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Chapter 4: Forces and Newton’s Laws of Motion

• Forces• Newton’s Three Laws of Motion• The Gravitational Force• Contact Forces (normal, friction, tension)• Application of Newton’s Second Law• Apparent Weight• Air Resistance• Fundamental Forces• CQ: 16, 18.• P: 3, 5, 21, 37, 41, 63, 73, 87, 97, 147.

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Force Concept

Contact Forces

Ex: car on road, ball bounce

Non-Contact

Ex: magnetism, gravity

/

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units

• Force units (SI): newton, N

• 1N ≈ ¼ lb.

• 1N = (1kg)(1m/s/s)

• N/kg = m/s/s

s

sm

kg

N /

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Inertia

• is ‘resistance’ to change in velocity

• Measurement: Mass

• SI Unit: Kilogram (Kg)

• /

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Universal Law of Gravity

• all matter is weakly attracted

• attraction is inverse-square with distance

• G = 6.67x10-11 N·m2/kg2

• Example: Two 100kg persons stand 1.0m apart

221

r

mmGF

NF 72

11 1067.6)1(

)100)(100(1067.6

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g vs G

• G is universal

• g ~ Mass and Radius

• /

m

Fg g

2r

MG

mrMm

G 2

mgFw g

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Contact Forces• Surfaces in contact are often under

compression: each surface pushes against the other. The outward push of each object is called the Normal Force.

• If the objects move (even slightly) parallel to their surface the resistance force experienced is called the frictional force.

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Normal forces are?

1. Always vertically upward.

2. Always vertically downward.

3. Can point in any direction.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

41 42 43 44 45

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Tension & Compression

• Compressed objects push outward away from their center (aka Normal Force).

• Stretched objects pull toward their center. This is called the Tension Force.

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Force Label Notation

• F = general force

• FN = normal force

• f = frictional force

• w = mg = Fg = weight

• T = tension force

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1111

Net Force = change of motion

yxyxamamFF

vector sum of all forces acting on an object

amFnet

xxamF

yyamF

1212

constant velocity

Force Diagram

Fnet = 0

a = 0

Example: Net Force = 0, Ball rolls along a smooth level surface

table force

weight force

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Example: Net-force on 0.5kg

• Net-force = 4N: Acceleration = 4N/0.5kg = 8m/s/s

• 5N, Right; 3N Left; Net-force = 2NAcceleration = 2N/0.5kg = 4m/s/s

• Falling; Net-force = mgAcceleration = mg/m = g = 9.8m/s/s

• /

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1. An object maintains constant velocity when the Net-Force on it is zero.

3. Forces always occur in pairs equal in size and opposite in direction.

2. An object’s acceleration equals the Net-Force on it divided by its mass.

Newton’s Laws of Motion

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Force Diagrams

• Object is drawn as a “point”

• Each force is drawn as a “pulling” vector

• Each force is labeled

• Relevant Angles are shown

• x, y axes are written offset from diagram

• Only forces which act ON the object are shown

NF

w F

30

40

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Example of a Force Diagram for a Sled

net force equals the mass times its acceleration.

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g’s

• one “g” of acceleration = 9.8m/s/s

• “two g’s” = 19.6m/s/s, etc.

• Example: What is the net force on a 2100kg SUV that is accelerating at 0.75g?

• Net-force = ma = m(0.75g) = 0.75mg = ¾ weight of car.

• /

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Block on Frictionless Incline

• a = wx/m =mgsin/m

• = gsin.

• Fn = wy.

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Newton’s 3rd Law of Motion

• equal-sized oppositely-directed forces

• Independent of mass

• Pair-notation

x x

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Newton’s 3rd Law Pair Notation

• use “x” marks on forces that are 3rd Law pairs.

• Use “xx” for a different interaction, etc.

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Force Diagram each object. Which has greater acceleration when

released?SpringForce

SpringForce

x x

Acceleration= F/m

Acceleration= F/(2m)

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Friction• Static Friction “sticking force”

• Kinetic Friction “sliding force”

• Coefficients: 0 = min, 1 ~ max

• e.g. teflon around 0.05

• Rubber on concretearound 1.0

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Using Coefficients of Friction

• Ex. 10kg block. FN = weight = mg = 98N. Static coef. = 0.50; Kinetic coef. = 0.30.

Nss Ff max, Nkk Ff

N

Nf s49

)98)(50.0(max,

N

Nfk29

)98)(30.0(

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Applications

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A 3kg object sits on a frictionless table. Two horizontal forces act, one is 2N in the y-direction, the other 4N in the x-direction. A top-view diagram will be shown.

Fnet

What is the magnitude of the net-force acting?

4

22

2,

2, )()(|| ynetxnetnet FFF

490cos20cos4, xnetF

290sin20sin4, ynetF

NFnet 47.4)2()4(|| 22

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What direction does the 3kg mass accelerate?

parallel to Fnet by Newton’s 2nd Law.

),.(180tan,

,1 IIIIIquadsF

F

xnet

ynet

6.26

4

2tan 1

N

N

We are in Quadrant I since x and y are both +

Fnet

4

22

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The magnitude of the acceleration is:

ssmkg

N

m

Fa

net//49.1

3

47.4

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Two 1kg Blocks; a = 1m/s/s

• Fnet = F = (2m)a = (2kg)(1m/s/s) = 2N

• Fnet = T = ma = (1kg)(1m/s/s) = 1N

• /

F

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Two 1kg Blocks; F = 10N

• a = F/(2m) = 10N/2kg = 5 m/s/s

• T = ma = (1kg)(5m/s/s) = 5N

• /

F

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4 Summary

• Fnet = ma (Fnet = 0, v = constant)

• forces always occur in pairs of equal size and opposite direction

• various force types (& symbols)

• equilibrium problems (a = 0)

• dynamic problems (a ≠ 0)

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3090

6030

Mg, 300 deg.

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Inclined Plane Forces

• Fxnet = FNcos90 + mgcos300 = (0.02)(a)

• = 0 + (0.02)(9.8)(0.5) = (0.02)a

• accel = 4.9 m/s/s

• Fynet = FNsin90 + mgsin300 = (0.02)(0)

• FN + (0.02)(9.8)(-.866) = 0

• FN = 0.17N

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Fnet

acceleration

Ex: Newton’s 2nd Law

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Coefficients of FrictionEx: Block&Load = 580grams

NkgNkgmgFN 68.5)/8.9)(580.0(

If it takes 2.4N to get it moving and 2.0N to keep it moving

42.068.5

4.2max, N

N

F

f

N

ss

35.068.5

0.2

N

N

F

f

N

ks

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1. 3kg box on level frictionless surface. F=86N acts 60° below horizontal.

xy

300cos)270cos(90cos)300cos( FwFFF Nx

wFFwFFF NNy 866.0)270sin(90sin)300sin(

NF

F60w

Example:

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xy

0xa

0ya

xx maF

2/14

360cos86

60cos

sma

a

maF

x

x

x

yy maF

NF

F

wFF

N

N

N

8.103

)8.9(360sin86

060sin

1.(cont)

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Q1. What are ax and FN if angle is 30?NF

F30w

30cos)90cos(90cos)30cos( FwFFF Nx

wFFwFFF NNy 30sin)90sin()30sin(90sin

2/25

330cos86

30cos

sma

a

maF

x

x

x

NF

F

wFF

N

N

N

4.72

)8.9(330sin86

030sin

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Interaction Notation

• Since all forces are ‘pairs’, label as interactions, e.g. 1 on 2, 2 on 1, etc.

• F12 = “force of object 1 on object 2”

• F21 = “force of object 2 on object 1”

• F34 = “force of object 3 on object 4”

• Etc.

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Interaction Notation Symbols

• F12 – general force, 1 on 2

• N12 – normal contact force, 1 on 2

• f12 – frictional force, 1 on 2

• W12 – gravitational force, 1 on 2

• T12 – tension force, 1 on 2

• m12 – magnetic force, 1 on 2

• e12 – electrical force, 1 on 2

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Gravitational Force

• All masses attract via gravitational force

• Attraction is weak for two small objects

• Ex: Attraction between two bowling balls is so small it is hard to measure.

• Force is proportional to mass product

• Force is inversely proportional to the square of the distance between objects

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Example: Net Force = 0. Block on a surface inclined 30° from horizontal. Applied force F acts 40° below horizontal.

NF

w F

30

40

Net Force = 0

velocity = constant

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Diagrams with Interaction Notation

• If f21 exists, then f12 also exists, and is opposite in direction to f21.

• f21 and f12 act on different objects.

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A 10kg box is being pushed along a horizontal surface by a force of 15N. A frictional force of 5N acts against the motion. We will want to (a) Calculate the net-force acting and (b) calculate the acceleration of the box.

xx maNNNF 10515

0. yy maweightforceNormalF

The net-horizontal force determines its x-acceleration

The y-acceleration is known to be zero because it remains in horizontal motion, thus

The net-force is 10N horizontal (0 vertical)

The x-acceleration is: ssmkg

N

m

Fa xx //1

10

10

Example: