed on the floor of the plane; no net forces act on the ball.

Less obvious: a plane cruising at constant velocity is also an inertial reference frame!


Test for Inertial Reference FrameTest for Inertial Reference Frame

EXAMPLE: Airplane taking off. Ball placed on floor of plane; no net forces act on ball.

.

The ball rolls back (it Ob acservatio celeran: tes)

Newton's 1st Law is violated.

The accelerating plane is NOT an ine

Conc

rtia

lusion :

l fr

ame.


We live in a World of Approximations . . .We live in a World of Approximations . . .

Strictly speaking, an Inertial Reference Frame has zero acceleration with respect to the “distant stars”.

The Earth accelerates a little (compared to the distant stars) due to its daily rotation and its yearly revolutionaround the Sun. Nevertheless, to a good approximation, the Earth is an Inertial Reference Frame.


T = 1 day = 8.64 x 104 sec, R ~ RE = 6.4 x 106 meters .

Plug this in: aatl = 0.034 m/s2 ( ~ 1/300 g) Close enough to zero that we will ignore it. Atlanta is a pretty good IRF.

What is the acceleration (centripetal) of Atlanta?

Is Atlanta a good IRF?Is Atlanta a good IRF?

Is Atlanta accelerating? YES!

22 2 1atl

v Ra

R T R

Atlanta is on the Earth.The Earth is rotating.



2-D, 3-D Kinematics and Projectile Motion

Independence of x and y components

*Georgia Tech track and field example*Football example*Shoot the monkey


3-D Kinematics3-D Kinematics

The position, velocity, and acceleration of a particle in 3 dimensions can be expressed as:

adv

dt

d x

dt

2

2v

dx

dtx x(t )

ˆˆ ˆ

x y zv v i v j v k

We have already seen the 1-D kinematics equations:

ˆˆ ˆ

x y za a i a j a k

ˆˆ ˆ r xi yj zk


3-D Kinematics 3-D Kinematics

For 3-D, we simply apply the 1-D equations to each of the component equations.

ad x

dtx

2

2

vdx

dtx

x x(t )

ad y

dty

2

2

vdy

dty

y y t ( )

ad z

dtz

2

2

vdz

dtz

z z t ( )

2

2

dvdr

vdt

r rd

atr

dt dt

Which can be combined into the vector equations:



So for constant acceleration we can integrate to get:

0

210 0 2

constant

a

v v at

r r v t at

Aside: the “4th” kinematics equation can be written as

2 2

0 2 v v a r

(more on this later)



Most 3-D problems can be reduced to 2-D problems when acceleration is constant:Choose y axis to be along direction of accelerationChoose x axis to be along the “other” direction of

motion

ExampleExample: Throwing a baseball (neglecting air resistance)Acceleration is constant (gravity)Choose y axis up: ay = -gChoose x axis along the ground in the direction of the

throw


Uniform Circular MotionUniform Circular Motion

What does it mean?

How do we describe it?

What can we learn about it?


What is Uniform Circular Motion?What is Uniform Circular Motion?

Motion with

Constant Radius R

(Circular)

Constant Speed

(Uniform)

is constantv

R

x

y

(x,y)

v


How can we describeHow can we describe Uniform Circular Motion? Uniform Circular Motion?

In general, one coordinate system is as good as any other: Cartesian:

» (x,y) [position]

» (vx ,vy) [velocity] Polar:

» (R,) [position]

» (vR ,) [velocity]

In uniform circular motion: R is constant (hence vR = 0). (angular velocity) is constant. Polar coordinates are a natural way to describe Polar coordinates are a natural way to describe

Uniform Circular Motion!Uniform Circular Motion!

R

x

y

(x,y)

v


Polar CoordinatesPolar Coordinates

R

x

y

(x,y)

2 3/2 2

-1

1

0

cossin

Conversion from polar toCartesian coordinates:

cos

sin

x

y

R

R

2 2

arctan

R x y

y x

And back:


R

x

y

(x,y)

v

s

Angular MotionAngular Motion

The arc length s (distance along the circumference) is related to the angle in a simple way:

s = R, where is the angular displacement. units of are called radians.

For one complete revolution:

2R = Rc

c = 2

has a period 2.

1 revolution = 2radians


In Cartesian coordinates, we say velocity vx = dx/dt.

x = vxt (vx constant) In polar coordinates, angular velocity d/dt = .

= t ( constant) has units of radians/second.

Displacement s = v t.

but s = R = Rt, so:

v = R x

t R

x

y

(x,y)

v

s

Angular MotionAngular Motion


Aside: Period and FrequencyAside: Period and Frequency

Recall that 1 revolution = 2 radians

(a) frequency ( f ) = revolutions / second

(b) angular velocity ( ) = radians / second By combining (a) and (b)

(rad/s) = [2 rad/rev] x f (rev/s)= 2f

Realize that:period (T) = seconds / revolutionSo T = 1 / f = 2/

= 2 / T = 2f

x

R

x

y

(x,y)

v

s


SummarySummary

xt

R

x

y

(x,y)

v

s

cos

sin

x

y

R

R

2 2

arctan

R x y

y x

d

dtt

Relationship between Cartesian and Polar coordinates:

d

dtv

s

s

vs t

s R

v R

AngularLinear Links

displacement:

velocity:

constant velocity:

Angular motion:


Polar Unit VectorsPolar Unit Vectors

We are familiar with the Cartesian unit vectors:

Now introduce “polar unit-vectors” and : points in radial directionpoints in tangential direction (counterclockwise)

ˆ ˆ ˆ, , i j k

r r

R

x

y

j

i

r


Acceleration in Uniform Circular MotionAcceleration in Uniform Circular Motion

Even though the speed is constant, velocity is not constant since the direction is changing: acceleration is not zero!

v

2v 1v

R

2v

1v

t

Notice that (hence )

points at the origin!

v

/v t

In the limit 0 ,t /a dv dt

and points in the

direction.

a

r

/ava v t Consider average acceleration in time t


This is called This is called Centripetal Acceleration. Now let’s calculate the magnitude:

vv

RR

Similar triangles:

But R = v t for small t

So:vt

vR

2 v

vv tR

Acceleration in Uniform Circular MotionAcceleration in Uniform Circular Motion

v

2v 1v

RR

2v

1v

R2v

aR


Centripetal AccelerationCentripetal Acceleration

Uniform Circular Motion results in acceleration:Magnitude: a = v2 / RDirection: - r (toward center of circle)^

R

a


Centripetal AccerationCentripetal Acceration(in terms of (in terms of ))

We know that and v = RavR

2

2Ra

R

Substituting for v we find that:

a = 2R


Example:Example:Uniform Circular MotionUniform Circular Motion

A fighter pilot flying in a circular turn will pass out if the centripetal acceleration he experiences is more than about 9 times the acceleration of gravity g. If his F18 is moving with a speed of 300 m/s, what is the approximate diameter of the tightest turn this pilot can make and survive to tell about it ?

(1) 500 m

(2) 1000 m

(3) 2000 m


avR

g 2

9

2km

Example:Example:SolutionSolution

.

2

2

2

m2s

ms

90000

9 9 9 81

vR

g

m1000m819

10000R

.2 2000D R m


Example: Propeller TipExample: Propeller Tip

The propeller on a stunt plane spins with frequency f = 3500 rpm. The length of each propeller blade is L = 80 cm. What centripetal acceleration does a point at the tip of a propeller blade feel?

f

L

what is a here?


ExampleExample

so 3500 rpm means = 367 s-1

Now calculate the acceleration. a = 2R = (367s-1)2 x (0.8m) = 1.1 x 105 m/s2 = 11,000 g

direction of acceleration points towards the propeller hub.

-11 min

1 1 2 0.105 0.105 min 60

rev rad radrpm x x s

s rev s

First calculate the angular velocity of the propeller:


Example: Newton & the MoonExample: Newton & the Moon

What is the acceleration of the Moon due to its motion around the Earth?

What we know (Newton knew this also):T = 27.3 days = 2.36 x 106 s (period ~ 1 month)R = 3.84 x 108 m (distance to

moon)RE = 6.35 x 106 m (radius of earth)

R RE


MoonMoon

So = 2.66 x 10-6 s-1.

Now calculate the acceleration. a = 2R = 0.00272 m/s2 = 0.000278 gdirection of acceleration points is towards the

center of the Earth.

6 -11 1 2 2.66 10

27.3 86400

rev day radx x x s

day s rev

Calculate angular velocity:


So we find that amoon / g = 0.000278 Newton noticed that RE

2 / R2 = 0.000273

This inspired him to propose that FMm 1 / R2

MoonMoon

R RE

amoong


The Space Shuttle is in Low Earth Orbit (LEO) about 300 km above the surface. The period of the orbit is about 91 min. What is the acceleration of an astronaut in the Shuttle in the reference frame of the Earth? (The radius of the Earth is 6.4 x 106 m.)

(1) 0 m/s2

(2) 8.9 m/s2

(3) 9.8 m/s2

ExampleExampleCentripetal AccelerationCentripetal Acceleration


First calculate the angular frequency :

RO = RE + 300 km = 6.4 x 106 m + 0.3 x 106 m = 6.7 x 106 m

RO

300 km

RE

-11 1 min 2 0.00115

91min 60

rev radx x s

s rev

Realize that:



Now calculate the acceleration:

a = 2R

a = (0.00115 s-1)2 (6.7 x 106 m)

a = 8.9 m/s2


RO

300 km

RE


Circular Orbits

Fictitious forces



SEEM to be very different,BUT they have the same free body diagram . . .


Orbital Motion is Projectile Motion!


( )w mg downward

Flat earth approximation

( )w mg center

Spherical Earth, near Earth

( )netFa g center

m

Orbiting projectile is in free fallOrbiting projectile is in free fall

h

2orbit

r

va g

r

orbit Ev rg R g


Fictitious forcesFictitious forces in in Non-InertialNon-Inertial Frames of Reference Frames of Reference

A car and driver move a constant speed. Suddenly, the driver brakes

Outside observer: if the seat is frictionless,the driver continues forward at constantspeed and collides with the front window.

Inside observer: a force F “throws” the driver against the front window. F

F is fictitious; the observer is in a non-inertial (accelerated) frame whereNewton’s laws do not apply. There is notrue “push” or “pull” from anything.


Fictitious forcesFictitious forces in in Non-InertialNon-Inertial Frames of Reference Frames of Reference

car

frictionless bench seat

bookv

path of book(Newton’s 1st Law)

● Car turns● Book continues on straight line● In driver’s reference frame, apparent (fictitious) force moves book across the bench seat


A car is passing over the top of a hill at (non-zero)speed v. At this instant,

1 n > w2 n = w3 n < w4 Can’t tell without knowing v


1

23

4A ball on a string swings in a vertical circle. The

string breaks when the string is horizontal and

the ball is moving straight up. Which trajectory

does the ball follow thereafter?


Roller-CoasterRoller-Coaster


2

r r

mvF n w ma

r

Bottom:

2

app

mvn w

rw

Top:

2

r r

mvF n w ma

r

2

app

mvn w

rw

Lose contact : critical0 rw

n v rgm



Full Disclosure

mg

mg

N

top bottom and force is

purely radial (centripetal).

v v

NN

mg

Except at the top and bottom,

0 since there is a non-zero

component of the weight in the

tangential direction. Since is

entirely tangential, the speed

must speed up or slow down

on the sidewalls

Ta

v

v

mg

is not purely

radial or purely tangential

acceleration

aa

N


Example Example


Example Example

Treat horizontal motion and vertical motion separately

Then just add the results: Principle of Superposition

Horizontal Motion: 0xx

dva

dt constantxv

0 10 m/sx xv v in +x direction

Velocity at highest point:


Example Example

Vertical Motion: ya g

20 0 0

20x yv vv v

0y yv v gt 21

0 0 2yy y v t gt

Need to determine magnitude of take-off velocity:

210 20 h hyh t tv g

00 y hv tg 210 20 oy oyv v

y g gh v g

?0yv

0h yt v g

210 0 2yy y v t gt

20 2yv gh

OR, use

.2 20 2y yv v g y

0y yv v gt

20 2yv gh, 0yy h v When

max , 0yy y h v We know when


Example Example

2 20 0 0 0x yv v v v

Magnitude of take-off velocity:

20 2xv gh

. .2

2m ms s

10 2 9 81 0 25 m

. ms10 2


Example Example

Two footballs are thrown from the same point on a flat field. Both are thrown at an angle of 30o above the horizontal. Ball 2 has twice the initial speed of ball 1. If ball 1 is caught a distance D1 from the thrower, how far away from the thrower D2 will the receiver of ball 2 be when he catches it?

(1) D2 = 2D1 (2) D2 = 4D1 (3) D2 = 8D1


The horizontal distance a ball will go is simply x = (horizontal speed) x (time in air) = v0x t

210 0 2yy y v t gt

To figure out “time in air”, consider the equation for the height of the ball:

When the ball is caught, y = y0

(time of throw)

(time of catch)

Example Example

210 2 0yv t gt

10 2 0yv gt t

two solutions

02

0

yvt

g

t


Ball 2 will go 44 times as far as ball 1!

So the time spent in the air is:

Example Example

The range, R, is thus : 0R x RR x t v t

02R yt v g

Notice: For maximum range, sin 45 2 1 for max. rangeo

v0

sin0oyv v

cos0oxv v

0 02 x yv v

g sin cos2

02v

g

sin2

0 2v

g


Shooting the MonkeyShooting the Monkey(tranquilizer gun)(tranquilizer gun)

Where does the zookeeper aim if he wants to hit the monkey?

( He knows the monkey willlet go as soon as he shoots ! )


Shooting the MonkeyShooting the Monkey

If there were no gravity, simply aim

at the monkey0

r r

0r v t



With gravity, still aim at the monkey!

Dart hits the monkey!0

21

2

v t gtr

021

2

r gtr


x = xx = x00

yy = -1/2 gg t2

This may be easier to think about.

It’s exactly the same idea!!

xx = = vv0 0 tt

yy = -1/2 gg t2


Documents

Lecture 07, Page 1 Physics 2211 Spring 2005 Dr. Bill Holm Physics 2211: Lecture 07 l A Gallery of Forces l Newton’s 2nd Law of Motion l Newton’s 1st Law