Chapter 4 The Laws of Motion. Kinematics Math of HOW things move ○ Position, velocity,...

Chapter 4The Laws of Motion

The Laws of Motion Kinematics

Math of HOW things move○ Position, velocity, acceleration

Dynamics WHY do things move? What causes a body to accelerate?

○ Forces => Acceleration

The properties of force and the relationships between force and acceleration are given by Newton’s Three Laws of Motion

The Laws of Motion

The First Law describes the natural state of motion of a body on which no forces are acting. The other two laws describe the behavior of a body under the influence of forces.

Early 1600’s, theories of object’s tendency to be at rest

Laws of Motion

Galileo GalileiDeveloped first correct ideas of motion

○ Gravity and constant acceleration○ Forces acting on bodies

Sir Isaac Newton1687 Principia Mathematica

○ Laws of Motion○ Law of Universal Gravitation

Invented calculus to further describe speed, acceleration

Forces Force

A central concept in all of physicsA vector quantity

○ Magnitude○ Direction

Force is used to describe push or a pullForces on objects

○ springs○ rubber bands○ ropes○ Cables

Forces (cont.)

ForceBouyant Forces

○ liquidsFriction

○ Surfaces

All examples above are known as “contact forces”

Newton’s First Law Newton’s First Law

“An object moves with a velocity that is constant in magnitude and direction, unless acted on by a nonzero net force.”

The net force on an object is defined as the vector sum of all external forces exerted on the object.

Often called the Law of inertiaTendency of object in motion to stay in motion

ConcepTest 4.2ConcepTest 4.2 Cart on Track ICart on Track I

1) slowly come to a stop

2) continue with constant acceleration

3) continue with decreasing acceleration

4) continue with constant velocity

5) immediately come to a stop

Consider a cart on a

horizontal frictionless

table. Once the cart has

been given a push and

released, what will

happen to the cart?

ConcepTest 4.2ConcepTest 4.2 Cart on Track ICart on Track I

1) slowly come to a stop

2) continue with constant acceleration

3) continue with decreasing acceleration

4) continue with constant velocity

5) immediately come to a stop

Consider a cart on a

horizontal frictionless

table. Once the cart has

been given a push and

released, what will

happen to the cart?

After the cart is released, there is no longer a forceno longer a force in

the x-direction. This does not mean that the cart stops This does not mean that the cart stops

moving!!moving!! It simply means that the cart will continuecontinue

moving with the same velocitymoving with the same velocity it had at the moment of

release. The initial push got the cart moving, but that

force is not needed to keep the cart in motion.

Newton’s Second Law

Newton’s Second Law “The acceleration a of an object is directly

proportional to the net force acting on it and inversely proportional to its mass.”

What it means:

a = ΣF / m

or conversely,

ΣF = ma


ΣF = ma

A LAW of nature! Precise definition of FORCE a and F are in the same direction

after F is completely summed

Newton’s Second Law (cont.)

ΣF = ma

ΣFΣFx = max

ΣFy = may

ΣFz = maz

No net force, means acceleration is zeroVelocity is constant

Newton’s Second Law (cont.) Units

Mass○ Kilograms kg

Acceleration○ m/s2

Force ○ Newtons 1N = 1 kg x m/s2 (m x a)○ Pounds 1N = 0.225lb

○ Pound is defined as F = ma = slug x ft/s2


Definition of Mass (physics)

A measure of resistance a body offers to changes in its velocity (acceleration)

Standard is kilogram Masses can be compared by balances Mass vs. Weight

Newton’s 2nd Law proves that different masses accelerate to the earth at the same rate, but with different forces.

We know that objects with different masses accelerate to the ground at the same rate.

However, because of the 2nd Law we know that they don’t hit the ground with the same force.

F = maF = ma

98 N = 10 kg x 9.8 m/s/s98 N = 10 kg x 9.8 m/s/s

F = maF = ma

9.8 N = 1 kg x 9.8 9.8 N = 1 kg x 9.8 m/s/sm/s/s

Newton’s Second Law (cont.) Example

A mass of 0.2kg slides along the table with a velocity of v = 2.8m/s. It stops in 1.0 m. What force is acting on the mass (neglect friction)?

Newton’s Third Law

Newton’s Third Law “If object 1 and object 2 interact, the force

F12 exerted by object 1 on object 2 is equal in magnitude but opposite to the force F21 exerted by object 2 on object 1.”

What it means:○ “for every action, there is an equal and

opposite reaction”


Action-Reaction Pair

F21 = -F12

Forces in nature always act in pairs

No single isolated force The mutual actions of two

bodies upon each other are ALWAYS equal and directed contrary to one another


Action-Reaction PairsNewton’s Law uses the forces

acting on an objectn and F are both acting on the

objectn is referred as to the normal

force and is the force exerted by the TV stand on the TV

Applications of Newton’s Laws

An object in equilibrium has no net external force acting on it, and the second law, in component form, implies that

ΣFx = 0and

ΣFy = 0for such an object. These two equations are useful for solving problems where the object is at rest or moving at constant velocity.


An object under acceleration requires the same two equations, but with the acceleration terms included:

ΣFx = max

and

ΣFy = may


AssumptionsObjects behave as particles

○ can ignore rotational motion (for now)Masses of strings or ropes are negligible

○ No stretching – constant lengthInterested only in the forces acting on the

object○ can neglect reaction forces

Pulleys are massless and frictionless○ Used to change direction


Free Body Diagram Diagram representing all the forces applied

to an object

Represent the object as a dotIdentify all the forces acting on the object,

not exerted Choose appropriate coordinate systemIncorrect FBD means incorrect solution


The force T is the tension acting on the box

n and F are the forces exerted by the earth and the ground

Only forces acting directly on the object are included in the free body diagramReaction forces act on other

objects and so are not included

Newton’s Third Law Free Body Diagram

ΣFy = N – Fg = n – mg = may

but the TV is not moving, so ay = 0

n – mg = 0

Therefore,

n = mg

n

Fg

Solving Newton’s Second Law Problems Read the problem at least once Draw a picture of the system

Identify the object of primary interest Indicate forces with arrows

Label each force Use labels that bring to mind the physical quantity

involved Draw a free body diagram

If additional objects are involved, draw separate free body diagrams for each object

Choose a convenient coordinate system for each object Apply Newton’s Second Law

The x- and y-components should be taken from the vector equation and written separately

Solve for the unknown(s)

ConcepTest 4.11ConcepTest 4.11 On an InclineOn an Incline

1) case A

2) case B

3) both the same (N = mg)

4) both the same (0 < N < mg)

5) both the same (N = 0)

Consider two identical blocks,

one resting on a flat surface,

and the other resting on an

incline. For which case is the

normal force greater?

1) case A

2) case B

3) both the same (N = mg)

4) both the same (0 < N < mg)

5) both the same (N = 0)

N

WWy

x

y

f

ConcepTest 4.11ConcepTest 4.11 On an InclineOn an Incline

Consider two identical blocks,

one resting on a flat surface,

and the other resting on an

incline. For which case is the

normal force greater?

In Case ACase A, we know that NN = =

WW. In Case BCase B, due to the angle

of the incline, NN < < WW. In fact,

we can see that N = W cos().

Newton’s Law review 1st – Law of Inertia

Objects at rest or motion will stay that way unless acted on by a force○ Sailboat○ Target

2nd – F = ma The acceleration is proportional to the force and inversely

proportional to the mass

3rd – Action-Reaction if a force is acted on to an object, the object will exert an

equal and opposite force○ Recoil

1) a net force acted on it

2) no net force acted on it

3) it remained at rest

4) it did not move, but only seemed to

5) gravity briefly stopped acting on it

ConcepTest 4.1c Newton’s First Law

You put your book on

the bus seat next to

you. When the bus

stops suddenly, the

book slides forward off

the seat. Why?

1) a net force acted on it

2) no net force acted on it

3) it remained at rest

4) it did not move, but only seemed to

5) gravity briefly stopped acting on it

The book was initially moving forward (since it was

on a moving bus). When the bus stopped, the book

continued moving forwardcontinued moving forward, which was its initial state initial state

of motionof motion, and therefore it slid forward off the seat.

ConcepTest 4.1c Newton’s First Law

You put your book on

the bus seat next to

you. When the bus

stops suddenly, the

book slides forward off

the seat. Why?

Follow-up:Follow-up: What is the force that usually keeps the book on the seat? What is the force that usually keeps the book on the seat?

A block of mass m rests on the

floor of an elevator that is

accelerating upward. What is

the relationship between the

force due to gravity and the

normal force on the block?

1) 1) N > mgN > mg

2) 2) N = mgN = mg

3) 3) N < mg (but not zero)N < mg (but not zero)

4) 4) N = 0N = 0

5) depends on the size of the 5) depends on the size of the elevatorelevator

ConcepTest 4.9b Going Up II

m

a

The block is accelerating upward, so

it must have a net upward forcenet upward force. The

forces on it are NN (up) and mgmg (down),

so NN must be greater than mgmg in order

to give the net upward forcenet upward force!

A block of mass m rests on the

floor of an elevator that is

accelerating upward. What is

the relationship between the

force due to gravity and the

normal force on the block?

1) 1) N > mgN > mg

2) 2) N = mgN = mg

3) 3) N < mg (but not zero)N < mg (but not zero)

4) 4) N = 0N = 0

5) depends on the size of the 5) depends on the size of the elevatorelevator

FF = = NN –– mgmg = = mama > 0 > 0

NN > > mgmg

m a > 0

mg

N

ConcepTest 4.9b Going Up II

Follow-up:Follow-up: What is the normal force if What is the normal force if the elevator is in free fall downward?the elevator is in free fall downward?


EquilibriumAn object either at rest or moving with a

constant velocity is said to be in equilibriumThe net force acting on the object is zero

(since the acceleration is zero)

ΣF = 0Should use components

ΣFx = 0 ΣFy = 0


Equilibrium Example - FBD


Equilibrium Example – FBDChoose the coordinate

system with x along the incline and y perpendicular to the incline

Replace the force of gravity with its components


Equilibrium Example?


Equilibrium – Multiple ObjectsWhen you have more than one object, the

problem-solving strategy is applied to each object

Draw free body diagrams for each objectApply Newton’s Laws to each objectSolve the equations

Applications of Newton’s Laws Example A traffic light

weighing 1.00x102 N hangs from a vertical cable tied to two other cables that are fastened to a support, as in the figure to the right. The upper cables makes angles of 37° and 53° with the horizontal. Find the tension in each of the three cables.


Example:An object with a mass m1 = 5.00kg rests on a frictionless horizontal table and is connected to a cable that passes over a pulley and is then fastened to a hanging object with mass m2 = 10.0 kg, as shown in Figure P4.30. Find the acceleration of each object and the tension in the table.

Forces of Friction An object moving on a surface encounters

resistance as it interacts through its surroundings. This resistance is called friction.

Friction is a force The force of friction is opposite of motion Two types of friction – static and kinetic Friction is proportional to the normal force

Friction Examples Car on the road

Forces of Friction Static Friction, ƒs

Static friction acts to keep the object from moving

If F increases, so does ƒs

If F decreases, so does ƒs

ƒs µ n

Forces of Friction

• Kinetic Friction, ƒk

– The force of kinetic friction acts when the object is in motion

– ƒk = µ n

• Variations of the coefficient with speed will be ignored

Forces of Friction

Friction Examples

Forces of Friction Friction Example

Axes are rotated as usual on an incline

The direction of impending motion would be down the plane

Friction acts up the planeOpposes the motion

Apply Newton’s Laws and solve equations

Static Friction...

FFapplied

mgg

NN

xx

y y

fS

We want to know how it acts in fixed or “static” systems: the force provided by friction depends on the forces applied on the system (magnitude: fs ≤ sN)

Opposes motion that would occur if s were zero

Static Friction... If a = 0.

x : Fapplied fS = 0

y: N = mg

FFappliedapplied

mgg

NN

xx

y y

fS

While the block is static: fS Fapplied (unlike kinetic friction)

Static Friction...

FF

mgg

NN

xx

y y

fS

The maximum possible force that the friction between two objects can provide is fMAX = SN, where s is the “coefficient of static friction”. So fS S N.

As one increases F, fS gets bigger until fS = SN and the object “breaks loose” and starts to move.

Static Friction... S is discovered by increasing FF until

the block starts to slide:x : FMAX SN = 0

y : N = mg S FMAX / mg

FFMAX

mgg

NN

xx

y y

Smg

Additional comments on Friction:

The force of friction does not depend on the area of the surfaces in contact (a relatively good approximation if there is little surface deformation)

GenerallyS > K for any system

Kinetic Friction Dynamics:

x-axis max = F KN

y-axis may = 0 = N – mg or N = mg

so F Kmg = m ax

maaxx

FF

mgg

NN

x

y y

K mg

vvfk

fk

Frictionless inclined plane

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

ma

Angles of the inclined plane

max = mg sin

mg

N

Case 1 - Frictionless inclined plane...

Use a FBD and consider x and y components separately:

FFxx max = mg sin ax = g sin FFy y may = 0 = N – mgcos N = mg cos

mgg

NN

mg sin

mg cos

maaxx

xx

yy

Case 2 - Inclined plane...static friction

Use a FBD and consider x and y components separately:\

FFxx max = 0 = mgsin - f 0 = g sin - f

FFy y may = 0 = N – mgcos N = mg cos

mgg

NN

mg sin

mg cos

maaxx= 0= 0

xx

yy

FrictionForce

Special case:

At the breaking point

f = s N = s mg cos

g sin = f = s mg cos

Example A roller coaster reaches the top of the

steepest hill with a speed of 6.0 km/hr. It then descends the hill, which is at an average of 45º and is 45.0m. What will its speed be at the bottom? Assume µk = 0.18.

Air Resistance and Drag

So far we’ve “neglected air resistance” in physics Can be difficult to deal with

Affects projectile motion Friction force opposes velocity through medium Imposes horizontal force, additional vertical forces Terminal velocity for falling objects

Dominant energy drain on cars, bicyclists, planes

Drag Force Quantified

With a cross sectional area, A (in m2), coefficient of drag of 1.0 (most objects), sea-level density of air, and velocity, v (m/s), the drag force is:

D = ½ C A v2 c A v2 in Newtons

c = ¼ kg/m3 ○ Increases as v increases

In falling, when D = mg, then at terminal velocity Example: Bicycling at 10 m/s (22 m.p.h.), with projected area of

0.5 m2 exerts ~30 Newtons Requires (F v) of power 300 Watts to maintain speed Minimizing drag is often important

“Free” Fall Terminal velocity reached when

Fdrag = Fgrav (= mg)

D = ¼Av2 ≈ mg v ≈ √4mg/A For 75 kg person with a frontal area of 0.5 m2,

vterm 50 m/s, or 110 mph

which is reached in about 5 seconds, over 125 m of fall

Trajectories with Air Resistance

Baseball launched at 45° with v = 50 m/s:Without air resistance, reaches about 63 m

high, 254 m rangeWith air resistance, about 31 m high, 122 m

range

Vacuum trajectory vs. air trajectory for 45° launch angle.

‘Free’ Fall Indoor sky-diving

Chapter 4 The Laws of Motion. Kinematics Math of HOW things move ○ Position, velocity,...

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Transcript of Chapter 4 The Laws of Motion. Kinematics Math of HOW things move ○ Position, velocity,...