1

4 Topics

• force and net force

• inertia and 1st law

• acceleration and 2nd law

• g notation

• force pairs and 3rd law

• force diagrams

• equilibrium

• friction

2

Force Concept

Force = push or pull

Contact Forces – requires touch

Ex: car on road, ball bounce

Non-Contact – does not require touch

Ex: magnetism, gravity

3

Net Force

FFFFnet

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vector sum of all forces acting on an object

4

Example Motion Diagram when Fnet = 0

Newton’s First Law: An object maintains an unchanged constant velocity unless or until it is acted on by a non-zero Net Force.

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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 = 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 Second Law: When a Net External Force acts on an object with mass m, the resulting acceleration of the object is parallel to the net external force and has magnitude of

m

Fa

Example Motion Diagrams when Fnet ≠ 0

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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?

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Inertia• The ‘resistance’ to a change in velocity

• Ex: accelerating a ping pong ball• Ex: accelerating a train

• Measurement: Mass

• SI Unit: Kilogram (Kg)

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Newton’s Third Law: Whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body

attraction

repulsion

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Motion of Ball

Force on Ball Force on Wall

Acceleration of BallAcceleration of Wall

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

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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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upward (decreasing) velocity

Fnet

acceleration

Ex: Newton’s 2nd Law

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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.

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Contact Forces

• Normal Force – perpendicular to surface

• Frictional Force – parallel to surface Static (no sliding) Kinetic (sliding)

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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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Static Friction• Objects usually “stick” when at rest

• Must be “budged” to get them moving

• “Budging force” = fs,max.

• fs ranges from 0 to fs,max.

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Kinetic Friction• After objects breaks free, friction

decreases

• fk = force that just keeps object moving at a steady speed

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Coefficients of Friction

N

ss F

f max, dimensionless (no units)

N

kk F

f dimensionless (no units)

Ex. tire on dry road, static coeff. ~ 1.0

Ex. tire on dry road, kinetic coeff. ~ 0.8

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4 Summary

• Newton’s Laws of Motion

• force-diagrams and net force

• contact forces: normal, frictional, other

• equilibrium

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Newton’s 2nd Law Examples

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Block on Frictionless Incline

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Two Connected Blocks

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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

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Example:

29

30

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

NF

w F

30

40

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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

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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?