REVISIONNEWTON’S LAW

NEWTON'S UNIVERSAL GRAVITATION LAW

Each body in the universe attracts every other body with a force that is directly proportional to the

product of their masses and inversely proportional to the square of the distance between their centres

1 22

Gm mF

r

WEIGHTLESSNESS

1 22 g

Gm mF F mg

r

In the absense of any contact forcesOnly gravitation (non-contact force) acts on the object

There is a gravitational force F between objects A and B

at a distance R from each other.

What will the gravitational force be if the mass of A is

doubled and the distance of separation

made three ti 2

9mes greater ? F

RATIO PROBLEMS

RATIO PROBLEMS

22 2

2 2

2

3

2

32

9

A B Ai f

A B

i

km m k m mF F

R R

km m

R

F

363  , 517    a

Three spheres li

nd 154 

e stationary in a straight line.

The masses of the spheres are

Determine the magnitude and direction of the net gravitatio ...

Sphere A

nal force

5.7

onA B Cm kg m kg m kg

a

5

5

5

10 right

3.5 10 right  Sphere B

Sph 9 10 er lefe C t

b

c

N

N

N

NET GRAVITATIONAL FORCE

NET GRAVITATIONAL FORCE

2

11

2

11 5

2

11

2

11 6

between A & B

?

363

517 6.67 10 363 517

0.5 0.5

6.67 10 5 10

between A & C

?

363

154 6.67 10 363 154

0.75 0.7

take right

5

6.67 10 6.6 10

AB A BAB

ABA

B

AB

AB A CAC

ACA

C

AC

F

F Gm mF

rm

m

r

G N

F

F Gm mF

rm

m

r

G N

5 6

5

:

5 10 6.6 10

5.7

as positive

10 right

net AB ACA F F Fa

N

NET GRAVITATIONAL FORCE

2

11

2

11 5

2

11

2

11 5

between A & B

?

363

517 6.67 10 363 517

0.5 0.5

6.67 10 5 10

between B & C

?

517

154 6.67 10 517 154

0.25 0.2

take right

5

6.67 10 8.5 10

AB A BAB

ABA

B

AB

AB B CBC

BCB

C

BC

F

F Gm mF

rm

m

r

G N

F

F Gm mF

rm

m

r

G N

5 5

5

:

8.5 10

as p

5 10

3.5 10 righ

osit ve

t

i

net BC ABB F F Fb

N

NET GRAVITATIONAL FORCE

2

11

2

11 6

2

11

2

11 5

between A & C

?

363

154 6.67 10 363 154

0.75 0.75

6.67 10 6.6 10

between B & C

?

517

154 6.67 10 517 154

0.25 0.25

6.67 10 8.5

r

10

take

AB A CAC

ACA

C

AC

AB B CBC

BCB

C

BC

F

F Gm mF

rm

m

r

G N

F

F Gm mF

rm

m

r

G N

6 5

5

5

:

6.6 10 8.5 10

9 10

9 10 l

ight as positi

e

v

t

e

f

net AC BCC F F Fc

N

N

the force or the component of a force which a surface exerts on an object with which it is in contact, and which is perpendicular to

the surface

NORMAL FORCE

WEIGHTgravitational force the Earth exerts on any object on or near its

surface

the force that opposes the motion of an object acts parallel to the surface.

– Perpendicular on normal force– independent of the area of

contact – independent of the velocity of

motion

FRICTIONAL FORCE

STATIC FRICTIONAL FORCE the force that opposes the tendency

of motion of a stationary object relative to a surface

KINETIC FRICTIONAL FORCE the force that opposes the motion of a

moving object relative to a surface

Nf F

• Is usually given• It depends on the 2 surface areas that are in

DIRECT contact• This is the relationship between the normal force

and the friction force of an object (the coefficient of friction therefor has no unit of measure)

COEFFICIENT OF FRICTION

NF

f

NEWTON'S FIRST LAW OF MOTION

A body will remain in its state of rest or motion at constant velocity unless a non-zero resultant/net force

acts on it.

NEWTON'S SECOND LAW OF MOTION

When a resultant/net force acts on an object, the object will accelerate in the direction of the force at an acceleration directly

proportional to the force and inversely proportional to the mass of the object.

Fres=ma

NEWTON'S THIRD LAW OF MOTION

When one body exerts a force on a second body, the second body exerts a force of equal magnitude in the

opposite direction on the first body

FORCE DIAGRAMS

FORCE DIAGRAMS

normal force

F Gravitation (weight)

friction

F Applied force

N

g

T

F

f

sin cos

cossin

cossin

y x

TT

x Ty T

t a

s sF F

FF

F FF F

normal force

F Gravitation (weight)

friction

F Vertical komponent of pplied force

F Horizontal komonent of pplied force

N

g

y

x

F

f

a

a

FORCE DIAGRAMS

||

normal force

friction

F Vertical komponent of weight

F Vertical komponent of weight

N

g

g

F

f

||

||

sin cos

sin cos

sin cos

g g

g g

g g g g

t a

s sF F

F F

F F F F

o

A block of mass 1 kg is connected to another block of mass 4 kg by a light inextensible string.

The system is pulled up a rough plane inclined at 30 to the horizontal, by means of a constant

40 N force parallel to the plane as shown in the diagram above.

The magnitude of the kinetic frictional force between the surface and the 4 kg block is 10 N.

The coefficient of kinetic friction between the 1 k

g block and the surface is 0,29.

State Newton's third law in words.

Draw a labelled free-body diagram showing ALL the forces acting on the 1 kg block as it

moves up the incline.

Calculate the magnit

a

b

ude of the:

Kinetic frictional force between the 1 kg block and the surface

Tension in the string connecting the two blocks

c

d

||

||

o o

Vir die 1kg blok

sin cos

sin cos

sin cos

1 9.8 sin 30 1 9.8 cos30

g g

g g

g g g g

t a

s sF F

F F

F F F F

||

||

o o

Vir die 4kg blok

sin cos

sin cos

sin cos

4 9.8 sin 30 4 9.8 cos30

g g

g g

g g g g

t a

s sF F

F F

F F F F

o

||

o

sien left

0.29 8.48take up as positive

2.46 up

0

1 9.8 cos30

8.48

for 1kg block

-

take up as positive

F

40 2.45 1 9.8 sin 30 1

32.6 1

for 4kg blo

N

res

N g

N g

res

T g

a

y axis f Fb

F ma N

F F

F F

N

c

x axis

ma

F f T F ma

T a

T a

||

o

2

ck

-

neem 'op' as positief

F

10 4 9.8 sin 30 4

29.6 4

T=T Susbtitute a in 29.6 4

32.6 1 29.6 4 29.6 4 0.6

3.03 5 32.03

0.6 .

res

g

x as

ma

T f F ma

T a

T a

let T a

a a

a N

a m s

||

||

o o

Vir die 1kg blok

sin cos

sin cos

sin cos

1 9.8 sin 30 1 9.8 cos30

g g

g g

g g g g

t a

s sF F

F F

F F F F

||

||

o o

Vir die 4kg blok

sin cos

sin cos

sin cos

4 9.8 sin 30 4 9.8 cos30

g g

g g

g g g g

t a

s sF F

F F

F F F F

o

see left

0.29 8.48take up as positive

2.46 down

0

1 9.8 cos 30

8.48

N

res

N g

N g

a

y axis f Fb

F ma N

F F

F F

N

||

o

for 1kg block

-

take up as positive

F

40 2.45 1 9.8 sin 30 1

32.6 1

for 4kg block

res

T g

c

x axis

ma

F f T F ma

T a

T a

||

o

for 4kg block

-

take 'up' as positive

F

10 4 9.8 sin 30 4

29.6 4

res

g

c

x axis

ma

T f F ma

T a

T a

2

T=T Susbtitute a in 29.6 4

32.6 1 29.6 4 29.6 4 0.6

3.03 5 32.03

0.6 .

let T a

a a

a N

a m s

NEWTON'S THIRD LAW

ACTION-REACTION PAIRS1. Same magnitude, opposite direction2. Simultaneous3. Only two objects

REVISIONWORK

WORKWork done on an object is the project of the

force and the displacement in the direction of the force.

Positive net workEnergy used to enhance motion

Negative net workEnergy used to prevent motion

cosW F x

WORK-ENERGY THEOREMThe net/total work done on an object is

equal to the change in the object's kinetic energy

OR the work done on an object by a

resultant/net force is equal to the change in the object's kinetic energy.

net kW E

TYPES OF FORCESCONSERVATIVE FORCE

a force for which the work done in moving an object between two points is independent of the path taken. Examples are gravitational force, the elastic force in a

spring and electrostatic forces (coulomb forces).

NON-CONSERVATIVE FORCEa force for which the work done in moving an object

between two points depends on the path taken. Examples are frictional force, air resistance, tension in a

chord

CONSERVATION OF MECHANICAL ENERGY

The total mechanical energy (sum of gravitational potential energy and

kinetic energy) in an isolated system remains constant

nk k pW E E

POWERthe rate at which work is done or energy is

expended.

WP

t

ave aveP Fv

o

A man pulls a box with a mass of 25kg from rest across a horizontal surface.

He applies a constant force of 40N, with an angle of 40 with the horizontal axis.

While in motion, the box experiences a c

onstant friction force of 10N

Draw a free body diagram and show ALL forces acting on the box

Give a reason why none of the vertical forces perform any work on the box

Determine the net work done on

.

a

b

c

the box when it moves from A to B (5m apart)

Use the work-energy theorem to determine the speed of the box when it reached B

The man now applies the same force, but with a smaller angle with the horiz

d

e ontal.

How does the work performed now (with the smaller angle) relate to the answer in c)

Give a reason for your answer (no calculation necess

ary)

WORK WITH RESULTANT FORCE

oo

sin cos

cossin

cossin

40cos 4040sin 40

y x

Tt

x Ty T

t a

s sF F

FF

F FF F

normal force normal force

F Gravitation (weight) F Gravitation (weight)

friction friction

F Vertical komponent of pplied force F Applied force

F Horizontal komone

The ver

nt of pplied force

N N

g g

y T

x

ofF

fa

F

f

a

a

b

otical forces are all 90 to the direction of motion, so that W=F cos 0x J

oo oo

determine W determine W determine W

cos cos cos

5 cos 010 5 cos180 5 cos 40

153.250 153.2

50 153.250 153.2

103.2103.2

40 cos 40 40

x x

T

x T

f F F

f f Fx x F T

net f F net f F

W F x W F x W F x

JJ J

W W W OR W W W

JJ

or

o

o

40cos 40 10

20.64 regs

determine F determine W

take right as positive cos

F 20.64 5 cos 0

103.2

res net

net res

res x

N

W F x

c F f

J

2 2

22

2

2

1

=

1 1

2 21 1

103.2 25 25 02 21

103.2 25 02

8.256

2.9 .

As die hoek kleiner word, sal cos toeneem

daarom sal die netto arbeid ook toeneem

T

T

net K

f i

f

f

f

f

net f F

F T

d W E

mv mv

v

v

v

v m s

e W W W

W F x

oMegan slides down a 10m slide that has a 30 angle of elevation.

If she weighs 60kg and a frictional force of 26N is present...

Draw a free body diagram

Determine the work done by friction

Determine t

a

b

c

he work done by gravity

Determine the total work done on Megan

Determine her velocity at the bottom of the slide

d

e

WORK WITHOUT RESULTING FORCE

|

o

|

normal force normal force

friction friction

F weight F Vertical component of weight

F Vertical component of weight

cos

26 10 cos180

260

N N

g g

g

f fW

F F

f

J

fa

xb F

|| ||

o o

||

||

o

o

sin60 9.8

588 sin

sin

588sin

cos cos

10 cos 60 10 cos 0

30

588 58

2940 29 0

n 3

4

8si 0

g g

g

g

g

g g

F g F gW F x

F m

W F x

g

J J

tcsF

NF

F F

of

o

2 2

22

2

2

1

||

o260 2940

2680

cos

268 10 cos 0

2680

=

1 1

2 21 1

2680 60 60 02 21

268

588sin 3

0 60 02

0 26

26

89.3

9.5 .

8

gnet f F

net res

net K

f i

f

re

f

f

s g

f

W W W

J

W F x

J

W E

mv mv

v

of F F f

d

N

v

v

v m s

e

A stationary 5kg block is released from a

height of 5m. It slides down the frictionless

slope to point P. It then moves across the

frictionless horizontal plane PQ en finally

moves up a second rough s

1

urface QR where

it comes to rest at R, 3m above ground.

It experiences a constant friction force of 18N

while in motion. The speed at Q is 9.89

alculate the angle θ of the slope QR

94m.s

a C

WORK WITH NON-CONSERVATIVE FORCES

2 2

o

2 2

cos 1 1

2 218 cos180 5 9.8 3 5 9.8 01 118 1475 0 5 9.89942 2

24518

18 245 147

nc k p

f p f ik f i

nc f

nc k p

W E E

W f x E mgh mghE mv mvx

x NW W

Jx

W E E

x

5.44x m

o

in the triangle

sin =

3sin

5.44

33.5

t

s

5.44