Physics 101Lecture 5

Newton`s LawsAssist. Prof. Dr. Ali ÖVGÜNEMU Physics Department

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The Laws of Motionq Newton’s first lawq Forceq Massq Newton’s second lawq Newton’s third lawqFrictional forcesq Examples Isaac Newton’s work represents one of the greatest

contributions to science ever made by an individual.

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Dynamicsq Describes the relationship between the motion

of objects in our everyday world and the forces acting on them

q Language of Dynamicsn Force: The measure of interaction between two

objects (pull or push). It is a vector quantity – it has a magnitude and direction

n Mass: The measure of how difficult it is to change object’s velocity (sluggishness or inertia of the object)

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Forcesq The measure of interaction

between two objects (pull or push)

q Vector quantity: has magnitude and direction

q May be a contact force or a field forcen Contact forces result from

physical contact between two objects

n Field forces act between disconnected objects

n Also called “action at a distance”

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Forces

q Gravitational Forceq Archimedes Forceq Friction Forceq Tension Forceq Spring Forceq Normal Force

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Vector Nature of Forceq Vector force: has magnitude and directionq Net Force: a resultant force acting on object

q You must use the rules of vector addition to obtain the net force on an object

......321 +++==å FFFFFnet!!!!!

2 21 2

1 1

2

| | 2.24N

tan ( ) 26.6

F F FFF

q -

= + =

= = - !

"

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Newton’s First Lawq An object at rest tends to stay at rest and an

object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force

q An object at rest remains at rest as long as no net force acts on itq An object moving with constant velocity continues to move with

the same speed and in the same direction (the same velocity) as long as no net force acts on it

q “Keep on doing what it is doing”q When forces are balanced, the acceleration of the object is zero

n Object at rest: v = 0 and a = 0n Object in motion: v ¹ 0 and a = 0

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Mass and Inertiaq Every object continues in its state of rest, or uniform

motion in a straight line, unless it is compelled to change that state by unbalanced forces impressed upon it

q Inertia is a property of objects to resist changes is motion!

q Mass is a measure of the amount of inertia.

q Mass is a measure of the resistance of an object to changes in its velocity

q Mass is an inherent property of an objectq Scalar quantity and SI unit: kg

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Newton’s Second Lawq The acceleration of an object is directly

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

mF

mF

a net

!!!

==å

amFFnet!!!

==å

SI unit of force is a Newton (N)2smkg1N1 º

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More about Newton’s 2nd Lawq You must be certain about which body we are

applying it toq Fnet must be the vector sum of all the forces that act

on that bodyq Only forces that act on that body are to be included

in the vector sumq Net force component along an

axis gives rise to the accelerationalong that same axis

xxnet maF =, yynet maF =,

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Example1:q One or two forces act on a puck that moves over frictionless ice

along an x axis, in one-dimensional motion. The puck's mass is m = 0.20 kg. Forces F1 and F2 and are directed along the x axis and have magnitudes F1 = 4.0 N and F2 = 2.0 N. Force F3 is directed at angle q = 30° and has magnitude F3 = 1.0 N. In each situation, what is the acceleration of the puck?

xxnet maF =,

21

1

m/s20kg2.0N0.4

)

===

=

mFa

maFa

x

x

221

21

m/s10kg2.0

N0.2N0.4)

=-

=-

=

=-

mFFa

maFFb

x

x

223

3,32,3

m/s7.5kg2.0

N0.230cosN0.1cos

cos )

-=-

=-

=

==-!

mFFa

FFmaFFc

x

xxx

q

q

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Gravitational Forceq Gravitational force is a vectorq Expressed by Newton’s Law of Universal

Gravitation:

n G – gravitational constantn M – mass of the Earthn m – mass of an objectn R – radius of the Earth

q Direction: pointing downward

2RmMGFg =

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q The magnitude of the gravitational force acting on an object of mass m near the Earth’s surface is called the weight w of the object: w = mg

q g can also be found from the Law of Universal Gravitationq Weight has a unit of N

q Weight depends upon location

Weight

mgFw g ==

R = 6,400 km

22 m/s 8.9==RMGg

2RmMGFg =

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Normal Forceq Force from a solid

surface which keeps object from falling through

q Direction: always perpendicular to the surface

q Magnitude: depends on situation

mgFw g ==

yg maFN =-

mgN =ymamgN =-

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Tension Force: Tq A taut rope exerts forces

on whatever holds its ends

q Direction: always along the cord (rope, cable, string ……) and away from the object

q Magnitude: depend on situation

T1

T2T1 = T = T2

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

exerted by object 1 on object 2 is equal in magnitude but opposite in direction to the force exerted by object 2 on object 1

q Equivalent to saying a single isolated force cannot exist

BonAon FF

!!-=

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q When an object is in motion on a surface or through a viscous medium, there will be a resistance to the motion. This resistance is called the force of friction

q This is due to the interactions between the object and its environment

q We will be concerned with two types of frictional forcen Force of static friction: fs

n Force of kinetic friction: fk

q Direction: opposite the direction of the intended motionn If moving: in direction opposite the velocityn If stationary, in direction of the vector sum of other forces

Forces of Friction: f

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q Magnitude: Friction is proportional to the normal forcen Static friction: Ff = F £ μsNn Kinetic friction: Ff = μkNn μ is the coefficient of

frictionq The coefficients of friction

are nearly independent of the area of contact

Forces of Friction: Magnitude

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Static Frictionq Static friction acts to keep

the object from movingq If increases, so does q If decreases, so does q ƒs £ µs N

n Remember, the equality holds when the surfaces are on the verge of slipping

F!

F!

ƒs!

ƒs!

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Kinetic Frictionq The force of kinetic

friction acts when the object is in motion

q Although µk can vary with speed, we shall neglect any such variations

q ƒk = µk N

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Explore Forces of Frictionq Vary the applied forceq Note the value of the

frictional forcen Compare the values

q Note what happens when the can starts to move

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Free Body Diagramq The most important step in

solving problems involving Newton’s Laws is to draw the free body diagram

q Be sure to include only the forces acting on the object of interest

q Include any field forces acting on the object

q Do not assume the normal force equals the weight

F hand on book

F Earth on book

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Hints for Problem-Solvingq Read the problem carefully at least onceq Draw a picture of the system, identify the object of primary interest,

and indicate forces with arrowsq Label each force in the picture in a way that will bring to mind what

physical quantity the label stands for (e.g., T for tension)q Draw a free-body diagram of the object of interest, based on the

labeled picture. If additional objects are involved, draw separate free-body diagram for them

q Choose a convenient coordinate system for each objectq Apply Newton’s second law. The x- and y-components of Newton

second law should be taken from the vector equation and written individually. This often results in two equations and two unknowns

q Solve for the desired unknown quantity, and substitute the numbers

xxnet maF =, yynet maF =,

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Example2:

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Example3:

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Objects in Equilibriumq Objects that are either at rest or moving with

constant velocity are said to be in equilibriumq Acceleration of an object can be modeled as

zero: q Mathematically, the net force acting on the

object is zeroq Equivalent to the set of component equations

given by

0=åF!

0=a!

0=å xF 0=å yF

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Equilibrium, Example 1q A lamp is suspended from a

chain of negligible massq The forces acting on the

lamp aren the downward force of gravity n the upward tension in the

chainq Applying equilibrium gives

0 0= ® - = ® =å y g gF T F T F

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Equilibrium, Example 2q A traffic light weighing 100 N hangs from a vertical cable

tied to two other cables that are fastened to a support. The upper cables make angles of 37° and 53° with the horizontal. Find the tension in each of the three cables.

q Conceptualize the traffic lightn Assume cables don’t breakn Nothing is moving

q Categorize as an equilibrium problemn No movement, so acceleration is zeron Model as an object in equilibrium

0=å xF 0=å yF

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Equilibrium, Example 2q Need 2 free-body diagrams

n Apply equilibrium equation to light

n Apply equilibrium equations to knot

NFTFTF

g

gy

10000

3

3

==

=-®=å

NFTFTF

g

gy

10000

3

3

==

=-®=å

NTTNT

TTT

NTT

TTTF

TTTTF

yyyy

xxx

8033.1 60

33.153cos37cos

010053sin37sin

053cos37cos

121

112

21

321

2121

===

=÷÷ø

öççè

æ=

=-+=

++=

=+-=+=

åå

!

!

!!

!!

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Accelerating Objectsq If an object that can be modeled as a particle

experiences an acceleration, there must be a nonzero net force acting on it

q Draw a free-body diagramq Apply Newton’s Second Law in component form

amF !!=å

xx maF =å yy maF =å

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Accelerating Objects, Example 1q A man weighs himself with a scale in an elevator. While

the elevator is at rest, he measures a weight of 800 N.n What weight does the scale read if the elevator accelerates

upward at 2.0 m/s2? a = 2.0 m/s2

n What weight does the scale read if the elevator accelerates downward at 2.0 m/s2? a = - 2.0 m/s2

q Upward:

q Downward:

mamgNFy =-=å

mg

N

kg6.81m/s8.9N800

)(

2 ===

+=+=

gwm

agmmamgN

mgN >

N5.636)8.90.2(6.81 =+-=N

mgN <mg

NN9.962)8.90.2(6.81 =+=N

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Example4:

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Inclined Planeq Suppose a block with a

mass of 2.50 kg is resting on a ramp. If the coefficient of static friction between the block and ramp is 0.350, what maximum angle can the ramp make with the horizontal before the block starts to slip down?

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q Newton 2nd law:

q Then

q So

0cos

0sin

=-=

=-=

åå

q

µq

mgNF

NmgF

y

sx

qcosmgN =

å =-= 0cossin qµq mgmgF sy

350.0tan == sµq!3.19)350.0(tan 1 == -q

Inclined Plane

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Multiple Objects

q A block of mass m1 on a rough, horizontal surface is connected to a ball of mass m2 by a lightweight cord over a lightweight, frictionless pulley as shown in figure. A force of magnitude F at an angle θ with the horizontal is applied to the block as shown and the block slides to the right. The coefficient of kinetic friction between the block and surface is μk. Find the magnitude of acceleration of the two objects.

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Multiple Objectsq m1:

q m2: amamgmTF yy 222 ==-=å

0sin

cos

1

11

=-+=

==--=

åå

gmFNF

amamTfFF

y

xkx

q

q

qsin1 FgmN -=)sin( 1 qµµ FgmNf kkk -==

amgamFgmF k 121 )()sin(cos =+--- qµq

21

12 )()sin(cosmm

gmmFa kk

++-+

=µqµq

)(2 gamT +=

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Newton’s LawsI. If no net force acts on a body, then the

body’s velocity cannot change.II. The net force on a body is equal to the

product of the body’s mass and acceleration.

III. When two bodies interact, the force on the bodies from each other are always equal in magnitude and opposite in direction.

Force is a vectorUnit of force in S.I.:

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Example1:

Example2:

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Problem1:

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Problem2:

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Problem3:

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Problem4:

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Problem5:

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Problem6:

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Problem7:

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Example3:

Example4:

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