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4 Newton’s Laws

• Force, net-force, mass & inertia

• Newton’s Laws of Motion

• Weight, Contact Forces

• Labeling & Diagramming

• Hk: 37, 49, 53, 57, 59, 61, 65, 67.

22

Force Concept

Contact Forces

Ex: sliding, bouncing

Non-Contact

Ex: magnetism, gravity

/

33

Inertia

• is ‘resistance’ to change in velocity

• Measurement: Mass

• SI Unit: Kilogram (Kg)

• /

44

units

• Force units (SI): newton, N

• 1N ≈ ¼ lb.

• 1N = (1kg)(1m/s/s)

• N/kg = m/s/s

s

sm

kg

N /

5

Net Force

FFFFnet

21

vector sum of all forces acting on an object

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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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Weight Force

mgr

MmGFg

2

2211 /1067.6 kgmNG

s

sm

kg

N

r

MGg

/8.98.9

2

mgFweight g

8

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.

9

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.

1010

Force Label Notation

• F = general force

• FN = normal force

• f = frictional force

• w = mg = Fg = weight

• T = tension force

• /

1111

Net Force = change of motion

yxyxamamFF

vector sum of all forces acting on an object

amFnet

xxamF

yy

amF

1212

Problem Solving Template: Two Equations – Two Unknowns

xxamF

____________

yyamF

____________

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velocity

Example: Ball rolls along a smooth level surface

Force Diagram

table force

weight force

xxamF

0

0/0 max

ynyamwFF

wFa ny zero, is Since

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Example of a Force Diagram for a Sled

net force equals the mass times its acceleration.

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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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upward (decreasing) velocity

Fnet

acceleration

Ex: Ball rolling up & slowing down (Use PHET Vector Addition for net-force)

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xy

60cos)270cos(90cos)60cos( FwFFF Nx

wFFwFFF NNy 866.0)270sin(90sin)60sin(

NF

F60w

Ex. m=3kg, F=86N, 60° below horizontal.

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2/14

360cos86

60cos

sma

a

maFF

x

x

xx

weight) the times3(about

8.103

)8.9(360sin86

060sin

NF

F

wFFF

N

N

Ny

Ex. Continued

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Block on Inclined Plane

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Ex. Calculate Acceleration of Block on a Frictionless Plane inclined 30°

xx mamgF 30sin

Plane toparallel axis x Choose

mass) ofnt (independe

//9.430sin ssmag x

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Ex. Calculate Normal Force on a Block on a Plane inclined 30°

)0(30cos

Plane tonormal axisy Choose

mmgFF ny

mass)on depends (value

30cosmgFn

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Velocity Acceleration Net Force

+ +

– +

+ –

– –

Complete the table below for the sign of the net force. Sketch a motion diagram for each case.

2323

Newton’s 3rd Law of Motion

• equal-sized oppositely-directed forces

• Independent of mass

• Pair-notation

x x

2424

Newton’s 3rd Law Pair Notation

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

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

2525

Force Diagram each object. Which has greater acceleration when

released?SpringForce

SpringForce

x x

Acceleration= F/m

Acceleration= F/(2m)

26

Motion of Ball

Force on Ball Force on Block

Acceleration of BallAcceleration of Block

Newton’s Second and Third Laws in Operation: Ball hits a large block on a smooth level surface.

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Solving Two Body Problems

• Force diagram each object & system (usually with one axis parallel to the acceleration). Use clockwise coordinates for problems with pulleys.

• System has a force-pair that cancels out• Solve simplest diagram first, then use this

information in another diagram• “ma” is not a force• /

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Two Connected Blocks

)1)(100( TFleftx

NT 100

)1)(200( TTFFbothx

)1)(100( TFFrightx

NNTF 200100

NF 2000

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4 Summary

• Zero net-force; constant velocity

• Acceleration = net-force/mass

• All forces are pairs

• Labeling & diagramming

• Solving problems using x, y force template

• Solving two body problems

• /

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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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Newton’s 2nd Law Examples

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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 in?

Its acceleration is parallel to Fnet by Newton’s 2nd Law. So we need to determine the direction of Fnet.

),.(180tan,

,1 IIIIIquadsF

F

xnet

ynet

6.26

4

2tan 1

N

N

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

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What is the magnitude of the acceleration?

ssmkg

N

m

Fa

net//49.1

3

47.4

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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 x

x //110

10

Example:

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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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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

39

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2) 3kg box at rest on frictionless 30° inclined plane. F acts 40° below horizontal.

NF

w F

30

40

41

NF

w F

30

40

NNx FFFFF 5.0766.0120cos)40cos(

wFFwFFF NNy 642.0866.0)40sin(120sin

0 yx aa

05.0766.0 NFF

04.29642.0866.0 FFN

FFN 532.104.29642.0)532.1(866.0 FF

NF

F

FF

0.43

4.29684.0

4.29642.0326.1

NFFN 8.65532.1

xy

42

0

)8.65(5.0)0.43(766.0

?05.0766.0

NFF

Check of Previous Problem:

0

4.29)0.43(642.0)8.65(866.0

?04.29642.0866.0

FFN

xF

yF

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Q2. 3kg box at rest on frictionless 30° inclined plane. F acts horizontally. Calculate F and Fn.

NF

w F

30

NNx FFFFF 5.030sin

wFwFF NNy 866.030cos

05.0 NFF

0866.0 wFN

NwFN 9.33866.0/4.29866.0/

NFF N 97.165.0

xy

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3. Three boxes are pushed by force F along a horizontal frictionless surface.

F=26N

3kg5kg

2kg

Force diagram object 1 (left box)

F12, surface reaction force

NF

F

w 3kg

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F21, surface reaction force

NF

w

5kg

F23, surface reaction force

F32, surface reaction force

NF

w

2kg

Diagram object 2:

Diagram object 3:

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1212 26 FFFFx

Object1: 3kg

NF

w

21F23FObject2: 5kg

2321 FFFx

NF

w

32F

Object3: 2kg

32FFx

Object1+2+3: 3kg+5kg+2kg

)10(2626 32232112 xxx amaFFFFF 2/6.210/26 smax

NF

F

w

12F

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8.7)6.2(326 12 FFx3kg NF 2.1812

)6.2(52.18 232321 FFFFx5kg NF 2.523

)6.2(232 FFx2kg NF 2.532

Summary:

Stimulus=26N Reactions: 18.2N, 5.2N

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Q3. Recalculate problem3 with order switched to 5kg, 3kg, 2kg.

F=26N3kg

5kg2kg 6.210/26/ mFa xx

)6.2(232 FFx2kg NF 2.532

)6.2(32.5212321 FFFFx3kg NF 1321

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4. Modified Atwood Machine with frictionless plane

sin11 gmTFCW

TgmFCW 22

TgmgmTFCW 21

21 sin

21

21

21

21sin

mm

gmgm

mm

Fa CW

CW

Let m1 = 1kg, m2 = 2kg, = 30°.

2/9.421

)2(30sin)1(sm

ggaCW

)9.4(2)2(22 TgTgmFCW

NgT 8.98.9)2(

solve for a and T in terms of m1, m2:

sin1 gm

cos1 gm

T T

gm2

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Q4. Recalculate problem4 with m1 = 6kg m2 = 1kg.

CWCW ammTgmgmTF )(sin 212121

21

21 sin

mm

gmgmaCW

2/8.216

)1(30sin)6(sm

ggaCW

)8.2(1)1(22 TgTgmFCW

NgT 6.12)8.2()1(

Note that T > (m2)g

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2. Block stays at same place on frictionless wedge.

a) Draw a force diagram for the block with the forces to correct relative scale.

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b) Use sum of vertical forces to calculate the size of Fn.

c) Use Fn to calculate the size of the acceleration in m/s/s.

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1. A 0.88 kg block projected up plane. Acceleration is 5.5m/s/s directed down the plane. Sliding friction is present.

Name(s):___________________________________________

a) Draw a force diagram for the block after projection and moving up the plane. Label each force clearly.

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b) Calculate the kinetic frictional coefficient.

c) The block is projected down the plane. Draw a force diagram for the block after projection and moving down the plane. Label each force clearly.

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d) Calculate the net force acting down the plane in newtons.

e) Calculate the acceleration of the block in m/s/s.

f) Is the acceleration i) up the plane, or ii) down the plane?