11 Knowledge Area : mechanics : Term 1 Newton’s Laws 12fs1.beta.obami.com/Portal/eb3e6362-5a36-4dc6-b58c-a14700...What applied upward force ( upthrust ) enables a 102,04 kg rocket

11 Knowledge Area : mechanics : Term 1 Newton’s Laws 12TOPIC Newton’s First Law

Inertia • inertia is defined as the resistance of a body to changing its velocity

• we do not know what causes inertia • inertia can be described as the reluctance to get going or to stop going• inertia can be described as the tendency to remain in equilibrium

Static equilibrium is when an object remains at rest, dynamic equilibrium is when an object moves at a constant speed.Newton’s First LawAn object continues in a state of rest or uniform velocity unless it is acted upon by an unbalanced ( net or resultant )external force.

Fappf

( only showing horizontal forces )

constant speed

applied force or driving force

friction force

At constant speed in a straight line, F app = f

Vertical motion with a helicopter. Horizontal motion with a truck.

Fapp = m g The forces are in equilibrium when the helicopter• hovers stationary in the air or• moves upwards or downwards at a constant speed

m g = Fgrav = W ( weight )

Fappupward applied force or upthrust of air on blades Fapp

m g

box

100 Nforce

–100 Nforce

Invisible forces can have two effects on the motion of objects :1. Two twins exert equal forces on an object in opposite directions and their forces effectively cancel out or balance. The motion ( or state of rest ) of the box cannot change. This state is called equilibrium.

100 Nforce

–100 Nforce

2. One of the opposing forces is bigger than the other and this produces an unbalanced force that changes the motion ( or state of rest ) of the box.

Fnet = 0

box

180 Nforce

–100 Nforce

180 Nforce

–100 Nforce

Fnet = 80 N

There is no opposition to this net or resultant or unbalanced force.The box has to accelerate in the direction of the bigger force.This is Newton’s Second Law.

This is Newton’s First Law.

Uniform velocity means moving at a constant velocity.

Fapp f

11 Knowledge Area : mechanics : Term 1 Newton’s Laws 13The importance of seat beltsA man sits in a car travelling at a constant velocity, v.Both man and car have the same velocity.( The windscreen and top of the car have not been drawn in )

The car then hits a stationary train, and stops. Why ? Because an exter nal resultant force ( or net force ) is exerted on the car by the train, which stops the car.

car

But there is no resultant force acting on the man, as the train does not touch him. He continues to move forwards, due to his inertia, with the original velocity that the car had before the impact.He would then smash into the windscreen.How do we stop this ?We must attach him to the car with a safety belt that is bolted to the car and thus becomes part of the car.

Fnetcar seat train

FnetFnet

If the car now hits a stationary train, the net force exerted on the car by the train acts on the whole car, including the safety belt which is part of the car.The seat belt exerts a force on him to stop him.

The safety belt transfers this net force to the man and and his forward velocity stops when the car stops.

v

v

Consider a vehicle that is stationary or moving horizontally in a straight line at constant speed

Fapplied

Ffriction = f

Picture the applied force and the friction force as if they were people who are pushing on a vehicle.Fnormal = N

Fgravity = W

N

f

W

Force diagram

N

W

free-body diagramIf we try to find the horizontal component of a vertical force N that the surface exerts on the vehicle :

N

90º horizontal component = N cos 90º = N ( 0 ) = 0

N

There is no component at 90º

Fapplied f Fapplied

Fnet = 0

For objects that are in equilibrium ( at rest or moving in a straight line at constant speed ) :• all forces along the plane of the motion must add up to zero ( in this case the plane of motion is horizontal )• all forces in the direction perpendicular to the plane of the motion must add up to zero

Superposition enables us to add up two forces of equal magnitude that are in opposite directions, so that they cancel.

Newton’s First Law deals with balanced forces where objects continue in their state of rest or uniform linear motion.

Newton’s Second Law deals with unbalanced forces that change an object’s speed or direction of motion.Newton’s Second Law ( Newton 2 )When a net force, Fnet, is applied to an object of mass, m, it accelerates in the direction of the net force. The acceleration, a, is directly proportional to the net force and inversely proportional to the mass.

Fnet = m a

– 6 N15 N

– 6 N 15 N

Example

9 NFnet

The vehicle only moves horizontally, not vertically, so ignore the self-cancelling vertical forces.

9 N direction of accelerationdirection of

net force

equilibrium

v

11 Knowledge Area : mechanics : Term 1 Newton’s Laws 14

Fapplied

Ffriction = f f f Fapplied

Fapplied

Case 1 : Fapplied > f

Fnet = Fapplied + f Use positive values for the force symbols

= 200 – 60 Take the original direction of motion as positive = 140 N East The vehicle speeds up as Fnet is in the direction of motion.

200 N – 60 N Force diagram free-body diagram

Fnet

140 N

ExampleGiven a vehicle moving horizontally Eastwards with an applied driving force of 200 N and a friction force of – 60 N.

v

Superposition enables us to add up two forces of different magnitudes that are in opposite directions, to get a net or unbalanced force.

Fapplied

FN = N

f

Fg = W

Force diagram

N

W

free-body diagram

fFapplied

FappliedFnet

The vertical forces are self-cancelling.

Superposition enables us to add up two vertical forces of the same magnitude that are in opposite directions, to get a net force of zero in the vertical direction. The vertical forces are in equilibrium.Thus for horizontal motion we only need to consider the non-equilibrium horizontal forces.

non-equilibrium

f

Horizontal motion Example 1

An applied force of 15 N East acts on a 3 kg toy car moving to the East. 6 N of friction acts on the car.Determine the acceleration of the car.

– 6N

f

15N

Fapplied

Take original direction of motion as positive

If Fapp > f it speeds up, a = 3 m••s–2 is taken as positive.

If Fapp < f it slows down, a = – 3 m••s–2 is taken as positive.

Fapp + f = m a

15 – 6 = 3 a

∴ a = 3 m••s–2

Tip 1 :

Tip 2 :

– 6 N15 N

F net = m a

Fapplied

Ffriction = f f

f Fapplied

Fapplied

Case 2 : Fapplied < f

Fnet = Fapplied + f Use positive values for the force symbols

= 40 – 60 Take the original direction of motion as positive = –20 N East = 20 N West The vehicle slows down as Fnet is against the direction of motion.

40 N – 60 N Force diagram free-body diagram

Fnet

–20 N

ExampleGiven a vehicle moving horizontally Eastwards with an applied driving force of 40 N when a friction force of – 60 N acts on it.

v+

+

+

+

11 Knowledge Area : mechanics : Term 1 Newton’s Laws 15

Horizontal motion Example 2 A 3 kg toy car moving East experiences 18 N of friction.Calculate the applied force if it slows down at 4 m••s–2?

– 18 N

f Fapp

Fapp + f = m a

Fapp – 18 = 3 ( – 4 )

Fapp = 18 – 12 = 6 N East

– 18 NFapp

Take East as being positive.

F net = m a

+

Vertical motion Example 1 What upward applied force enables a 100 kg man to accelerate upwards at 5 m • s–2 in a lift?

mg = (100)(9,8) = 980 Ntrue weight of man

Fapp

Fapp + m g = m a

Fapp + ( 100 )(–9,8 ) = ( 100 )( + 5 )

Fapp = 980 + 500 = 1 480 N up( apparent weight )

If Fapp > mg it accelerates up, a = 5 m • s–2 (positive).

Fapp < mg it accelerates down, a = – 5 m • s–2 (negative).

Fapp + W = m a

Fg = W

cable

lift

Fapplied

a = 5 m • s–2+

W = Fg = m g

Weight depends on the downward force W = mg acting on us. But when we are in the air, as in diving, we do not feel our weight. Weight is thus the sensation that we feel when the upward reverse force from a firm surface acts on our feet as it opposes the downwards force mg acting on us.

1 480 N is the upward force the man experiences underneath his feet.This upwards force creates the sensation of weight. But his true weight is 980 N.So the 1 480 N is his apparent weight as he accelerates up, feeling heavier.

apparent weight

Vertical motion Example 2 What upward applied force, exerted by a cable on a lift, enables a 100 kg man to accelerate downwards at 5 m • s–2?

mg = (100)(9,8) = 980 Ntrue weight of man

Fapp

Fapp + m g = m a

Fapp + ( 100 )(–9,8 ) = ( 100 )( – 5 )

Fapp = 980 – 500 = 480 N up( apparent weight )

Fapp + W = m a

W

cable

lift

Fapplied

a = –5 m • s–2

+

W = m g

480 N is the upward force the man experiences underneath his feet.This upwards force creates the sensation of weight. But his true weight is 980 N.So the 480 N is his apparent weight as he accelerates down, feeling lighter.

apparent weight

Take up as positive for a lift or rocket.

Always take up as positive in a lift or rocket.

Take up as positive. g = –9,8 m • s–2

g = –9,8 m • s–2

Whether the lift goes up or down, the cable exerts an upward force on the lift.When the lift moves at a constant speed, up or down, then the upward Fapp = m g = WWhen the lift accelerates up, then the upward Fapp > m gWhen the lift accelerates down, then the upward Fapp < m g

100 kg

100 kg

11 Knowledge Area : mechanics : Term 1 Newton’s Laws 16Vertical motion Example 3 : Fire a rocket vertically upwardsWhat applied upward force ( upthrust ) enables a 102,04 kg rocket to accelerate upwards at 10 m•s–2 from rest, subject to a constant average resistance from air friction of 500 N. Assume the mass of the rocket remains constant. Take vertically upwards as positive.

Fapplied

W = Fg

Ffriction = f

Fapplied

W

Fair friction

force diagram

free-body diagram

Fnet = Fapplied + f + W m a = Fapp + f + W

W = = m g = ( 102,04 ) ( ─ 9,8 ) = ─ 1 000 N

Consider only the vertical forces as we only want to

add the forces that are in the same straight line.

f = – 500 N air friction acts down if rocket goes up

( 102,04 )( 10 ) = Fapp – 500 – 1 000

1 020,4 + 500 + 1 000 = FappFapp = 2 520,4 N upwards

Vertical motion Example 4 : A rocket slowly accelerates downwardsWhat applied upward force ( upthrust ) enables a 102,04 kg rocket to accelerate downwards at 1 m•s–2, subject to a constant average resistance from air friction of 100 N, as it descends onto its landing pad. Assume the mass of the rocket remains constant. Take vertically upwards as positive.

f

W

Fapp

Fapplied

W

Ffriction = f

Fapplied

Fg

Fair friction

force diagram

free-body diagram

Fnet = Fapplied + f + W m a = Fapp + f + W

W = m g = ( 102,04 ) ( ─ 9,8 ) = ─ 1 000 N

Consider only the vertical forces as we only want to

add the forces that are in the same straight line.

f = 100 N air friction acts upwards if rocket goes down

( 102,04 )( –1 ) = Fapp + 100 – 1 000

–102,04 – 100 + 1 000 = FappFapp = 797,96 N upwards

f

W

Fapp

+

g = –9,8 m•s–2

+

Two-mass systemsIf two objects are linked by a taut string then they can be regarded as one mass.

Example 1 : given the following frictionless system

5 kg 10 kg Fnet = 45 NF1

• tension acts both ways in the string and thus is a pseudoscalar, as it doesn’t act in one direction• " tension " describes two forces pulling equally hard in opposite directions, in the string• we say that the string is in tension

The 10 kg mass exerts a forward force F1on the 5 kg mass.

5 kg 10 kg Fnet = 45 NF2

The 5 kg mass exerts a backwards force F2on the 10 kg mass.

( ii ) The force in the string acts away from a body towards the other body that it pulling on the string.

( i ) If the string has no mass ( is light ) and does not stretch ( inextensible ) then the magnitude of the force in the string is the same. F1 = F2 = F

( iii ) The magnitude of the acceleration is the same for all objects that are linked together

5 kg 10 kg Fnet = 45 NFF

g = –9,8 m•s–2

11 Knowledge Area : mechanics : Term 1 Newton’s Laws 17

If there is more than one object, a free-body diagram must be drawn for each object and Newton 2 must be applied to each object separately.

Two-mass system Example 1 : given the following frictionless system, calculate the magnitude of a and F.

5 kg 10 kg Fnet = 45 NF F

5 kgF Choose to the right as positive

10 kgF 45 N

Fnet = m aF = 5 a

Fnet = m a45 – F = 10 a

substitute 45 – 5 a = 10 a45 = 15 aa = 45 = 3 m•s–2

15F = 5 a = ( 5 )( 3 ) = 15 N

substitute

Two-mass system Example 2 : Given the following frictionless system, calculate the magnitude of a and F.

5 kg

10 kg

F

F

If the string is light and does not stretch and the pulley is frictionless and has no mass, then the force F is transmitted through the string without its magnitude changing.The weight of the suspended 10 kg mass is the force that pulls the 10 kg mass downwards and the 5 kg mass towards the right.The force exerted by the string on the 10 kg is upwards as it is caused by the 5 kg pulling backwards on the 10 kg.NOTEThe forces caused by the string always act towards a frictionless pulleybecause each mass is pulled towards the other mass.m g

Draw a free-body diagram for each separate mass.

When objects accelerate, the equation Fnet = m a must be applied separately in the x and y directions.

5 kgF

Fnet = m aF = 5 a

10 kgF

Fnet = m a98 – F = 10 a

substitute98 – 5 a = 10 a98 = 15 aa = 98 = 6,53 m•s–2

15F = 5 a = ( 5 )( 6,53 ) = 32,65 N

substitute

Choose to the right as positive for the horizontal.

Choose down as positive for the vertical.

( To the right and down were the original directions of motion ).

m g = ( 10 )( 9,8 ) = 98 N

m g is not a net force as the 5 kg mass pulls backwards on the 10 kg mass.Pretend that the string is cut and they are thus two separate masses, but with the force F pulling on both masses as shown.This is a convenient fiction but it works very neatly as the horizontal and vertical directions can be calculated separately.

Check :

10 kg32,65 N

98 N

Fnet = ma98 – 32,65 = 10 a65,35 = 10 aa = 6,535 m•s–2

5 kgF Fnet = ma

32,65 = 5 aa = 6,53 m•s–2

Both linked objects have the same magnitude of acceleration.

11 Knowledge Area : mechanics : Term 1 Newton’s Laws 18

Forces always occur in pairs if two objects interact• if A pushes B then B pushes A at the same time with a reverse force or oppositely directed force• Newton’s Third law tells us how two objects simultaneously interact with each other

Newton’s Third law ( Newton 3 )When object A exerts a force on object B, object B simultaneously exerts an oppositely directed force of equal magnitude on object A. Example : If A pulls B with a 7 N force ...

A B7 N

A– 7 N

B7 N

If

then ... then B also pulls A in the opposite direction with 7 N.Both forces occur simultaneously. There is no action - reaction!

Topic : Newton's Third Law ( Newton 3 or N3 )

force

reverse force

Pairs of interacting objects exert forces, that are equal in magnitude but opposite in direction, on each other.

Each pair of the two Newton 3 forces do not cancel as they act on different objects.

A– F

BF

F

attractattract

Example A ball falling towards the Earth. The Earth exerts a non-contact force on the ball. Somewhere in the Earth the ball exerts a force of equal magnitude in the opposite direction. These are Newton 3 pairs.

ball

inside the Earth

force exerted by the Earth on the ball

force exerted by the ball on the Earth ( this force is somewhere inside the Earth )

( reverse the objects in the name of the force )

– F

Earth

• we can only add forces that act on the same object

These are Newton 3 pairs. ( N3)

force exerted by the table on the block

force exerted by the block on the table

force exerted by the block on the Earth

force exerted by the Earth on the block

ExampleIdentify the two " Newton 3 force pairs " when a block rests on a table.Clue : reverse the names of the two objects in the names of the forces. The forces act in opposite directions.Two forces act on the block, the force exerted by the table on the block and the force exerted by the Earth on the block. Each of these two forces has its own reverse Newton 3 force.

floor

( Newton 3 )

contact forces

non-contact forces

Either force can be regarded as " the force " and the other one automatically becomes the " reverse force" of the Newton 3 pair.

Earth

Inertia is the resistance of a body to a change in its state of motion.The net force acting on a body is the vector sum of all the forces acting on a body.Or The net force acting on a body is the product of the mass of the body and its acceleration in the direction of the force. Fnet = m a

N3

N3

• Newton 3 force pairs have the same magnitude• they act in opposite directions• they do not cancel• they act simultaneously

11 Knowledge Area : mechanics : Term 1 Newton’s Laws 19

Example ( considering only the horizontal forces )Consider a donkey pulling a cart. Regard the rods and the harness that is wrapped around the donkey as part of the cart. A Newton 3 pair is the force ( F2 ) exerted by the donkey on the cart and the force ( – F2 ) exerted by the cart on the donkey. These two forces are equal in magnitude and opposite in direction. Whether or not the cart moves has nothing to do with the force that the cart exerts on the donkey ( – F2 ). The donkey pushes backwards against the ground ( – F1 ), causing the reverse Newton 3 force of the ground to push the donkey forward ( F1 ). Part of this forward force F1 is needed to move the donkey forward. The remaining part of F1 is used to produce the force F2. If F2 is greater than the friction F3 between the cart wheels and the ground, ( which acts in such a way as to resist the cart's motion ) then the cart will start to move forwards, only if F2 > F3.

When the cart is moving at a steady velocity then F2 = F3 in magnitude.

A large mass has a small acceleration, a small mass has a large acceleration

force exerted by the bat on the ball

force exerted by the ball on the bat

Let the bat and the person holding it have a combined larger mass M, and the ball has a smaller mass m.

• bat’s large mass M has small acceleration “ a ”• ball’s small mass m has large acceleration “ A ”

F

– F

F bat on ball = – F ball on bat

M • a = – m • A

( N 3 )

( N 2 )

The force can accelerate the small mass of the ball.The same size force is less successful in accelerating the larger mass of the bat and the batsman who is holding the bat.

During the contact time, a shock wave goes into the Earth at the speed of sound and that portion of the ground which is affected by the shock wave also adds its mass to the batsman. Hence a batsman is not much good on a slippery pitch where he has poor contact with the ground. Even a small batsman like the Australian Donald Bradman, using a light bat, could hit the ball very hard. His test average was 99 runs.

mmball

These two forces do not cancel as they act on different objects

• tension acts both ways in a connecting rod or string• " tension " describes two forces pulling equally hard in opposite directions, in a rod or string• we say that the rod or string is in tension

force exerted by the donkey on the ground,

force exerted by the ground on the donkey,

friction F3

force exerted by the donkey on the cart

force exerted by the cart on the donkey

harn

ess

in tension rod

cart

donkey

F2 – F2

bat

F1 – F1

11 Knowledge Area : mechanics : Term 1 Ticker tapes 20

ticker timertape

string

falling mass

AIM:To investigate the relationship, if any, between the acceleration of masses by a constant force.

METHODA constant force, due to a falling mass, is applied to a dynamics trolley.Slope the runway until the trolley just begins to move at a constant speed. This compensates for friction losses.A ticker tape attached to the trolley runs through a ticker timer of frequency 50 Hz as the trolley is moving on the surface.The above procedure is repeated 4 times, each time using the same force, but varying the mass of the trolley as follows:

Tape 1: 6,25 kg Tape 2: 3,57 kg Tape 3: 2,27 kg Tape 4: 1,67 kg

• 14 • 18 mm • 22 mm • 26 mm • 30 mm •

• 11 • 18 mm • 25 mm • 32 mm • 39 mm •

• 14 • 25 mm • 36 mm • 47 mm • 58 mm •

Tape 1 6,25 kg

Tape 2 3,57 kg

• 10 • 25 mm • 40 mm • 55 mm • Tape 4 1,67 kg

Tape 3 2,27 kg

ANALySIS OF TICkER TAPES f = 50 Hz T = 1 = 0,02 s 5 T = ( 5 )( 0,02 ) = 0,1 s 50Tape 1vi = ∆s = 0,014 = 0,14 m•s−1 vf = ∆s = 0,030 = 0,30 m•s−1

∆t 0,1 ∆t 0,1

∴ a = ∆v = vf – vi = 0,30 – 0,14 = 0,16 = 0,4 m•s−2 ∆t ∆t ( 4 )( 0,1 ) 0,4

Tape 2vi = ∆s = 0,011 = 0,11 m•s−1 vf = ∆s = 0,039 = 0,39 m•s−1

∆t 0,1 ∆t 0,1

∴ a = ∆v = vf – vi = 0,39 – 0,11 = 0,28 = 0,7 m•s−2 ∆t ∆t ( 4 )( 0,1 ) 0,4

Tape 3vi = ∆s = 0,014 = 0,14 m•s−1 vf = ∆s = 0,058 = 0,58 m•s−1

∆t 0,1 ∆t 0,1

∴ a = ∆v = vf – vi = 0,58 – 0,14 = 0,44 = 1,1 m•s−2 ∆t ∆t ( 4 )( 0,1 ) 0,4

Tape 4vi = ∆s = 0,010 = 0,10 m•s−1 vf = ∆s = 0,055 = 0,55 m•s−1

∆t 0,1 ∆t 0,1

∴ a = ∆v = vf – vi = 0,55 – 0,10 = 0,45 = 1,5 m•s−2 ∆t ∆t ( 3 )( 0,1 ) 0,3

pulley

11 Knowledge Area : mechanics : Term 1 Ticker tapes 21

m (kg) a (m•s–2) 1 ( kg−1 )m

1 x 10−1 ( kg−1 )m

m a

1 6,0 0,4 0,17 1,7 2,402 3,4 0,7 0,29 2,9 2,383 2,2 1,1 0,45 4,5 2,424 1,6 1,5 0,63 6,3 2,40

ANALySIS OF RESULTS

Graph of acceleration against massScale : x-axis 5 mm → 1 kg y-axis 25 mm → 1 m•s−2

1,51,41,31,21,11,00,90,80,70,60,50,40,30,20,10

a ( m•s−2

0 1 2 3 4 5 6 7 mass (kg)

The graph is a hyperbola.

equation of hyperbola graph is x y = k ( a constant )

OR m a = kThe control variable that was kept constant was the force, F. ∴ m a = F

This means that as mass increases, acceleration decreases.

Acceleration is inversely proportional to mass.

independent variable

dependent variable

We can write : a ∝ 1 m

But we cannot be sure of this curved graph as it may just turn out to be part of a hyperbola if we had taken more readings.

The only graph that we can really trust is a straight line through the origin.This means that y is directly proportional to x.In other words, y depends on x. If we double x, we automatically double y.

How can we modify a hyperbola to turn it into a straight line passing through the origin?

y

x

y ∝ x

x y = k So y = k or y = k 1 ∴ y ∝ 1 x x x

y = m x“m” is a constant in Maths

or, y = k x“k” is a constant in Science

Thusy = k x means y ∝ x.So " ∝ ” means “ = k ”

We write :

1,51,41,31,21,11,00,90,80,70,60,50,40,30,20,10

a ( m•s−2

0 1 2 3 4 5 6 7 1 x 10–1 (kg–1)mass

To check if this is a hyperbola, we draw a graph of acceleration vs inverse of mass a vs 1 m

For a constant net force applied to an object, the acceleration is inversely proportional to the mass.

11 Knowledge Area : mechanics : Term 1 Ticker tapes 22

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A fixed mass is accelerated by 1, 2 and 3 rubber band forces. f = 50 Hz. Draw a graph of F vs a.

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11 Knowledge Area : mechanics : Term 1 Ticker tapes 23A trolley is accelerated by three different forces. f = 50 Hz

1. f = 50 Hz T = 1 = 0,02 s 10 T = ( 10 )( 0,02 ) = 0,2 s 50Tape : 1 forcevi = ∆s = 0,118 = 0,59 m•s−1 vf = ∆s = 0,15 = 0,75 m•s−1

∆t 0,2 ∆t 0,2

∴ a = ∆v = vf – vi = 0,75 – 0,59 = 0,16 = 0,2 m•s−2 ∆t ∆t ( 4 )( 0,2 ) 0,8

Tape : 2 forcesvi = ∆s = 0,162 = 0,81 m•s−1 vf = ∆s = 0,21 = 1,05 m•s−1

∆t 0,2 ∆t 0,2

∴ a = ∆v = vf – vi = 1,05 – 0,81 = 0,24 = 0,4 m•s−2 ∆t ∆t ( 3 )( 0,2 ) 0,6

Tape : 3 forcesvi = ∆s = 0,202 = 1,01m•s−1 vf = ∆s = 0,25 = 1,25 m•s−1

∆t 0,2 ∆t 0,2

∴ a = ∆v = vf – vi = 1,25 – 1,01 = 0,24 = 0,6 m•s−2 ∆t ∆t ( 2 )( 0,2 ) 0,4

Force ( units ) a ( m•s−2 )1 0,22 0,43 0,6

Graph of force vs acceleration

0,6

0,5

0,4

0,3

0,2

0,1

01 32

a ( m•s−2 )

Force ( units )

F ∝ a

A straight line through the origin means “F is directly proportional to a” or “F depends on a”. In other words, if we double “F” we will automatically double “a”.

We write F ∝ a or alternatively F = k a where “k” is a constant.In this experiment the mass ( m ) of the trolley was kept constant. ∴ F = m a

11 Knowledge Area : mechanics : Term 1 Gravitation 24TOPIC : Newton’s Law of Universal Gravitationgravitational energyEnergy is invisible and seems to be transferred around by invisible energy fields.Earth has a property called mass which enables it to set up a gravity field which is a source of energy that is able to produce an attractive force on any other object that also has mass.

gravitational mass is a property that is difficult to describe.Gravitational mass is the property that enables an object to be attracted towards the centre of the Earth.

Mass is a scalar and has no direction.The unit for mass is the kilogram ( kg ).

9,8 N downwards.

The Earth exerts a downwards gravitational force of 9,8 N on a mass of 1 kg, that is on or near the surface of the Earth.

The main effect of the Earth’s gravitational field on an object is to produce “ weight ”.

mass piece

The force with which a mass piece is attracted downwardstowards the Earth is called weight.

weight = 9,8 N

mass = 1 kg

Weight is the gravitational force acting on an object and thus it is a vector.

The symbol for weight is Fgravity = Fg = W

The unit for weight is the newton ( N ).Weight always acts downwards, perpendicular to the Earth’s surface

Fg = weight = W

W = m g = ( 1 )( 9,8 ) = 9,8 N

Inertial massInertial mass is the property of an object which makes it reluctant to change its velocity or state of motion. If an object is stationary it tends to stay at rest ( a state of zero motion ). If an object is moving at a constant speed in a straight line, it tends to continue with this state of motion.

We do not understand what causes this inertia but we know it depends on the object’s mass.Fortunately, the gravitational mass and the inertial mass have the same value when we measure them so we often just use the word “mass” to describe either of them.

Thus the Earth is surrounded by a gravitational energy field that acts downwards towards the centre of the Earth. Scientists are still struggling to understand how this gravitational force actually works. The energy is the property of the field.

We do not know what this invisible energy is, but we can recognise its effects on visible matter.Gravitational energy makes objects fall downwards.

energy field

A field is a region of space in which an object experiences a force.

Mass is a property of matter i.e. the atoms.

Weight is a property of the energy of the Earth’s gravitational field.

g = 9,8 m•s−2 is the gravitational acceleration of an object in free fall near the Earth ( this is an approximate value )

Weight is the gravitational force that the Earth exerts on any object that is on or near its surface.

The weight of a 30 kg object isW = m g = ( 30 )( 9,8 ) = 294 N

11 Knowledge Area : mechanics : Term 1 Gravitation 25

Newton’s Law of Universal Gravitation F = G m1 m2

d 2Every particle attracts every other particle in the universe with a force that isdirectly proportional to the product of their masses and inversely proportional to the square of their distance apart.

• the force drops off rapidly with increasing distance• the law only works for point masses as large objects will be distorted• spheres behave like point masses at their centres

F

F F

m1 m2

F

d

d

m2

“ G ” is the force of attraction between two 1 kg masses that are 1 m apart.

1 kg 1 kg

1 m

G G

m1

G is called “the universal constant” and its value is always G = 6,7 x 10–11 N•m2•kg–2

1 is an inverse square law which means that the gravitational force decreases quickly with increasing distance. d 2• double the distance between two masses and their attractive force reduces to one quarter

1

2

3

point source of light

intensity of illumination, per tile

distance from point source ( d )

• the gravitational force exerted by the earth obeys the same law that gover ns the intensity of light falling on tiles

91

41

1

F ∝ 1 d 2

F = 8N 8NExample 1 Given two masses that are 2 m apart and their attractive force is 8 N.

Example 2 Given two masses that are 2 m apart and their attractive force is F = 8 N.

2 m

32N

1 m

2NF ’

2N4 m

If the distance between them is doubled , calculate the new force of attraction.

If the distance between them is halved the new force F ’ = 1 x F = 4 F = ( 4 )( 8 ) = 32 N ( ½ )²

Moving objects further apart reduces the force

F ’ ∝ 1 = 1 22 4

F ’ = 1 x 8 N = 2 N 4

Moving objects closer increases the force. 32N

Moon Earth

tile

4 tiles

30 000 km•h–1 inertia

Earthorbit

rocket

Example 3 Calculate the force of attraction between two spherical objects A and B whose centres are 340 cm apart.A has a mass of 8 x 10 4 kg. B has a mass of 2 x 10 3 kg.

F = G m1 m2 = ( 6,7 x 10–11)(8 x 10 4)( 2 x 10 3 )

d 2 ( 3,4 ) 2

= 0,000927 N or 9,27 x 10– 4 N

Example 4 The force of attraction between two spherical objects A and B is 9,27 x 10–4 N.How far apart are their centres ?A has a mass of 1,6 x 10 5 kg.B has a mass of 1 x 103 kg.

d 2 = G m1 m2 = ( 6,7 x 10–11)(1,6 x 10 5 )(1 x 10 3 )

F 9,27 x 10– 4

= 11,56∴ d = 3,4 m

11 Knowledge Area : mechanics : Term 1 Gravitation 26

• let the new acceleration due to planet's gravity pull be g' g ' = G ( 4 M )

( 3 d ) 2 = 4 G M

9 r 2

= 4 g 9

All objects fall with gravitational acceleration “g” near the Earth• when an object falls near the Earth, its mass can be ignored• thus all objects fall at the same speed

m

r

When a mass m falls :G M m = F = m g

d 2

• mass m cancelsAcceleration ( g ) due to Earth’s gravitational pull is independent of the mass ( m ) of the object

∴ gEarth = G MEarth

d 2Earth

m = mass of objectr = radius of EarthM = mass of Earth

M

Example 5 Calculate “ g ” on a planet that has 4 times Earth's mass , M , and 3 times Earth's radius, r.

g = G M d 2

Example 6 Calculate the weight of a 100 kg astronaut on the moon if gmoon = 1,6 m•s–2.WMoon = m gMoon = ( 100 )( 1,6 ) = 160 N

Example 7 Calculate the value of gEarth if mass of Earth = 6 x 1024 kg and radius of Earth = 6,4 x 103 km.

gEarth = G MEarth

d 2Earth

= ( 6,7 x 10–11 )( 6 x 1024 ) = 9,81 m•s–2 ( 6,4 x 106 )2

earth

rocketFire a rocket 400 km upwards.

The gravitational force exerted by the Earth pulls the rocket back down to the Earth’s surface.

The Earth pulls the rocket down but it also moves tangentially sideways due to its inertia. At this sideways speed the rocket is in an exquisite balance which enables it to keep falling freely towards the Earth at the same rate that the Earth curves away from it. An astronaut in the rocket perceives himself as weightless due to being in free fall in the rocket’s orbit.In free fall you do not feel your weight. you only feel your weight when you stand on a firm sur-face that can exert an upward force on you. The astronaut has zero effective weight.

If we fired a rocket vertically upwards at 40 000 km•h–1 ( the escape velocity ) then it would break the bonds of Earth’s gravitational force and escape from the Earth. If it has a smaller vertical velocity it will always fall back to Earth, however high it gets.To put a rocket in orbit it must also have a horizontal velocity of 30 000 km•h–1.

weightEarth