Classical mechanics describes the motion of object and the forces acting on them. Classical mechanics is very accurate as long as we do not try to study.

Classical mechanics describes the motion of object and the forces acting on them.Classical mechanics is very accurate as long as we do not try to study something as small as an atom, or something moving close to the speed of light.

A force is an action exerted on an object which may change the object’s state of rest or motion.

EQ: What is a force?

Forces can cause a massive object to accelerate.

The SI unit of force is the newton, N.

Forces can act through contact or at a distance.

What happened to the door?

Newton’s Contributions Calculus Light is composed of

rainbow colors Reflecting Telescope

Your objectives are to learn.

Three Laws of Motion Theory of Gravitation

What is a force?

Here are some ways of describing forces: A push A pull A stretch A squeeze A catch A twist

We can’t see forces but we can see the effects of a force.

Acceleration

TensionCompression-DeformationDe-accelerationTorque

Any change in motion

Forces can make things:

Speed up Slow down

Change direction Change shape

Forces can act through contact or at a distance.

Forces can be labeled into two categories: field force and contact force.

N= kg * m s2

lb = slug * ft s2

X =

Mass vs. weight: Do you have the same amount of mass on the Moon as

you do the Earth?

Force = Mass*Acceleration: Units

Mass vs. weight: Do you have the same amount of mass on the Moon as

you do the Earth?

Mass- a measure in the amount of matter in an object.

Weight: a measure of the amount of gravitational force acting on the mass of an object.

EX:

Wf= the weight of an object due to a gravitational fieldm = mass of the objectg = acceleration due to gravity

mgWf

Wf = 195lbgE = 32.2ft/s2

m = ?

Wf = mg m = Wf

g

Wf = mg m = 195lb * s2

32.2ft= 6.055 but what are the units?

Weight

W = mg weight is a force weight = mass x acceleration due to gravity

Units N = kg x m/s2

weight in Newtons = mass in kg x 9.81 m/s2

a 1 kg mass weighs 9.81 N

Wf = mg m = 195lb * s2

32.2ft

Derived Units

m = 195slug * ft * s2

s2 32.2ft

: Pause for a Cause

Your Turn: Calculate your mass

NOTE: MASS and WEIGHT are NOT the same thing. MASS never changesWhen an object moves to a different planet.

Pause for a Cause

What is the weight of an 85.3-kg person on earth? On Mars=3.2 m/s2?

NW

NWmgW

MARS 96.272)2.3)(3.85(

94.835)8.9)(3.85(

Force Diagrams The effect of a force depends on both magnitude

and direction. Thus, force is a vector quantity.

Diagrams that show force vectors as arrows are called force diagrams, or Free Body Diagrams.

Force diagrams that show only the forces acting on a single object are called free-body diagrams.

Force Diagrams

In a force diagram, vector arrows represent all the forces acting in a situation.

A free-body diagram shows only the forces acting on the object of interest—in this case, the car.

Force Diagram Free-Body Diagram

The sum of all the forces is called the Net Force.

The key to analyzing problems:pictorial representation of forces complete with labels.

W1,Fg1 or m1g

•Weight(mg) – Always drawn from the center, straight down•Force Normal(FN) – A surface force always drawn perpendicular to a surface.•Tension(T or FT) – force in ropes and always drawn AWAY from object.•Friction(Ff)- Always drawn opposing the motion.

m2g

T

T

FN

Ff

mg

Fn

Ff

mg sinθmg cosθ

θ

Practice A

A women is pulling on her suitcase with a force of 70.0 N directed at an angle of +30.0° to the horizontal. Find the x and y components of this force.

Given:F = 70 NΘ = 30.0°

Force in the X-axisFx = F(cos θ) =

Force in the Y-axisFy = F(sin θ) =

Force in the X-axisFx = 70.0 N(cos 30.0°) =

Force in the Y-axisFy = 70.0 N(sin 30.0°) =

60.6 N

35.0 N

soh cah toa

X-axis

Y-axisHyp

Essential Question EQ: A fan blows on two balls, a bowling ball and a balloon.

Describe what you think will happen.

bowling ball

balloon

Thought Experiment: a Pro Golfer tees up a bowling ball and strikes it with his driver. Describe what you think will happen?

I)Inertia:

II)F = ma

III)Action-reaction

An object at rest will stay at rest, and an object in motion will stay in motion at constant velocity, unless acted upon by an unbalanced force.

Force equals mass times acceleration.

For every action there is an equal and opposite reaction.

Inertia: Video Inertia is the tendency of an object to resist being

moved or, if the object is moving, to resist a change in speed or direction.

Newton’s first law is often referred to as the law of inertia because it states that in the absence of a net force, a body will preserve its state of motion.

Mass is a measure of inertia.

Why?

An object in motion remains in motion in a straight line and at a constant speed OR an object at rest remains at rest, UNLESS acted upon by an EXTERNAL (unbalanced) Force.

There are TWO conditions here and one constraint.

Condition #1 – The object CAN move but must be at a CONSTANT SPEEDCondition #2 – The object is at RESTConstraint – As long as the forces are BALANCED!!!!! And if all the forces are balanced the SUM of all the forces is ZERO.

The bottom line: There is NO ACCELERATION in this case AND the object must be at EQILIBRIUM ( All the forces cancel out). 00 Facc

Pause for a Cause: If acceleration = 0, then the sum of the forces must = ?

Newton’s 1st Law of Motion

W

Ry

if Ry = Wthen resultant force = 0if v = 0 and F = 0STATIC EQUILIBRIUM

W

Ry

FpFr

Fr = resistive forceFp = propulsive forceif v = 0 and F = 0DYNAMIC EQUILIBRIUM

V

Since the Fnet = 0, a system moving at a constant speed or at rest MUST be at EQUILIBRIUM.

TIPS for solving problems• Draw a FBD• Resolve anything into COMPONENTS• Write equations of equilibrium• Solve for unknowns

A 10-kg box is being pulled across the table to the right at a constant speed with a force of 50N.

a) Calculate the Force of Friction

b) Calculate the Force Normal

mg

Fn

Fa

Ff

NFF fa 50

NFmg n 98)8.9)(10(

Suppose the same box is now pulled at an angle of 30 degrees above the horizontal.

a) Calculate the Force of Friction

b) Calculate the Force Normal

mg

Fn

Fa

Ff30

NFF

NFF

axf

aax

3.43

3.4330cos50cos

Fax

Fay

NF

FmgF

mgFF

mgF

N

ayN

ayN

N

73

30sin50)8.9)(10(

!

If an object is NOT at rest or moving at a constant speed, that means the FORCES are UNBALANCED. One force(s) in a certain direction over power the others.

THE OBJECT WILL THEN ACCELERATE.

Inertia

The sum of all forces acting on an object is called net force. F

Unbalanced Forces

Unequal opposing forces produce an unbalanced force

causing motion

Balanced ForceEqual forces in opposite

directions produce no motion

What is the net force of the truck on bottom?

Force 1 (F1) and force 2 (F2) are applied to the ball at the same time with the same magnitude. Draw a new arrow describing the combined direction of F1 & F2 on the ball. This is called the vector sum.

Graphing assignment

F1 = F2F1 = F2

Vector Sum of Force

Graphing Assignment Two horizontal ropes are attached to a post

that is stuck in the ground. The ropes pull the post producing the following vector forces

Determine the direction and magnitude the support cable must have to prevent pole failure.

F1 = (5, 5) F2 = (-3, 6)

F1 = (5Nx, 5Ny) F2 = (-3Nx, 6Ny)

(X, Y) (X, Y)

Graphing Assignment

If F1 + F2 = -F3 diagram the direction and magnitude that the pole pulls back (F3) on the ropes?

Hint: -F3XY = (F1X + F2X) (F1y + F2y)

Graphing Assignment

What is the angle between F2 & F3?

What is the angle between F1 & F3?

What is the angle between F1 & F2?

Hint: tan(θ) = Opposite / Adjacent •θ = tan -1 (Opposite/Adjacent) or •(Y/X) •(-1) means (1/tan) or inverse.

Fnet = maF1

F2

Fnet = F1 + F2

F1xy + F2xy = maF1 = Fx + Fy F2 = Fx + Fy

F1

F2

Fnet = F1 + F2

F1 = F1x + F1y

F2 = F2x + F2y

F1x = F1 cos θ1x = F2x = F2 cos θ2x =

F1y = F1 sin θ1y = F2y = F2 sin θ2y =

F1x = 6.00E2 cos 30.0˚ = F2x = 6.00E2 cos -45.0˚ =

Given:Fnet = ma

a = ?m = 2.00E3 KgFh = 6.00E2 Nθ1 = 30.0˚θ2 = 45.0˚

5.02E2 N4.24E2 N

ΣFx = 9.44E2 N

F1y = 6.00E2 sin 30.0˚ = F2y = 6.00E2 sin -45.0˚ =

3.00E2 N-4.24E2 N

ΣFy = -1.24E2 N

Net Force

ΣFx = F1x + F2x

ΣFy = F1y + F2y

F1

F2

Given:a = ?m = 2.00E3 KgFh = 6.00E2 Nθ1 = 30.0˚θ2 = 45.0˚

ΣFx = 9.44E2 N

ΣFy = -1.24E2 N

F = maax = Fx

m ay = Fy

m

ax = 9.44E2 N = 2.00E3 Kg

2D: Accelerationa2 = ax

2 = ay2

a = √ax2 = ay

2

ay = -1.24E2 N = 2.00E3 Kg

0.472 m/s2

-0.062 m/s2

a = √(0.472)2 + (-0.062)2 =

a = √ax2 = ay

2

0.476 m/s2

Magnitude of Acceleration

F1

F2

Given:a = 0.476 m/s2

m = 2.00E3 KgFh = 6.00E2 Nθ1 = 30.0˚θ2 = 45.0˚

ΣFx = 9.44E2 N

ΣFy = -1.24E2 N

-0.062 m/s2

0.472 m/s2= -7.48˚

Direction of AccelerationThe direction will be the ratio of the acceleration in the x-axis & the y-axis

ay sin θ tan θax cos θ

= ay ax

θ = tan-1 ay ax

θ = tan-1

F1

F2

Fa

a F am

a k F

m 1

1

where k1 = constant of proportionality

F k ma where kk

` 2

1

1

m m

The acceleration of an object is directly proportional to the NET FORCE and inversely proportional to the mass.

maFm

Fa

maFa

NETNET

NET

1 FFNET

Tips:•Draw an FBD•Resolve vectors into components•Write equations of motion by adding and subtracting vectors to find the NET FORCE. Always write larger force – smaller force.•Solve for any unknowns

zz

yy

x

maF

maF

F

xma

A 10-kg box is being pulled across a frictionless table to the right by a rope with an applied force of 50N. Calculate the acceleration of the box.

mg

FNFa

2/5

0.100.50

sma

a

maFNet

In which direction, is this object accelerating?

The X direction!

So N.S.L. is worked out using the forces in the “x” direction only

A 10-kg box is being pulled across the table to the right by a rope with an applied force of 50N. Calculate the acceleration of the box if a 12 N frictional force acts upon it.

mg

FNFa

Ff

2/8.3

101250

sma

a

maFF

maF

fa

Net

In which

direction, is this object accelerating?

The X direction!

So N.S.L. is worked out using the forces in the “x” direction only

Slope

F

a

Slope = rise / run

F

a

Force Acceleration

Mass

5 N 2 m/s2

10 N 4 m/s2

15 N 6 m/s2

Slope

F

a

Slope = rise / run = F / a, the slope is equal to the mass. Or, think of y = m x + b, like in algebra class. y corresponds to force, m to mass, x to acceleration, and b (the

y-intercept) is zero.

F

a

Friction- A non-conservative force that opposes motion- Acts parallel to the surfaces in contact.- µ is the coefficient of friction.

- It is unique for each material- It is determined experimentally

What about surface area? Which would have a greater friction force?

Friction force does not depend on the area of contact. Same FN and µ!

stress = F a

TWO types of Friction Static – Friction that keeps an object at rest

and prevents it from moving Kinetic – Friction that acts during motion

-A force that opposes motion

-Acts parallel to the surfacesin contact.

Force of Friction

The Force of Friction is directly related to the Force Normal.

Mostly due to the fact that BOTH are surface forces

Nkkf

Nssf

Nf

FF

FF

FF

friction oft coefficien

alityproportion ofconstant

Note: Friction ONLY depends on the MATERIALS sliding against each other, NOT on surface area.

The coefficient of friction is a unitless constant that is specific to the material type and usually less than one.

Pause for a CauseA 1500 N crate is being pushed

across a level floor at a constant speed by a force F of 600 N at an angle of 20° below the horizontal as shown in the figure.

a) What is the coefficient of kinetic friction between the crate and the floor?

mg

FNFa

20

Ff

Fay

Fax

NN

N

FF

F

FF

F

F

F

FF

k

Ny

x

Ny

f

N

fk

Nkf

150020sin600

20cos600

sin

cos

kk

= 0.33

Pause for a CauseIf the 600 N force is instead pulling the

block at an angle of 20° above the horizontal as shown in the figure, what will be the acceleration of the crate. Assume that the coefficient of friction is the same as found in (a)

mg

FN

Ff

20

Fa

Fax

Fay

m

Fa

maF

Net

Net

a

a

9.152

57.4288.563

81.91500

)20sin6001500(331.020cos600

gF

FFF

Net

yNx )sin(cos

m

FFF yfx )(

= 0.884 m/s2

FN = mg

Newton’s Third Law“For every action there is an EQUAL and

OPPOSITE reaction. This law focuses on action/reaction pairs (forces) They NEVER cancel out

All you do is SWITCH the wording!•PERSON on WALL•WALL on PERSON

N.T.LThis figure shows the force during a collision between a truck and a train. You can clearly see the forces are EQUAL and OPPOSITE. To help you understand the law better, look at this situation from the point of view of Newton’s Second Law.

TrainTrainTruckTruck

TrainTruck

aMAm

FF

There is a balance between the mass and acceleration. One object usually has a LARGE MASS and a SMALL ACCELERATION, while the other has a SMALL MASS (comparatively) and a LARGE ACCELERATION.

N.T.L ExamplesAction: HAMMER HITS NAILReaction: NAIL HITS HAMMER

Action: Earth pulls on YOUReaction: YOU pull on the earth

TensionIn physics, tension is the pulling force exerted by a string, cable, chain, or similar solid object on another object.

m1g

m2g

T

T

FN

A mass, m1 = 3.00kg, is resting on a frictionless horizontal table is connected to a cable that passes over a pulley and then is fastened to a hanging mass, m2 = 11.0 kg as shown below. Find the acceleration of each mass and the tension in the cable. amTgmm

amTm

maFNet

222

11

14

)8.9)(11(

)(

21

2

122

mm

gma

mmagm

amamgm

amamgm

122

212

Acceleration = 7.7 m/s2

amT

amTgm

maFNet

1

22

NT 1.23)7.7)(3(

Run

RiseSlope

ma

FmaF NET

Net

A traffic light weighing 100 N hangs from a vertical cable tied to two other cables that are fastened to a support at the angle illustrated. Find the tension in each of the three cables.

T3 =>T3 - Fg

T3 – Fg = 0equilibrium

T3 = Fg

T3 = 100 N

ΣFx = T1x - T2x

ΣFy = T1y + T2y = T3

ΣFx = T1x cosθ - T2x cosθ = 0

ΣFy = T1y sinθ + T2y sinθ – T3= 0

Hints:ΣFx there are two forces in the x-axis and they are opposing each other

ΣFy there three forces in the y-axis T1 & T2 are opposing T3

1) ΣFx = T1x cosθ - T2x cosθ = 0

2) ΣFy = T1y sinθ + T2y sinθ – T3= 0

There two equations & two unknows

Solve equation 1 for T2

T2x = T1x * cos37˚

cos53˚T2x = 1.33T1x

T1 sinθ + T2 sinθ – T3= 0

T1 sinθ + 1.33T1 sinθ – T3= 0

T1 sinθ + 1.33T1 sinθ = T3

T1 (sinθ + 1.33sinθ) = T3

T1 = _____T3_______

(sinθ + 1.33sinθ)

T1 = _____100 N_____

(sin37˚ + 1.33sin53˚)T1 = 60.1 N

T2 = 79.9 N

T1x cos37˚ - T2x cos53˚ = 0T2x cos53˚ = T1x cos37˚

Inclines

cosmg

sinmg

mg

FNFf

Tips•Rotate Axis•Break weight into components•Write equations of motion or equilibrium•Solve

Pause for a Cause

m2

m1

Two packing crates of masses 10.0 kg and 5.0 kg are connected by a light string that passes over a frictionless pulley. The 5.00 kg crate lies on a smooth incline of angle 40.0˚. Find the acceleration of the 5.00 kg crate and the tension in the string.

m1g

m2g

FNT

T

Ff

40

40

m2gcos40

m2gsin40

amTgmm

amTgmm

222

111

sin)2

)1

maFNET

Pause for a Cause

m2

m1

m1g

m2g

FNT

T

Ff

40

40

m2gcos40

m2gsin40

amTgmm

amTgmm

222

111

sin)2

)1

maFNET

)(

)sin(

)()sin(

sin

sin)2

)1

12

12

1212

1212

21122

111

mm

mmga

mmammg

termslikefactor

amamgmgm

amamgmgmm

amgmTm

44.4)0.50.10(

)0.1040sin0.5(8.9

a

Solve 1) for T

Sub 1) for T into 2) for T

Pause for a Cause

gmamTamgmT

maFNET

1111)1

m2

m1

Masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a frictionless pulley. As shown in the diagram, m1 is held at rest on the floor and m2 rests on a fixed incline of angle 40 degrees. The masses are released from rest, and m2 slides 1.00 m down the incline in 4 seconds. Determine (a) The acceleration of each mass (b) The coefficient of kinetic friction and (c) the tension in the string.

m1g

m2g

FNT

T

Ff

40

40 amTFgm f 22 )(sin)2 m2gcos40

m2gsin40

cos

cos

2

2

gmF

gmF

FF

kf

N

Nkf

Example

cos

sin

cossin

cossin

sin

)(sin

sin

2

2112

22112

21122

2112

2112

22

gm

amgmamgm

gmamgmamgm

amgmamgmgm

amgmamFgm

amgmamFgm

amTFgm

k

k

k

Nk

f

f

2

2

2

/125.0

)4(2101

21

sma

a

attvx ox

gmamTamgmT

maFNET

1111

amTFgm f 22 )(sin

NT 7.39)8.9(4)125(.4

235.057.67

125.12.395.07.56

k

The End