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Chap. 4: Newton’s Law

of Motion

Force; Newton’s 3 Laws; Mass and Weight

Free-body Diagram (1D)

Free-body Diagram (1D, 2 Bodies)

Free-body Diagram (2D)

Equilibrium

Frictional Force

Circular Motion / Rotation

1

And Chap.5 Applying Newton’s Laws (more examples)

F

Who Wanted “Force”?

2

What is Force?

an interaction between two objects

a push or a pull on an object; either a contact force

or a long-range force

Push up on an apple by hand; contact force

Pull down on an apple by earth; long-range force

a vector

it has both a magnitude and a direction

causes an acceleration on an object

2

2

m/s 8.9

; ˆ)m/s 8.9(ˆ)0(

gg

jig

jmgiF ˆ)(ˆ)0(g

y

g

g

v

v

4

Newton’s Laws of Motion

Newton’s 3 Laws

Newton’s 1st Law

Law of inertia

“Mass” (kg) is a measure of the inertia of a body.

Newton’s 2nd Law

Dynamical analysis using Free-Body Diagram (FBD)

Newton’s 3rd Law

Action-reaction pair

m

Fa

5

Newton’s Laws of Motion

Kinematics

(r, v, a)

Structure of Newtonian Mechanics

Inertial Reference Frame

(Newton’s 1st Law)

Action-Reaction

(Newton’s 3rd Law)

Mass

(m)

The Nature of Force The Nature of Object The Nature of Motion

F = m a

(Newton’s 2nd Law)

Kinematics

(r, v, a)

Kinematics

(r, v, a)

Kinematics

(r, v, a)

6

Newton’s Laws of Motion

WHAT HAPPEN?

9

Newton’s Laws of Motion

The force on a hokey puck

causes the acceleration

If the net force on a hokey

puck is zero (equilibrium),

the acceleration is zero.

0

0

a

F

11

Force: Acceleration/Equilibrium Acceleration

Kinetic Equations

(see Chap. 2 &3)

Newton’s Laws of Motion

12

Diagram to Understand Forces

The force of the starting

block on the runner has a

vertical component that

counteracts her weight and a

large horizontal component

that accelerate her.

Vector Nature

(see Chap. 1)

Newton’s Laws of Motion

The two cases are identical as far as the acceleration of

the box is concerned. This demonstrates the VECTOR

NATURE of the force. FFFFF

321R

13

Force: Vector

Newton’s Laws of Motion

1300 N

1300 N

90o

FFFF

21R

14

Similar to Fig. P4.37

Vector direction x-y coordinates

Find the magnitude of F2 and its direction relative

to F1.

Newton’s Laws of Motion

1300 N

1300 N

90o

FFFF

21R

15

Similar to Fig. P4.37 x

y

yyy

xxx

FFR

FFR

21

21

?

?

01300

13000

Find the magnitude of F2 and its direction relative

to F1.

Newton’s Laws of Motion

Find the tensions in each of three chains when the

weight of a car engine is W.

0321O)Point R(at

TTTF

17

Fig. 5.3

Also see Fig. E4.2

Newton’s Laws of Motion

What is 2nd Law?

m

Fa

m

Fa

m

Fa

m

Fa

z

z

y

y

xx

18

Newton’s Laws of Motion

y

x Impulsive Force “F ”

(for a short time

interval)

Top View

vk

(1) A hockey puck is initially

at rest on a flat ice surface.

(2) Then, it receives

a horizontal kick in

a direction of the

red arrow.

F = m a Change in velocity for non-zero F

[Quick Quiz] The motion of the puck

right after the kick is:

(a) Motion with constant velocity

(b) Motion with constant acceleration

(c) Motion with constant deceleration

(d) None of above 19

Newton’s Laws of Motion

Which of the path 1-5 below would the puck most

closely follow after receiving the force?

y

x

Top View

A hockey puck is sliding

at constant velocity

on a flat ice surface

5 4 3

2

1

v0

Impulsive Force “F ”

(for a short time

interval)

21

Newton’s Laws of Motion

vx2 = v0x

2 + 2 ax (x – x0)

x – x0 = 55 m

v0x = 28 m/s

Was: What is the acceleration?

25

Newton’s Laws of Motion

Fxnet = m ax

x – x0 = 55 m

m = 1500 kg

Fxnet?

v0x = 28 m/s

Was: What is the acceleration?

Now: What is the net force?

26

vx2 = v0x

2 + 2 ax (x – x0)

Newton’s Laws of Motion

vx2 = v0x

2 + 2 ax (x – x0)

Fxnet = m ax

x – x0 = 55 m

m = 1500 kg

Fxnet?

v0x = 28 m/s

Was: What is the acceleration?

Now: What is the net force?

vx2 = v0x

2 + 2 ax (x – x0) ax = 7.1 m/s2

SFx = m ax (1500 kg) x (7.1 m/s2)

27

• If you exert a force on a body, the body always exerts a force (the “reaction”) back upon you.

• Figure 4.25 shows “an action-reaction pair.”

• A force and its reaction force have the same magnitude but opposite directions. These forces act on different bodies. [Follow Conceptual Example 4.8]

What is 3rd Law?

Newton’s Laws of Motion

Which force is greater?

From Giancoli 3rd ed.

29

Newton’s Laws of Motion

What is 3rd Law? (II)

Push forward

on a rocket

by gas

Push backward

on gas

by a rocket

Action and reaction forces have the same magnitude

but are opposite in direction.

30

Newton’s Laws of Motion

[Quick Quiz] A massive truck collides head-on

with a small sports car.

1) Which vehicle experiences the greater force of

impact?

31

2) Which vehicle experiences the greater

acceleration?

Newton’s Laws of Motion

A stonemason drags a marble block across

a floor by pulling on a rope attached to the

block. Why does the block move while the

stonemason remains stationary?

34

Newton’s Laws of Motion

FP-G

Force on

Rope exerted

By Person

FR-P =10 N

Force on

Box exerted

By Rope

Assume the rope is massless.

At rest

x

What is the tension here?

0? 10? 20? Or else?

Quick Quiz

FB-R =10 N

38

Newton’s Laws of Motion

Free-Body Diagram

jmgiF ˆ)(ˆ)0(g

y

g

g

v

v

Draw only force(s) on the apple

41

Newton’s Laws of Motion

A hockey puck is sliding at a

constant velocity across a flat

horizontal ice surface. Which is

the correct free-body diagram?

[A]

44

Quick Quiz

Newton’s Laws of Motion

Free-Body Diagram (1D)

46

Newton’s Laws of Motion

Free-Body Diagram (1D) M

oti

on

y

m = 10.0 kg

47

Find ay

Newton’s Laws of Motion

ay = 0

Weight in Elevator

y

ay = 4.9 m/s2

49

m = 60 kg

Fg = mg = 588 N

SFy= m ay FN = ?

Example 5.9

Newton’s Laws of Motion

53

Weight on Planet

Newton’s Laws of Motion

FBD (1B/1D 2B/1D) M

oti

on

y

54

Find ay

Motion

Find ax

Newton’s Laws of Motion

Free-Body Diagram (x-axis)

Motion

55

P4.43, P4.54

e.g., 2 boxes sliding on desk (no friction)

Find acceleration and tension.

Newton’s Laws of Motion

Find FP and FT

ax = 2.50 m/s2

FT FP

59

P4.43

Exercise

1) 1B 2B

2) Now a is given. Then you are asked to find forces.

Newton’s Laws of Motion

2) Apply Newton’s 2nd law:

e.g., 2 boxes sliding on desk (no friction)

Find acceleration and tension.

Free-Body Diagram (x-axis) The rope has a mass …

60

1) Draw F.B.D.

identify all forces exerted on box1, rope and box2

separately.

FP = 40.0 N

Newton’s Laws of Motion

Free-Body Diagram (y-axis)

65

Newton’s Laws of Motion

Example 1: One paint bucket (mass m1) is hanging by

a massless cord from another paint

bucket (mass m2), also hanging by a

massless cord. The two buckets are pulled

upward with an acceleration by the

upper cord. Draw the free-body

diagram for each bucket. Determine the

tension on each code if a = 3.00 m/s2.

a

a

y

66

P4.57

Newton’s Laws of Motion

Example 1: One paint bucket (mass m1) is hanging by

a massless cord from another paint

bucket (mass m2), also hanging by a

massless cord. The two buckets are pulled

upward with an acceleration by the

upper cord. Draw the free-body

diagram for each bucket. Determine the

tension on each code if a = 3.00 m/s2.

m1

m2 a

a

Fg1

Fg2

FT2

FT1

FT1

y

67

Newton’s Laws of Motion

Acceleration?

Tension at the midpoint of the rope?

71

P4.54

Newton’s Laws of Motion

72

Newton’s Laws of Motion

2)

e.g., Box sliding on desk (no friction)

Free-Body Diagram (2D)

1)

Motion

x

y

73

m = 10.0 kg

Newton’s Laws of Motion

2) Apply Newton’s 2nd law:

m ax =

m ay (=0) =

e.g., Box sliding on desk (no friction)

1) Draw F.B.D -identify all forces exerted on the box.

FN

FG

FP Motion

x

y

FP cos(30.0o)

FP sin(30.0o) + FN m g

74

Free-Body Diagram (2D)

Newton’s Laws of Motion

2) Apply Newton’s 2nd law:

10.0 ax =

10.0 (0) =

e.g., Box sliding on desk (no friction)

1) Draw F.B.D -identify all forces exerted on the box.

FN

FG

FP Motion

x

y

40.0 cos(30.0o)

40.0 sin(30.0o) + FN (10.0)(9.80)

75

Free-Body Diagram (2D)

Newton’s Laws of Motion

e.g., Box sliding on desk (no friction)

FN

FG

FP Motion

x

y

76

Free-Body Diagram (2D)

Try E4.4

Circular Motion

Free-Body Diagrams?