module 6 dynamics

8/9/2019 Module 6 Dynamics

1/116

2005 Pearson Education South Asia Pte Ltd

MECHANICS

FE1001 Physics I NTU - College of Engineering

1. Units, Physical Quantities

and Vectors

2. Motion Along A Straight Line

3. Motion in 2 or 3 Dimensions

4. Newtons Law ofMotion5.Applying Newtons Laws

6. Work and Kinetic Energy7. Potential Energy and Energy Conservation


2/116


3/116

4. Newtons LawsofMotion


ChapterObjectives

Use kinematic quantities displacement, velocity,and acceleration along with two new concepts,

force and mass, to analyze the principles of

dynamics.

Understand Newtons laws of motions

Apply Newtons First Law

Apply Newtons Second Law

Apply Newtons Third Law


4/116



ChapterOutline

1. Force and Interactions2. Newtons First Law

3. Newtons Second Law

4. M

ass andW

eight5. Newtons Third Law

6. Free-Body Diagrams

7. Concept Summary and Key Equations


5/116



4.1 Force andInteractions

For daily usage, a force is either a push or a pull. When the force involves direct contact between two

bodies, we call it a contact force. (e.g.: a rope

pulling on a block, friction that ground exerts on a

baseball player sliding home)

Long-range forces act even when the bodies are

separated by empty space. (one example is gravity)

The force of gravitational attraction that the earthexerts on your body is called your weight.


6/116




Force is a vector quantity, thus we need to describethe direction in which it acts as well as its

magnitude.

The SI unit of the magnitude of force is the newton,

abbreviated as N.


7/116




A common instrument for measuring force is thespring balance.

When two forces act at the same time at

pt A of a body, experiment shows that the effect on

the bodys motion is the same as the effect of asingle force equal to the vector sum of the

original forces:

This principle is called

the superposition of forces.

21 and FFTT

RT

21 FFRTTT

!


8/116




This principle allows us to replace a force by itscomponent vectors. This is shown in the figure.

Note that any force can be replaced by its

component vectors, acting at the same pt.


9/116




In figure below, a wriggly line is drawn throughthe force vector to show that weve replaced

it by its x- and y-components.

FT


10/116




This vector sum (resultant) of all the forces actingon a body is the NET FORCE acting on the body.

Where is read as the vector sum of theforces or net force.

The pair of component equations for Eqn 4.1 is

Where is the sum of the x-components, andso on.

!! 1.4321 FFFFRT

.TTTT

!! 2.4yyxx FRFR

FT

xF


11/116




OnceRx andRy is found, we can find themagnitude and direction of net force acting on the

body.

The magnitude is

And the angle between and the +x-axis can be

found from the relation: tan U =Ry/Rx.

The components may be +ve, -ve, or zero, and the

angle in any of the four quadrants.

In 3-D problems, we add the z-component to Eqn

4.2, and magnitude of net force is

22

yx RRR !RT

222

zyx RRRR !


12/116



Example 4.1 SuperpositionofForces

Three professional wrestlers are fighting over thesame champion's belt.

As viewed from above, they apply three horizontal

forces to the belt that are shown on the next slide,

where the belt is located at the origin.

The magnitudes of the forces areF1 = 250 N,

F2= 50 N, andF3 =120 N.

Find the x- and y-components of the net force on thebelt, and find the magnitude and direction of the net

force.


13/116



Example 4.1 SuperpositionofForces


14/116



Example 4.1 SOLN

Solution:Identify and set up:

This is just a problem in vector addition.We are

asked to find the components of the net force, so well

attack this problem using the component method.Our

target variables are the magnitude and direction of

the net force as well as its x- and y-components;

we'll find these using Eq. (4.2).

RT


15/116



Example 4.1 SOLN

Execute:From Fig. 4.5a, angles between the forces

and the +x-axis are U1 =18053 =127,U2 =0, and U3 = 270. The x- and y-components of the

3 forces are

321 and FF,FTTT

N120270sinN120

N0270cosN120

N00sinN50

N500cosN50

N200127sinN250

N150127cosN250

3

3

2

2

1

1

!r!

!r!

!r!

!r!

!r!

!r!

y

x

y

x

y

x

F

F

F

F

F

F


16/116



Example 4.1 SOLN

Execute:From Eq. (4.2) the net force has

components

The net force has a negative x-componentand a

positive y-component, so it pts to the left and toward

the top of the page in Fig. 4.5b (i.e., in the 2ndquadrant). The magnitude of the net force is

! FRTT

N80N120N0N200

N100N0N50N150

321

321

!!!

!!!

yyyy

xxxx

FFFR

FFFR

! FR TT

N128N80N100 2222 !!! yx RRR


17/116



Example 4.1 SOLN

Execute:To find the angle between the net force and the +x-

axis,we use the relation tan U =Ry/Rx,or

The 2 possible solutions are U =39 orU =39 +

180 =141. Since the net force lies in the 2ndquadrant, as mentioned earlier, the correct answer is

141 (see Fig. 4.5b).

80.0arctanN100

N80arctanarctan !

!!

x

y

RRU


18/116



Example 4.1 SOLN

Evaluate:In this situation, the net force is notzero, and you can

see intuitively that wrestler1 (who exerts the largest

force, on the belt) is likely to walk away with the

belt at the end of the struggle.

In Section 4.2, we will explore in detail what happens

in situations in which the net force is zero.

1FT


19/116



4.2 Newtons First Law

We agree that if a body is at rest, and if no net forceacts on it, that body will remain at rest.

What if there is net force of zero acting on a body in

motion?


20/116




Imagine sliding a hockey puck along a table, waxedfloor and air-hockey table and observe that the

distance the puck slides depends on the friction

between it and the surface.

Friction is a form of interaction between the undersurface of the puck and the sliding surface.

If we eliminate friction completely, the puck will

never slow down and no force is needed to keep

the puck moving once its has started.


21/116


22/116




Suppose a hockey puck rests on a horizontalsurface with negligible friction, such as an air-

hockey table or a slab of wet ice.

If puck is initially at rest, and single horizontal force

acts on it, the puck starts to move.1FT

2F

T


23/116




Or if the puck is in motion initially, the force exertedchanges its speed, its direction, or both.

The net force is equal to , which is not zero.

Suppose we apply a 2nd force , equal in

magnitude to but opposite in direction. The 2

forces are negatives of each other, , and

their vector sum is zero:

The results show that in Newtons first law, zero net

force is equivalent to no force at all.

1FT

2FT

1FT

12 FFTT

!

0!!! 1121 FFFFTTTT

F


24/116




This is really the principle of superposition that westudied in section 4.1.

When a body is acted on by no forces, or by

several forces such that the vector sum (resultant)

is zero, the body is in equilibrium.

In equilibrium, a body is either at rest or moving in a

straight line with constant velocity.

F

or such a body in equilibrium, net force is zero:

34.0!FT


25/116




For Eqn 4.3 to be true, each component of the netforce must be zero, so

When Eqn 4.4 is satisfied, the body is in

equilibrium.

Assumption made here is the body can be

represented as a pt particle. If body has finite size,we will need to consider where on the body the

forces are applied. This is covered in Chapter11.

4.400 !! yx FF


26/116



ConceptualExample 4.2 Zeronetforce meansconstantvelocity

In the classic 1950 science fiction filmRocketshipX-M,a spaceship is moving in the vacuum of outer space, far

from any planet, when its engine dies.

As a result, the spaceship slows down and stops.What does

Newtons first law say about this event?

Solution:

In this situation there are no forces acting on the spaceship, so

according to Newtons first law, it will notstop. It continues to

move in a straight line with constant speed. Some science

fiction movies have made use of very accurate science; thiswas not one of them.


27/116



ConceptualExample 4.3 Constantvelocity means zeronetforce

You are driving a Porsche 911 Carrera on a straight testingtrack at a constant speed of150 km/h.

You pass a 1971 Volkswagen Beetle doing a constant 75 km/h.

For which car is the net force greater?

Solution:

The key word in this question is net. Both cars are in

equilibrium because their velocities are both constant;

therefore the netforce on each car is zero.


28/116



ConceptualExample 4.3 SOLN

This conclusion seems to contradict the commonsense idea that the faster car must have a greater

force pushing it.

It's true that there is a forward force on both cars, and

it's true that the forward force on your Porsche ismuch greater than that on the Volkswagen (thanks to

your Porsches high-power engine).

But there is also a backwardforce acting on each car

due to friction with the road and air resistance.


29/116




The only reason these cars need their enginesrunning at all is to counteract this backward force so

that the vector sum of the forward and backward

forces will be zero and the car will travel with constant

velocity.The backward force on your Porsche is greater

because of its greater speed, which is why your

engine has to be more powerful than the

Volkswagen's.


30/116




Inertial frames of reference When we discussed the concept of relative velocity

in section 3.5, we also introduce the concept of

frame of reference.

This concept is CENTRAL to Newtons laws of

motion.

We need to find the correct frame of reference

before we can applyN

ewtons laws of motion.

Thus, an inertial frame of reference is one where

Newtons First Law is valid.


31/116




Inertial frames of reference Because Newtons first law is used to define what

we mean by an inertial frame of reference, it issometimes called the law of inertia.

The figure shows how we use Newtons first law todefine what you experience when riding in a vehiclethats accelerating.

The vehicle is initially at rest, then begins to

accelerate to the right.

A passenger on roller skates has almost no forceacting on her, and she tends to remain at restrelative to the inertial frame of the earth, according

toN

ewtons first law.


32/116




Inertial frames of reference


33/116




Inertial frames of reference As the vehicle accelerates around her, she moves

backward relative to the vehicle.

Similarly, a passenger in a vehicle that is slowing

down tends to continue moving with constant

velocity relative to the earth (Fig. 4.8b).

This passenger moves forward relative to the

vehicle.

A vehicle is also accelerating if it moves at a

constant speed but is turning (Fig. 4.8c).


34/116




Inertial frames of reference In each case discussed, you as the observer in the

vehicles frame of reference might by tempted toconclude that there is a net force acting on the

passenger, since her velocity relative to the vehiclechanges in each case.

This is WRONG! The net force on the passenger isindeed zero.

The mistake you committed is to applyN

ewtonsfirst law in the vehicles frame of reference, which isNOT an inertial frame and is not valid forNewtonsfirst law.


35/116




Inertial frames of reference The earths surface is not the only inertial frame of

reference.

If we have an inertial frame of reference A, where

Newtons first law is obeyed, then ANY secondframe of reference B will also be inertial is it moves

relative to A with constant velocity .

To prove this, we use the relative velocity Eqn(3.36) from section 3.5

AB /T

ABBPAP /// TTT

!


36/116




Inertial frames of reference Suppose that P is a body moving with constant

velocity with respect to an inertial frame A.

By Newtons first law, the net force on the body is

zero.

Velocity ofPrelative to anther frame B has a

different value,

If the relative velocity of the 2 frames isconstant, then is constant as well.

AP/T

ABAPBP /// TTT

!

AB /

T

BP/T


37/116




Inertial frames of reference Thus B is also an inertial frame, since velocity ofP

in this frame is constant, and net force on P is

zero, so Newtons first law is obeyed in B.

Observers in framesA and B will disagree aboutthe velocity ofP, but they will agree that Phas a

constant velocity (zero acceleration) and has zero

net force acting on it.

There is no single inertial frame of reference that is

preferred over the others for formulating Newtons

laws.


38/116




Inertial frames of reference If one frame is inertial, then every other frame

moving relative to it with constant velocity is also

inertial.

Thus, state of rest and state of motion withconstant velocity are not different; both occur when

vector sum of forces acting on body is zero.


39/116



4.3 NewtonsSecond Law

In Fig. 4.10, we see a hockey puck sliding to theright on ice with negligible friction.

There are no horizontal forces acting on the puck.

Downward force of gravity and upward contact

force exerted by ice surface sum to zero.

So, net force acting on the puck is zero, the

puck has zero acceleration, and its velocity is

constant.

FT


40/116





41/116




In Fig. 4.10b, we apply a constant horizontal forceto a sliding puck in same direction that the puck is

moving.

Then is constant and in same horizontal

direction as . During the time the force is acting, velocity of the

puck changes at a constant rate, i.e. the puck

moves with constant acceleration.

Speed of puck increases, so acceleration is in

the same direction as and .

FT

T

FT

T

aT


42/116




In Fig. 4.10c, we reverse the direction of the forceon the puck so that acts in the direction

opposite to .

The puck has an acceleration, it moves more and

more slowly to the right. If leftward force continues to act, the puck

eventually stops and begins to move more and

more rapidly to the left.

Acceleration is to the left, in same direction as

This experiment shows that acceleration is

constant if is constant.

FT

T

FT

aT

FT


43/116




We conclude that the presence of a net forceacting on a body cause the body to accelerate. The

direction of the acceleration is the same as that of

the net force.

If magnitude of net force is constant, then themagnitude of the acceleration is also constant.

These conclusions obtained can

also be applied to a body moving

along a curved path.


44/116




Another experiment doneshows the relationshipbetween acceleration andnet force acting on a body.

W

e apply a constanthorizontal force to a puckon a frictionless horizontalsurface, as shown inFig. 4.11, using a spring

balance. This horizontal force equals

the net force on the puck.


45/116




Doubling the net force doubles the acceleration,halving the force halves the acceleration, and so

on.

For a given body, the ratio of the magnitude

of the net force to the magnitude a = of theacceleration is constant, regardless of the

magnitude of the net force.

This ratio is called the inertial mass, or simply the

mass, of the body and denote it by m. That is,

FT

5.4ma!FT

a

T


46/116


47/116




Mass and Force The definition of newton is

One newton is the amount of net force that gives an

acceleration of one meter per second squared to a

body with a mass of one kilogram.

It is also related to the units of mass, length, and

time.

For Eqn 4.5 to be dimensionally sound, it must betrue that2

m/skg1N1 !


48/116




Mass and Force Eqn 4.5 can also be used to compare a mass with

a standard mass and thus measure masses.

We apply a constant net forceFto a body having a

known mass m1 and acceleration ofa1.We thenapply SAME force to another body having unknown

mass m2 and acceleration of magnitude a2.

Then according to Eqn 4.5

6.42

1

1

2

2211

a

a

m

m

amam

!

!


49/116




Mass and Force Figure 4.13 shows the inverse

proportionality between mass

and acceleration.

When 2 bodies with mass m1and m2 are fastened together,

the mass of the composite

body is m1+ m2.

This additive property of mass

may seem obvious, but it has

to be verified experimentally.


50/116




Newtons 2nd

law Eqn 4.5 relates the magnitude of the net force on a

body to the magnitude of the acceleration that it

produces.

Weve also seen that the direction of the net forceis the same as the direction of the acceleration,

whether the bodys path is straight or curved.

Newton wrapped up all these relationships and

experimental results in a single concise statement

the we now call Newtons second law of motion.


51/116




Newtons 2nd

lawIf a net external force acts on a body, the body

accelerates. The direction of acceleration is the

same as the direction of the net force. The net force

vector is equal to the mass of the body times theacceleration of the body.

In symbols,

7.4aF TT

m!


52/116




Newtons 2nd

law Or we say the acceleration of a body (rate of

change of its velocity) is equal to the vector sum

(resultant) or all forces acting on the body, divided

by its mass:

N

ote that the acceleration is in the same directionas the net force.

m

!

Fa

TT

f


53/116




Newtons 2nd

law Newtons second law is a fundamental law of

nature, the basic relation between force and

motion.

We need to highlight FOUR aspects of Eqn 4.7:

1. Eqn 4.7 is a vector equation.

Usually we use it in component form, with a

separate eqn for each component of force andthe corresponding acceleration:

8.4zzyyxx maFmaFmaF !!!

4 N t L f M ti


54/116




Newtons 2nd

law The set of component eqns is equivalent to the

single vector Eqn 4.7.

2. Newtons second law refers to external forces.

This refer to forces exerted on the body by

other bodies in its environment.

3. Eqns (4.7) and (4.8) are valid only when the mass

m is constant.

4. Newtons second law is valid only in inertial frame

of reference, just like the first law.

4 N t L f M ti


55/116




Note that the qty is not a force. All thatEqns (4.7) and (4.8) say is that the vector

is equal in magnitude and direction to the

vector sum of all the forces acting on

the body. Its incorrect to think of acceleration as a

force; rather, acceleration is a result of a

nonzero net force.

We always examine motion relative to inertialframes of reference only.

a

Tm

aT

m

FT

4 N t L f M ti


56/116




2.1.3 (Tension Change)

2.1.4 (Sliding on an Incline)

4 N t L f M ti


57/116



Example 4.4 Determining accelerationfrom force

A worker applies a constant horizontal force withmagnitude 20 N to a box with mass 40 kg resting on

a level floor with negligible friction.What is the

acceleration of the box?

Solution:

Identify:

This problem involves force and acceleration.

Whenever you encounter a problem of this kind, you

should approach it using Newtons second law.

4 N t L f M ti


58/116



Example 4.4 SOLN

Set up:In anyproblem involving forces, the first steps are to

(i) choose a coordinate system and (ii) identify all of

the forces acting on the body in question.

Its usually convenient to take one axis either alongor opposite to the direction of the bodys acceleration,

which in this case is horizontal (see Fig. 4.15a).

Hence we take the +x-axis to be in the direction of

the applied horizontal force (that is, the direction in

which the box accelerates) and the +y-axis to be

upward (Fig. 4.15b).

4 N t L f M ti


59/116



Example 4.4 SOLN

4 Newtons Laws of Motion


60/116



Example 4.4 SOLN

Set up:

In most force problems that youll encounter, the forcevectors all lie in a plane, so thez-axis isnt used.

The force acting on the box are (i) the horizontal force

exerted by the worker; (ii) the weight of the box,i.e., the downward force of the earths gravitationalattraction; and (iii) the upward supporting forceexerted by the floor.

As in section 4.

2, we call the force a normal forcebecause it is normal to the surface of contact.We aretold friction is negligible, so no friction force is present.

F

TwT

nT

nT



61/116



Example 4.4 SOLN

Set up:Since the box doesnt move vertically at all, the y-

acceleration is zero; ay = 0.Our target variable is the

x-component of acceleration ax.Well find it using

Newtons second law in component form as given byEqn (4.8).

Execute:

From Fig. 4.15b, only the 20-N force has a nonzero

x-component. Hence, the first relation in Eqn (4.8)

tells us thatN20!! FFx



62/116



Example 4.4 SOLN

Execute:

Hence the x-component of acceleration is

Evaluate:The acceleration is in the +x-direction, the samedirection as the net force. The net force is constant,so the acceleration is also constant. If we are given

the initial position and velocity of the box, we can findthe position and velocity at any later time from theeqns of motion with constant acceleration we derivedfrom chapter 2.

22

m/s50.0

kg40

m/skg20

kg40

N20!

!!!

m

Fa

xx



63/116



Example 4.4 SOLN

Evaluate:Notice that to determine ax, we didnt have to use the

y-component ofNewtons second law from Eqn (4.8),

.

By using this eqn, can you show the magnitude nof

the normal force in this situation is equal to the

weight of the box?

! yy maF



64/116

4. Newton s LawsofMotion


4.4 Mass and Weight

M

ass characterizes the inertial properties of abody.

The greater the mass, the greater the force is

needed to cause a given acceleration, as shown

in Newtons second law: The weight of a body is the force of the earths

gravitational attraction for the body.

The relationship between weight and mass can

be related back to Newtons second law.

aF

TTm!


65/116



66/116



4.4 Mass and Weight

G

enerally speaking, a body with mass m musthave weight with magnitude w given by

The weight of a body is a force, a vector qty, andwe can write Eqn (4.9) as a vector eqn:

Remember that g is the magnitude of , theacceleration due to gravity, so g is always a +venumber. Thus w, given by Eqn (4.9) is themagnitude of the weight and is always +ve.

9.4mgw !

10.4gw TT m!

gT



67/116



4.4 Mass and Weight

The weight of a body acts on the body all thetime, whether it is in free fall or not.

When a 10-kg flowerpot hangs suspended

from a chain, it is in equilibrium, and its

acceleration is zero. But its weight, given by Eqn (4.10), is still

pulling down on it.

In this case, the chain pulls up on the pot,

applying and upward force.

The vector sum of the forces is zero, and the

pot is in equilibrium.



68/116



4.4 Mass and Weight

2.9 (Pole-Vaulter Vaults)



69/116



ConceptualExample 4.5 Netforce and accelerationinfreefall

In Example 2.6 (section 2

.5) a one-euro coin wasdropped from rest from the Leaning Tower of Pisa. If

the coin falls freely, so that the

effects of the air are negligible,

how does the net force on thecoin vary as it falls?



70/116




In free fall, the acceleration of the coin is constantand equal to g. Hence by Newtons second law the

net force is also constant and equal

to ,which is the weight w of the coin (Fig. 4.18).

The velocity of the coin changes as it falls, but thenet force acting on it remains constant.

If this result surprises you, its because you still

believe in the erroneous common sense idea that

greater speed implies greater force.

If so, you should reread Conceptual Example 4.3.

a

T

aFTT

m!gT

m



71/116




The net force on a freely falling coin is constant evenif you initially toss it upward.

The force that your hand exerts on the coin is a

contact force, and it disappears the instant that the

coin loses contact with your hand.From then on, the only force acting on the coin is its

weight w.



72/116



4.4 Mass and Weight

Variation ofgwith location We use g= 9.80 m/s2 for problems on the earth.

In fact, the value of g varies somewhat from pt to

pt on the earths surface, from about 9.78 to 9.82

m/s2, because the earth is not perfectly sphericaland because of effects due to its rotation and

orbital motion.

At a pt where g= 9.80 m/s2, weight of a standard

kilogram is w = 9.80 N.

At a different pt where g= 9.78 m/s2,

weight w = 9.78N but the mass is still 1 kg.

4. Newtons Laws of Motion


73/116



4.4 Mass and Weight

Variation ofgwith location Weight of body varies from location to location,

but its mass does not.



74/116



4.4 Mass and Weight

Measuring mass and weight

The easiest way to measure the mass of a body is

to measure its weight, often by comparing with a

standard.

Because of Eqn (4.9), two bodies that have thesame weight at a particular location also have the

same mass.

The familiar equal-arm balance can determine

with great precision when the weights of two

bodies are equal and hence when their masses

are equal.



75/116



4.4 Mass and Weight


This method

doesnt work in

the apparent

zero-gravityenvironment of

outer space.

Instead Newtons

second law needsto be used directly.



76/116



4.4 Mass and Weight


The concept of mass plays 2 different roles in

mechanics.

The weight of a body is proportional to its mass;

we call the property related to gravitationalinteractions gravitational mass.

We call the inertial property that appears in

Newtons second law the inertial mass.

If the two quantities are different, the acceleration

due to gravity might well be different for different

bodies.



77/116

e to s a s o ot o


4.4 Mass and Weight

The SI units for mass and weight are oftenmisused in everyday life.

Incorrect expressions such as This box

weighs 6 kg are nearly universal.

What is meant is that the mass of the box,determined indirectly by weighting, is 6 kg.

In SI units, weight (a force) is measured in

newtons, while mass is measured in

kilograms.



78/116


Example 4.7 Mass and Weight

A 1.96v

104 N Lincoln Town Car traveling in the+x-direction makes a fast stop; the x-component ofthe net force acting on it is 1.50 v 104 N.What is its acceleration?

Solution:

Identify:

Again we will use Newtons second law to relate forceand acceleration. To use this relationship, we need toknow the cars mass. However, because the newtonis a unit for force, we know that 1.96 v 104 N is thecars weight,not its mass. So well also have to usethe relationship between a bodys mass and itsweight.



79/116


Example 4.7 SOLN

Set up:

Our target variable is the x-component of acceleration

of the car, ax (The motion is purely in the x-direction.)

We use Eqn. (4.9) to determine the cars mass from

its weight, then use the x-component ofNewtonssecond law from Eqn. (4.8) to determine ax.

Execute:

The mass m of the car is

kg2000

m/s80.9

m/skg1096.1

m/s80.9

N1096.12

24

2

4

!

v!

v!!

g

wm



80/116


Example 4.7 SOLN

Execute:

Then Fx = max gives

Evaluate:

The negative sign means that the acceleration vectorpoints in the negative x-direction. This makes sense:the car is moving in the positive x-direction and isslowing down.

2

244

m/s5.7kg2000

m/skg1050.1

kg2000

N1050.1

!

v!

v!!

m

Fa

xx



81/116


Example 4.7 SOLN

Evaluate:

Note that the acceleration can alternatively be

written as 0.77g. Its of interest that 0.77 is also

the ratio of1.50 v 104 N (the x-component of the

net force) to 1.96 v 104 N (the weight).Indeed, the acceleration of a body expressed as a

multiple ofg is always equal to the ratio of the net

force on the body to its weight.

Can you see why?



82/116


4.5 Newtons Third Law

A force acting on a body is always the result of itsinteraction with another body, so forces always

comes in pairs.

Example:When you kick a football, the forward

force exerted on the ball launches it into itstrajectory, but you also feel the force the ball

exerts back on your foot.

The force that is exerted on the other body is in

the opposite direction to the force that body exertson you.



83/116



Experiments show that whenever two bodiesinteract, the two forces that they exert on eachother are always equal in magnitude and oppositein direction.

This fact is called Newtons third law of motion.

In the figure,is the force applied bybody A (first subscript)on body B (2nd subscript),

and is the forceapplied by body Bon body A.

BonAFT

AonBFT



84/116



The mathematical statement ofN

ewtons third lawis

Expressed in words:

If body A exerts a force on body B (an action),

then body B exerts a force on body A (a

reaction). These two forces have the same

magnitude but are opposite in direction. These

two forces act on different bodies.

11.4AonBBonA FFTT

!



85/116



action and reaction are two opposite forces orwe also refer them as an action-reaction pair.

Note that this does not imply any cause-and-effect relationship.

W

e stress again that the two forces inN

ewtonsthird law acts on different bodies.

In Fig. 4.21, the action and reaction forces arecontact forces that are present ONLY when twobodies are touching.

But Newtons third law also applies to long-rangeforces that do not require physical contact. E.g.gravitational attraction.



86/116


ConceptualExample 4.8 Whichforceisgreater

After your sports car breaks down, you start to push itto the nearest repair shop.While the car is starting tomove, how does the force you exert on the carcompare to the force the car exerts on you?

How do these forces compare when you are

pushing the car along at a constant speed?Solution:

In both cases, the force you exert on the car is equalin magnitude and opposite in direction to the force thecar exerts on you.

Its true that you have to push harder to get thecar going than to keep it going. But no matter howhard you push on the car, the car pushes just as hardback on you.



87/116



Newtons third law gives the same result whether thetwo bodies are at rest, moving with constant velocity,or accelerating.

You may wonder how the car knows to push back

on you with the same magnitude of force that youexert on it.

Remember that the forces you and the car exert oneach other are really interactions between the atomsat the surface of your hand and the atoms at thesurface of the car.



88/116



These interactions are analogous to miniature springsbetween adjacent atoms, and a compressed springexerts equally strong forces on either of its ends.

Fundamentally, the reason why we know that objects

of different masses exert equally strong forces oneach other is that experiment tells us so.

Never forget that physics isn't merely a collection ofrules and equations; rather, its a systematicdescription of the natural world based on experimentand observation.



89/116


ConceptualExample 4.9 ApplyingNewtonsthirdlaw: Objects atrest

An apple sits on a table in equilibrium.

What forces act on it?

What is the reaction force to each of the forces acting on theapple?

What are the action-reaction pairs?

Solution:The figures on next slide shows the apple on the table, and theforces acting on the apple.

The apple is A, the table T, and the earth E.

In the diagram, is the weight of the apple; i.e

., thedownward gravitational force exerted bythe earth E (first

subscript) on the apple A (second subscript).AonEFT



90/116





91/116



Similarly, on A is the upward force exerted bythe tableT (first subscript) on the apple A (second subscript).

As the earth pulls down on the apple, the apple exertsan equally strong upward pull on the earth, as shown inFig. 4.22d. and are an action-reaction pair,representing the mutual interaction of the apple and the earth,so

Also, as the table pushes up on the apple with forcethe corresponding reaction is the downward force

exerted by the apple on the table (Fig. 4.22c). So we have

EonAFT

EonAFT

E

onAFT

AonEFT

AonEEonA FFTT

!

AonTTonA FFTT

!

AonTFT

TonAFT



92/116



The two forces acting on the apple are and

.

Are they an action-reaction pair? No, they aren't, despite beingequal and opposite. They do not represent the mutualinteraction of two bodies; they are two different forces actingon the same body.

The two forces in an action-reaction pair never act on thesame body.

Heres another way to look at it. Suppose we suddenly yankthe table out from under the apple (Fig. 4.22e). The two forces

and then become zero, but and arestill there (the gravitational interaction is still present).

AonEFT

Aon

TFT

EonAFT

AonEFT

AonTFT

TonAFT



93/116



Since is now zero, it cant be the negative ofand these two forces cant be an action-

reaction pair.AonEF

T AonT

FT

4. Newtons LawsofMotionC t l E l 4 10 A l i N t thi d l Obj t i


94/116


ConceptualExample 4.10 ApplyingNewtonsthirdlaw: Objectsin

motion

A stonemason drags a marble block across a floor bypulling on a rope attached to the block (Fig. 4.23a).

The block mayor may not be in equilibrium. How are

the various forces related? What are the action-

reaction pairs?



95/116



Figure 4.23b shows the horizontal

forces acting oneach body: the block (B), the rope (R),and the mason (M).We'll usesubscripts on all the forces to helpexplain

things.Vector represents the forceexerted bythe mason on the rope.Its reaction is the equal andopposite force exerted

bythe rope on the mason.

MonRFT

RonMFT



96/116



Vector represents the

force exerted by the rope onthe block. The reaction to itis the equal and opposite force

exerted by the block on therope:

Be sure you understand that theforces and are not aaction-reaction pair.

BonRRonB

RonMMonR

FF

FF

TT

TT

!

!

and

BonR

FT

RonBFT

RonBFT

RonMFT



97/116



Both of these forces act on the same body (the rope);an action and its reaction mustalways act on differentbodies.Furthermore, the forces andare not necessarily equal in magnitude. ApplyingNewtons second law to the rope, we get

If the block and rope are accelerating (speeding up orslowing down), the rope is not in equilibrium, and

must have different magnitude than .

roperopeaFFFT

TTTmRonBRonM !!

RonMFT

RonBFT

RonBFT

RonMFT



98/116



By contrast, the action-reaction forces andare always equal in magnitude, as are

and .Newtons third law holds whether or notthe bodies are accelerating.

In the special case in which the rope is in equilibrium,

the forces and are equal in magnitude.But this is an example ofNewton's firstlaw, not histhird.Another way to look at this is that in equilibrium,

in the eqn on the previous page.

Then because ofNewton's first orsecond law.

MonRFT

RonMFT

BonRFT

RonBFT

RonMF

T

RonBF

T

0!ropeaT

RonMRonB FFTT

!



99/116



This is also true if the rope is accelerating but hasnegligibly small mass compared to the block or themason. In this case, mrope = 0 in the eqn on theprevious page, so again .

Since always equals by Newtons

third law (they are an action-reaction pair), in thesesame special cases, also equals .

In other words, in these cases the force of the ropeon the block equals the force of the mason on the

rope, and we can then think of the rope astransmitting to the block, without change, the forcethe person exerts on the rope (Fig. 4.23c).

BonRFT

RonMFT

RonBFT RonMRonB

FFTT

!

BonRFT



100/116



This is a useful point of view, but you have toremember that it is valid onlywhen the rope hasnegligibly small mass or is in equilibrium.

If you feel as though youre drowning in subscripts atthis pt, take heart.Go over this discussion again,

comparing the symbols with the vector diagrams, untilyoure sure you see what's going on.



101/116


ConceptualExample 4.11 ANewtons Third Law

We saw in Conceptual Example 4.10 that the

stonemason pulls as hard on the rope-blockcombination as that combination pulls back on him.Why, then, does the block move while thestonemason remains stationary?

Solution:

The way out of this seeming conundrum is to keep inmind the difference between Newtons secondlawand his thirdlaw.

The only forces involved in Newtons second law are

those that act on that body.The vector sum of these forces determines how thebody accelerates (and whether it accelerates at all).



102/116



By contrast, Newton's third law relates the forces that

two differentbodies exert on each other. The third lawalone tells you nothing about the motion of eitherbody.

The reason why the stonemason doesnt move is that

the net force acting onhim is zero. It is thevector sum of theupward normal forcefrom the floor, his

weight acting downward,the rope pulling to theleft, and the friction forceof the floor pushing himto the right (Fig. 4.24).



103/116



Because the stonemason has shoes with non-skid

soles that don't slip on the floor, the friction force isstrong enough to exactly balance the pull of the rope.

If the floor were freshly waxed, so that there was littlefriction between the floor and the stonemason's

shoes, he would indeed start sliding to the left.

Four forces also act on the block-rope combination:normal force, weight, friction, and the force of thestonemason pulling to the right (see Fig. 4.24).

If the block is initially at rest, it begins to slide if the

stonemason's force is greaterthan the friction forcethat the floor exerts on the block. (The marble blockhas a smooth underside, which helps to minimizefriction.)


C SO


104/116



Then the net force on the block is not zero, and the

block accelerates to the right.Once the block is moving, the stonemason doesn'tneed to pull quite so hard; he need exert only enoughforce to exactly balance the friction force on the block.

Then the net force on the moving block is zero, andthe block continues to move at a constant velocity inaccordance with Newton's first law.

The moral of this example is that when analyzing themotion of a body, remember that only forces acting

on the body determine its motion.From this perspective, Newton's third law is merely atool that can help you determine what those forcesare.


4 5 N t Thi d L


105/116



A body (such as the rope in Fig. 4.23) that haspulling forces applied at its ends is said to be intension.

Tension at any pt is the magnitude of force actingat that pt.

In Fig. 4.23b, the tension at the right-hand end ofthe rope is the magnitude of (or of

) and tension at left-hand equalsmagnitude of (or of ).

If the rope is in equilibrium and if no forces actexcept at its ends, the tension is the same at bothends and throughout the rope.

RonMFT

MonRFT

RonBF

TBonRF

T


4 5 N t Thi d L


106/116



We emphasize once more a fundamental

truth: the two forces in an action-reaction pair

never act on the same body.

Remember this simple fact to help you avoid

confusion about action-reaction pairs andNewtons third law.


4 6 F B d Di


107/116


4.6 Free-Body Diagrams

Newtons first and second laws apply to a specific

body.

Whenever you use Newtons first law, ,

for an equilibrium situation orNewtons second

law, , for a non-equilibrium situation,you must decide at the beginning to which body

you are referring.

Only forces acting on the body matter. The sum

includes all the forces that act on the bodyin question.

0!FT

aF

TT

m!

FT


4 6 F B d Di


108/116



Dont get confused between the forces acting on

a body and the forces exerted by that body on

other bodies.

Free-body diagrams are essential to help identify

the relevant forces. A free-body diagram is a diagram showing the

chosen body by itself, free of its surroundings,

with vectors drawn to show the magnitudes and

directions of all the forces applied to the body bythe various other bodies that interact with it.


4 6 F B d Di


109/116



Be careful to include all forces acting on the body,

but be equally careful NOT to include any forces

that the body exerts on any other body.

The forces in an action-reaction pair must NEVER

appear in the same free-body diagram becausethey never act on the same body.

Also, forces that a body exerts on itself are never

included since these cant affect the bodys

motion.


4 6 F b d Di


110/116


4.6 Free-body Diagram

For a complete free-body diagram, you must

be able to answer for each force the

question, What other body is applying this

force?

If you cant answer that question, you maybe dealing with a nonexistent force.

Be careful NOT to avoid nonexistent forces

such as the force of acceleration or the

force, discussed in section 4.3.aT

m


4 6 F B d Di


111/116



For a problem with more than one body, you have

to take it apart and draw a separate free-body

diagram for each body.

Figure 4.23b shows a separate free-body diagram

for the rope in the case in which the rope isconsidered massless.

The figure also shows diagrams fro the block and

mason but are incomplete free-body diagrams, as

they dont show all the force acting on each body.


Concepts Summary


112/116


ConceptsSummary

Force is a quantitative measure of the interactionbetween two bodies. It is a vector qty.

When several forces act on a body, the effect onits motion is the same as when a single force,equal to the vector sum (resultant) of the forces,

acts on the body. Newtons first law states that when the vector sum

of all forces acting on a body (the net force) iszero, the body is in equilibrium.

If the body is initially at rest, it remains at rest; if itis initially in motion, it continues to move withconstant velocity.


Concepts Summary


113/116


ConceptsSummary

Newtons First Law is valid only in inertial frames

of reference.

The inertial properties of a body are characterized

by its mass.

The acceleration of a body under the action of agiven set of forces is directly proportional to the

vector sum of the forces (the net force) and

inversely proportional to the mass of the body.

This relationship is Newtons second law. LikeNewtons first law, this law is valid only in inertial

frames of reference.


Concepts Summary


114/116


ConceptsSummary

The unit of force is defined in terms of the units ofmass and acceleration.

In SI units, the unit of force is the newton (N),equal to 1 kgm/s2.

The weight of a body is the gravitational forceexerted on it by the earth (or whatever other bodyexerts the gravitational force).

Weight is a force and is therefore a vector qty.

The magnitude of the weight of a body at anyspecific location is equal to the product of its massm and the magnitude of the acceleration due togravity gat that location.


Concepts Summary


115/116


ConceptsSummary

While the weight of a body depends on its

location, the mass is independent of a location.

Newtons third law states that action equals

reaction. That is, when two bodies interact, they

exert forces on each other that at each instant areequal in magnitude and opposite in direction.

Each force in an action-reaction pair acts on only

one of the two bodies; the action and reaction

forces never act on the same body.


Key Equations


116/116

KeyEquations

34.0!FT !! 1.4

321 FFFFR

T

.

TTTT

7.4aF TT

m!

zz

yy

xx

maF

maF

maF

!

!

!

8.4

9.4mg!

11.4AonBBonA FFTT

!

module 6 dynamics

Documents