1

4 Topics

• force and net force

• types of forces

• Newton’s Laws & force diagrams

• Ch.4 Homework:

• 1, 3, 5, 6, 8, 13, 16, 23, 26, 34, 39, 45, 49, 62, 63, 66, 68, 69, 72, 81, 87, 90, 97, 99, 101, 105.

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

Force Label Notation

• Each force gets a distinctive label, and sketch & context supplies the interaction information

• F – general force

• FN – normal force

• f – frictional force

• W – weight

• T – tension force

4

Net Force

FFFFnet

21

vector sum of all forces acting on an object

5

constant velocity

Force Diagram

Fnet = 0

a = 0

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

table force

weight force

6

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.

7

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

8

Example of a Force Diagram for a Sled

net force equals the mass times its acceleration.

9

Newton’s Second Law: acceleration equals Net External Force (on object) divided by object mass:

mass

Fa

Example Motion Diagrams when Fnet ≠ 0

10

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?

11

units

• Force units (SI): newton, N

• 4.45N = 1lb.

• 1N = (1kg)(1m/s/s)

• N/kg = m/s/s

12

Inertia• is ‘resistance’ to change in velocity

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

• Measurement: Mass

• SI Unit: Kilogram (Kg)

3090

6030

Mg, 300 deg.

• 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

15

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

16

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.

17

upward (decreasing) velocity

Fnet

acceleration

Ex: Newton’s 2nd Law

18

Contact Forces

• Normal Force – perpendicular to surfaces

• Frictional Force – along surface. f ~ FN and to types of surfaces

19

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

20

Friction• Surfaces “stick” when at rest, this “static”

friction varies from 0 to “fs,max”

• Moving friction is called “fk” (~ indep.of v)

• Characterized by “coefficients”, “0” = frictionless, “1” is high value

• e.g. teflon around 0.05,

• Rubber on concretearound 1.0

21

Coefficient of Static Friction

• Ex. 10kg block sits on level surface with static coeff. frict. = 0.50. Force needed to budge = 0.50Fn

• = 0.50mg

• = 0.50(10kg)(9.8N/kg) = 49N.

N

ss F

f max, dimensionless (no units)

22

Coefficient of Sliding Friction

• Ex. 10kg moving on level surface with sliding frict. coef. 0.30. Force needed to keep it at const. vel. = 0.30Fn = 0.30mg

• =0.30(10kg)(9.8N/kg)= 29N.

N

kk F

f dimensionless (no units)

23

Velocity Acceleration Net Force

+ +

– +

+ –

– –

Complete the table below for the sign of the net force. Sketch a motion diagram for each case. (+) is rightward direction, (-) is leftward direction.

24

4 Summary

• if Fnet = 0, v = constant.

• Fnet = ma

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

• various forces (& symbols)

• equilibrium problems (a = 0)

• dynamic problems (a ≠ 0)

25

Block on Frictionless Incline

• a = wx/m =mgsin/m

• a = gsin.

• Fn = wy.

26

Two-Box Horizontal

27

One-Box Vertical

28

Two-Box Vertical

29

Force Diagrams: Free-fall vs. Terminal Velocity

30

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:

31

32

Two Connected Blocks

33

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

34

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 +

35

What is the magnitude of the acceleration?

ssmkg

N

m

Fa

net//49.1

3

47.4

36

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

37

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:

38

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)

39

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

40

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.

41

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

42

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

43

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

44

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.