Download Force - DCS Physics

Forces
By Neil Bronks
Force causes a body to change
velocity…….. accelerate
The unit is called the
Newton (N)
Distance, Speed
and Time
D
Speed = distance (in metres)
time (in seconds)
S
T
1) Dave walks 200 metres in 40 seconds. What is his speed?
2) Laura covers 2km in 1,000 seconds. What is her speed?
3) How long would it take to run 100 metres if you run at 10m/s?
4) Steve travels at 50m/s for 20s. How far does he go?
5) Susan drives her car at 85mph (about 40m/s). How long does it
take her to drive 20km?
6) Convert 450m/s into km/hr.
• A scalar quantity is a quantity that has
magnitude only and has no direction in
space
Examples of Scalar Quantities:
 Length
 Area
 Volume
 Time
 Mass
• A vector quantity is a quantity that has
both magnitude and a direction in space
Examples of Vector Quantities:
 Displacement
 Velocity
 Acceleration
 Force
Speed vs. Velocity
Speed is simply how fast you are travelling…
This car is travelling at a
speed of 20m/s
Velocity is “speed in a given direction”…
This car is travelling at a
velocity of 20m/s east
Scalar vs. Vector
Scalar has only magnitude….. mass
This car has a mass of
2000kg
Vector has magnitude and direction …….. Weight
This car has a Weight of
20000N
Distance and Displacement
Scalar- Distance
travelled 200m
VectorDisplacement
120m
• Vector diagrams are
shown using an arrow
• The length of the
arrow represents its
magnitude
• The direction of the
arrow shows its
direction

The resultant is the sum or the combined effect of
two vector quantities
Vectors in the same direction:
6N
4N
=
10 N
=
10 m
6m
4m
Vectors in opposite directions:
6 m s-1
10 m s-1
=
4 m s-1
6N
9N
=
3N
When two vectors are joined
tail to tail

Complete the parallelogram

The resultant is found by
drawing the diagonal



When two vectors are joined
head to tail
Draw the resultant vector by
completing the triangle
Vector Addition
Speed in still air 120m/s

Wind
50m/s
R2 = 1202 + 502 = 14400 + 2500
= 16900
Tan  = 50/120
R = 130m/s
 = 22.60
2004 HL Section B Q5 (a)
Two forces are applied to a body, as shown. What is the magnitude and
direction of the resultant force acting on the body?
Solution:


Complete the parallelogram (rectangle)
The diagonal of the parallelogram ac
represents the resultant force
The magnitude of the resultant is found using
Pythagoras’ Theorem on the triangle abc
Magnitude  ac  12  5
ac  13 N
2
a
2
12
Direction of ac : tan  
5
12
   tan 1  67
5
d
5
b

12 N
θ
5N

12
c
Resultant displacement is 13 N 67º
with the 5 N force

Here a vector v is resolved into
an x component and a y
component
y
• When resolving a vector into components
we are doing the opposite to finding the
resultant
• We usually resolve a vector into
components that are perpendicular to
each other
x
• Here we see a table being
pulled by a force of 50 N
at a 30º angle to the
horizontal
y=25 N
30º
x=43.3 N
•When resolved we see
that this is the same as
pulling the table up with a
force of 25 N and pulling
it horizontally with a force
of 43.3 N
• If a vector of magnitude v and makes an angle θ with the
horizontal then the magnitude of the components are:
• x = v Cos θ
• y = v Sin θ
y=v Sin θ

Proof:
x
Cos 
v
x  vCos
θ
x=vx Cos θ
y
Sin  
v
y  vSin
y
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown. What is the
horizontal
component of the force?
Vertical Component  y  15Sin 60  12.99 N
12.99 N
Horizontal Component  x  15Cos60  7.5 N
Vertical
Component
Solution:
60º
Horizontal
Component
7.5 N
H/W - 2003 HL Section B Q6
• A person in a wheelchair is moving up a ramp at constant speed.
Their total weight is 900 N. The ramp makes an angle of 10º with
the horizontal. Calculate the force required to keep the wheelchair
moving at constant speed up the ramp. (You may ignore the
effects of friction). (Stop here and freeze)
Solution:
If the wheelchair is moving at constant speed (no acceleration), then
the force that moves it up the ramp must be the same as the
component of it’s weight parallel to the ramp.
Complete the parallelogram.
Component of weight
10º
parallel to ramp:
 900Sin10  156.28 N
80º
10º
Component of weight
perpendicular to ramp:
 900Cos10  886.33 N
900 N
886.33 N
• If a vector of magnitude v has two
perpendicular components x and y, and v
makes and angle θ with the x component
then the magnitude of the components
are:
• x= v Cos θ
y=v Sin θ
y
θ
• y= v Sin θ
x=v Cosθ
Acceleration
V-U
Acceleration = change in velocity (in m/s)
(in m/s2)
time taken (in s)
A
T
1) A cyclist accelerates from 0 to 10m/s in 5 seconds. What is her
acceleration?
2) A ball is dropped and accelerates downwards at a rate of 10m/s2
for 12 seconds. How much will the ball’s velocity increase by?
3) A car accelerates from 10 to 20m/s with an acceleration of 2m/s2.
How long did this take?
4) A rocket accelerates from 1,000m/s to 5,000m/s in 2 seconds.
What is its acceleration?
Velocity-Time Graphs
V
V
t
1/.Constant Acceleration
t
2/.Constant Velocity
V
3/.Deceleration
t
Velocity-time graphs
1) Upwards line =
80
Constant Acceleration
Velocity
m/s
60
4) Downward line =
Deceleration
40
20
0
10 =
2) Horizontal line
Constant Velocity
20
303) Shallow
40 line50
=
Less Acceleration
T/s
80
60
Velocity
m/s
40
20
0
T/s
10
20
30
40
1) How fast was the object going after 10 seconds?
2) What is the acceleration from 20 to 30 seconds?
3) What was the deceleration from 30 to 50s?
4) How far did the object travel altogether?
50
80
60
Velocity
m/s
40
20
0
T/s
10
20
30
40
50
The area under the graph is the distance travelled by the object
80
60
0.5x10x20=100
Velocity
m/s
40
20
0.5x10x40=200
0
0.5x20x60
=600
40x20=800
10
20
30
40
Total Distance Traveled
=200+100+800+600=1700m
50
T/s
Motion Formula
v = u + at
A car starts from rest and accelerates for
12s at 2ms-2. Find the final velocity.
U=0 a=2 and t = 12 find v=?
Using V = U + at = 0 + 2x12 = 24m/s
v2 = u2 + 2as
A car traveling at 30m/s takes 200m to
stop what is it’s deceleration?
U=30 s=200 and v = 0 find a=?
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900/400=-2.25ms-2
Motion Formula
S = ut + 0.5at2
A train accelerates from rest at 10ms-2
for 12s find the distance it has traveled.
Using S = ut + 0.5at2 = 0x12 +0.5x10x144 =720m
Velocity and Acceleration
Dual
timer
t1
Light
beam
Air
track
Photogate
Pulley
l
Card
s
Slotted
weights
l
v
t2
l
u
t1
v u
a
2s
2
2
H/W
• LC Ord 2008
• Q1
Friction is the force that opposes
motion
The unit is
called the
Newton (N)
Lubrication
reduces friction
Friction is the
force between
two bodies in
contact.
Lubrication reduces friction
and separates the two bodies
Advantages and disadvantages of
Friction
• We can walk across a
surface because of
friction
• Without friction walking is
tough. Ice is a prime
example.
• It can also be a pain
causing unwanted heat
and reducing efficiency.
Friction
1) What is friction?
2) Give 3 examples where it is annoying:
3) Give 3 examples where it is useful:
4) What effect does friction have on the surfaces?
Recoil
m=2kg
Mass of canon=150kg
ub=400m/s
Momentum of Recoil = Momentum of the Shoot
Mass Canon x Velocity Canon = Mass of Ball x Velocity of Ball
150 x Uc = 2 x 400
V= 800/150 =
5.3m/s
Momentum
10m/s
V=? m/s
3kg
2kg
6kg
3kg
2 m/s
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10m/s = 3kg x (-2m/s) + 6kg x v
6v = 30 + 6
V = 6m/s
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF
CONSERVATION OF MOMENTUM
t1
l
Dual
timer
Photogate
Light
beam
Card
Air track
Vehicle 1
Velcro pad
Vehicle 2
1. Set up apparatus as in the diagram.
2. Level the air-track. To see if the track is level carry
out these tests:
a) A vehicle placed on a level track should not drift
toward either end.
Measure the mass of each vehicle m1 and m2
respectively, including attachments, using a balance.
4. Measure the length l of the black card in metres.
5. With vehicle 2 stationary, give vehicle 1 a gentle push.
After collision the two vehicles coalesce and move off
together.
6 Read the transit times t1and t2 for the card through
the two beams.
l
v
t2
l
u
t1
Calculate the velocity before the collision, and after the
collision,
momentum before the collision=momentum after the
collision,
m1u = (m1 + m2) v.
Repeat several times, with different velocities and
different masses.
H/W
• LC Ord
• 2007 Q1
Newton’s Laws
• 1 /. Every body stays in it’s state of rest
or constant motion until an outside force
acts on it
• 2/. The rate of change of momentum is
proportional to the applied force and in
the direction of the applied force.
• F=ma
• 3/. To every action there is an equal and
opposite reaction
Newton 2
force Rate of change of Momentum
mv  mu
force
t
m(v  u )
force
t
(v  u )
but  a 
t
Forcem.a
Or Force=k.m.a where k=constant
As this is the basic constant so we say k=1 and
Force=m.a
TO SHOW THAT a F
Dual
timer
t1
Light
beam
Air
track
Photogate
Pulley
l
Card
s
Slotted
weights
TO SHOW THAT a F
t1
Dual
timer
Photo
gate
Light
beam
t1 time for card to pass
first photo-gate
TO SHOW THAT a F
t1
t2
Dual
timer
Photo
gate
Light
beam
t2 time for card to pass
second photo-gate
Procedure
Set up the apparatus as in the diagram. Make sure the card
cuts both light beams as it passes along the track.
Level the air track.
Set the weight F at 1 N. Release the vehicle.
Note the times t1 and t2.
Remove one 0.1 N disc from the slotted weight, store this
on the vehicle, and repeat.
Continue for values of F from 1.0 N to 0.1 N.
Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s.
l
u
t1
F/N
1/.
t1/s
l
v
t2
t2/s
V/m/s
v u
a
2s
2
U/m.s
2
A/m/s2
Remember to include the following table to get full marks.
All tables are worth 3 marks when the Data has to be
changed. Draw a graph of a/m s-2 against F/N Straight
line though origin proves Newton's second law
Newton’s Laws on the
Internet
Balanced and unbalanced
forces
Reaction
Consider a camel standing on a road.
What forces are acting on it?
These two forces would be equal –
we say that they are BALANCED.
The camel doesn’t move anywhere.
Weight
Balanced and unbalanced
Reaction
forces
What would happen if we took the
road away?
Weight
Balanced and unbalanced
forces
What would happen if we took the
road away?
The camel’s weight is no longer
balanced by anything, so the camel
falls downwards…
Weight
Balanced and unbalanced forces
1) This animal is either
________ or moving
with _____ _____…
3) This animal is getting
_______….
2) This animal is getting
_________…
4) This animal is…
Let Go or Hang On?
A painter is high up on a ladder, painting
a house, when unfortunately the ladder
starts to fall over from the vertical.
Determine which is the less harmful
action for the painter: to let go of the
ladder right away and fall to the ground,
or to hang on to the ladder all the way to
the ground.
Reaction
Friction
Gravity
Engine force
Force and acceleration
If the forces acting on an object
are unbalanced then the object will
accelerate, like these wrestlers:
Force and acceleration
If the forces acting on an object
are unbalanced then the object will
accelerate, like these wrestlers:
Force (in N) = Mass (in kg) x Acceleration (in m/s2)
F
M
A
1) A force of 1000N is applied to push a
mass of 500kg. How quickly does it
accelerate?
2) A force of 3000N acts on a car to
make it accelerate by 1.5m/s2. How
heavy is the car?
3) A car accelerates at a rate of 5m/s2.
If it ‘s mass is 500kg how much
driving force is the engine applying?
4) A force of 10N is applied by a boy
while lifting a 20kg mass. How much
does it accelerate by?
Using F=ma
10=20xa
a=0.5m/s2
Using F=ma
1000=500xa
a=2m/s2
Using F=ma
3000=mx1.5
m=2000kg
Using F=ma
F=5x500
F=2500N
Net Force creates Acceleration
Fnet=200N
F=-100N
F=-200N
F=200N
Fnet=100N
F=200N
Fnet=0N
F=-200N
Fnet=-200N
H/W
• LC Ord
• 2004 Q6
Net Force creates Acceleration
800kg
F=-100N
F=200N
As net force causes acceleration F=m.aFnet=100N
100N = 800kg.a
a=100/800 = 0.125m/s2
Acceleration gives Net Force
900kg
Friction=?
Feng=5000N
As net force causes acceleration F=m.aa=3m/s2
Fnet = 900kg. 3m/s2
Fnet= 2700N
So Friction = Feng – 2700N
Friction=2300N
A car of mass 500kg has an engine that produces 3kN
of force what is the friction if the car is accelerating
at 1.1 m/s2 ?
If the engine stops how long before the car stops if it
is travelling at 20m/s when the engine cuts out?
Archimedes Principle
• A body in a fluid
experiences an
up-thrust equal
to the weight of
liquid displaced.
12N
20N
8N
Internet Diagram
Floatation
• A floating body
displaces its
own weight in
water.
Floatation
• A floating body
displaces its
own weight in
water.
=
10000t
10000t
Measuring Liquid Density
• A hydrometer is an
instrument used to
measure the specific
gravity (or relative
density) of liquids; that
is, the ratio of the
density of the liquid to
the density of water.
Terminal Velocity
Consider a skydiver:
1) At the start of his jump the air
resistance is _______ so he
_______ downwards.
2) As his speed increases his air
resistance will _______
3) Eventually the air resistance will be
big enough to _______ the
skydiver’s weight. At this point
the forces are balanced so his
speed becomes ________ - this is
called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the
air resistance suddenly ________,
causing him to start _____ ____.
5) Because he is slowing down his air
resistance will _______ again until
it balances his _________. The
skydiver has now reached a new,
lower ________ _______.
Velocity-time graph for terminal
Parachute opens –
Velocity
velocity…
diver slows down
Speed
increases…
Terminal
velocity
reached…
Time
New, lower terminal
velocity reached
Diver hits the ground
Weight vs. Mass
Earth’s Gravitational Field Strength is 9.8m/s2. In other
words, a 1kg mass is pulled downwards by a force of 9.8N.
W
Weight = Mass x acceleration due to gravity
(in N)
(in kg)
(in m/s2)
M
g
1) What is the weight on Earth of a book with mass 2kg?
2) What is the weight on Earth of an apple with mass 100g?
3) Dave weighs 700N. What is his mass?
4) On the moon the gravitational field strength is 1.6N/kg. What will
Dave weigh if he stands on the moon?
Weight vs. Mass
900kg
900kg
• Mass is the amount of matter in us
• Same on Earth and Space
9000 N
• Weight is the pull of gravity on us
• Different on Earth and Space
0N
Homework
• LC Ordinary Level
• 2002 Q6
Galileo’s Falling Balls
Gravity
all bodies have gravity we feel it only
from planet sized objects
T=0
• Acceleration due
to gravity is
9.81m/s2
• That means
every falling body
gets 9.81m/s
faster every
second
T=1s
v=0m/s
v=9.81m/s
T=2s
v=19.62m/s
T=3s
v=29.43m/s
Internet
• Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
Electromagnet
Switch
Ball bearing
h
Trapdoor
Electronic timer
When the switch opens the ball falls
The
timer
records
the time
from
when the
switch
opens
until trap
door
opens
Set up the apparatus. The millisecond timer starts when the
ball is released and stops when the ball hits the trapdoor
Measure the height h as shown, using a metre stick.
Release the ball and record the time t from the millisecond
timer.
Repeat three times for this height h and take the smallest
time as the correct value for t.
Repeat for different values of h.
Calculate the values for g using the equation . Obtain an
average value for g.
h  gt
1
2
2
Place a piece of paper between the ball bearing and the
electromagnet to ensure a quick release
Finding Drag
The sky diver accelerates at
2m/s2 what is his drag?
Force due to gravity=80.g
=80.(9.8)=784 N
Net Force=m.a=80.2=160N
Drag=784-160=624N
Drag
80kg
Newton’s
Cannon
Launching a satellite
The cannon ball is
constantly falling
towards the earth
but earth curve is
same as it’s path
The Moon orbits the
Earth. It is also in
free fall.
Newton's Law of Gravitation
• This force is always positive
• Called an inverse square law
F  m1m2
d2
Where
F = Gravitational Force
m1.m2 = Product of masses
d = Distance between their center
of gravity
Gravity Calculations
• To make an equation we add a
constant
• G The UNIVERSAL
GRAVITATIONAL CONSTANT
F = G m1 m2
d2
Example What is the force on a man of mass 100kg
standing on the surface of Mars.
Mars mass=6.6x1023 kg and radius=3.4x106m
G=6.67x10-11 Nm2kg-2
F = G m1 m2
d2
= 6.67x10-11 x 6.6x1023 x100
(3.4x106)2
= 380N
• 2010 Question 6 [Higher Level]
• (Radius of the earth = 6.36 × 106 m, acceleration due
to gravity at the earth’s surface = 9.81 m s−2
• Distance from the centre of the earth to the centre of
the moon = 3.84 × 108 m
• Assume the mass of the earth is 81 times the mass of
the moon.)
•
• State Newton’s law of universal gravitation.
• Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth,
which is twice the radius of the earth.
• Note that 2d above surface is 3d from earth’s centre
• A spacecraft carrying astronauts is on a straight line
flight from the earth to the moon and after a while its
engines are turned off.
• Explain why the spacecraft continues on its journey to
the moon, even though the engines are turned off.
• Describe the variation in the weight of the astronauts
as they travel to the moon.
• At what height above the earth’s surface will the
astronauts experience weightlessness?
• The moon orbits the earth every 27.3 days. What is its
velocity, expressed in metres per second?
• Why is there no atmosphere on the moon?
H/W
• LC Ord 2008
• Q6
Hookes Law
Force
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Extension
More force means more
Extension - they are
proportional
Hookes Law Calculation
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Force=0N
Length=5cm
Ext.=0cm
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Force=6N
Length=8cm
Ext.=3cm
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Force=12N
Length=11cm
Ext.=6cm
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Force
=24N
Length
=17cm
Ext.
=12cm
Hookes Law Example
Force =Constant (k) x Extension
Example a/. A mass of 3kg causes an extension of
0.3m what is the spring constant?
3x9.8 = k x 0.3
K=98N/m
B/. What is the extension if 40N is put on the same spring?
Force = Spring Constant x Extension
40 = 98 x s
S = 40/98 = 0.41 m
Homework
• LC Ord
• 2003 Q6
Work done
When any object is moved around work will need
to be done on it to get it to move (obviously).
We can work out the amount of work done in
moving an object using the formula:
Work done = Force x Distance Moved
in J
in N
in m
W
F
D
Kinetic energy
Any object that moves will have kinetic
energy.
The amount of kinetic energy an object has
can be found using the formula:
Kinetic energy = ½ x mass x velocity squared
in J
in kg
KE = ½ mv2
in m/s
Some example questions…
A 70kg boy is running at about 10m/s. What is his kinetic energy?
Using KE=½mv2=0.5x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car. The car
covered 50m before it stopped. How much work did the brakes do?
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed
of 5m/s?
Using KE=½mv2=0.5x0.1x5x5=1.25J
A crane is lifting a 50kg load up into the air with a constant speed. If
the load is raised by 200m how much work has the crane done?
Work Done = force x distance = 50x9.81x200 = 98100J
KE = ½ mv2
Potential energy
An object has potential energy because of it’s
position or condition.
That means it is high or wound up
The formula is for high objects:
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
PE at top=KE at bottom
At the
bottom the
bob has no
PE only KE
KE = ½ mv2
At the top of the
oscillation the
pendulum bob
stops. All it’s
energy is PE
PE = mgh
h
PE at top=KE at bottom
mgh = ½ mv2
mgh = ½
H=10cm
gh = ½
2
mv
2
v
v2 = 2gh
v2 = 2(9.8)0.1
v = 1.4m/s
Power
• The rate at which work
is done
• POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m. If the jet has
mass 200tonnes find the work done and the power?
Work Done = Force x Distance = 200x1000x9.81x4000
=7.8 x 109 Joules
Power = Work Done / Time = 7.8 x 109 Joules / 120
= 6.5 x 107 Watts
H/W
• LC Ord 2007
• Q6
Pressure
Pressure depends on two things:
1) How much force is applied, and
2) How big (or small) the area on which this force is
applied is.
Pressure can be calculated using the equation:
F
Pressure (in N/m2) = Force (in N)
Area (in m2)
P
A
Some example questions…
1) A circus elephant weighs 10,000N and can stand on one foot.
This foot has an area of 50cm2. How much pressure does he
exert on the floor (in Pa)?
2) A 50kg woman copies the elephant by standing on the heel of one
of her high-heeled shoes. This heel has an area of 1cm2. How
much pressure does she exert on the floor?
Pressure=Force/area = 500N/ 0.0001m2 = 5000000 Pa
Extension task:
Atmospheric pressure is roughly equivalent to 1kg pressing on every
square centimetre on our body. What does this equate to in
units called Pascals? (1 Pascal = 1N/m2)
Pressure – in Fluids
Pressure increases with depth
Pressure and Depth
As the frog
goes deeper
there is a
greater weight
of water above
it.
Atmospheric Pressure
• The earth is covered with
layer of Gas.
• We are at the bottom of a
gas ocean 200km deep.
• The effect of this huge
column of gas is 1 Tonne
of weight on our
shoulders.
• This is called
• ATMOSPHERIC
PRESSURE
Heavy!
Proving Atmospheric Pressure
Very full glass
of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the
atmospheric
Pressure
holds the
card in place
The Barometer
• The weight of the air
holds up the mercury.
• If we use water the
column is 10.4m high.
• 1 Atmosphere is
760mm of Hg.
The Altimeter
• As we go higher there is
less air above us.
• There is less
Atmospheric pressure
• We can measure the
altitude using a
barometer with a
different scale.
Aneroid Barometer
• Works on
changes in size of
small can.(Get it!)
Pressure and Volume in gases
This can be expressed using the equation:
Initial Pressure x Initial Volume = Final Press. x Final Vol.
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20N/m2. What will the
pressure be if the volume is reduced to 1.5m3?
2) A gas increases in volume from 10m3 to 50m3. If the initial
pressure was 10,000N/m2 what is the new pressure?
3) A gas decreases in pressure from 100,000 Pascals to 50,000
Pascals. The final volume was 3m3. What was the initial volume?
4) The pressure of a gas changes from 100N/m2 to 20N/m2. What is
the ratio for volume change?
Pressure and Volume in gases
Pressure
Volume
Pressure x
volume
20
10
200
200
1
200
4
50
200
Internet Demo
Boyles Law
Pressure is inversely
proportional to
volume
VERIFICATION OF BOYLE’S LAW
1. .
Volume
scale
Bicycle pump
Tube with volume of
air trapped by oil
Reservoir of oil
Valve
Pressure
gauge
Using the pump, increase the pressure on the air in
the tube. Close the valve and wait 20 s to allow the
temperature of the enclosed air to reach equilibrium.
Read the volume V of the air column from the scale.
Take the corresponding pressure reading from the
gauge and record the pressure P of the trapped air.
Reduce the pressure by opening the valve slightly –
this causes an increase the volume of the trapped air
column. Again let the temperature of the enclosed air
reach equilibrium.
Record the corresponding values for the volume V
and pressure P .
Repeat steps two to five to get at least six pairs of
readings.
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically, a smaller force on piston
A will produce a larger force on
piston B because the pressure of
the liquid is constant.
Magic!
1) If the area of the slave piston is ten times bigger than the master
piston what force will be needed to lift an object weighing 1000N?
Pressure in Slave = 1000/10=100Pa
Pressure in Master = Force/1 = 100Pa
Force in the master only 100N amazing
2006 Question 12 (a) [Higher Level]
Define pressure.
Is pressure a vector quantity or a scalar quantity? Justify
your answer.
State Boyle’s law.
A small bubble of gas rises from the bottom of a lake.
The volume of the bubble increases threefold when it
reaches the surface of the lake where the atmospheric
pressure is 1.01 × 105 Pa. The temperature of the lake is
4 oC. Calculate the pressure at the bottom of the lake;
Calculate the depth of the lake.
(acceleration due to gravity = 9.8 m s–2; density of water
= 1.0 × 103 kg m–3)
H/W
• LC Ord
• 2005 Q6
Center of Gravity
• Things stay standing
(STABLE) because
their Center of Gravity
acts through their
base.
• The perpendicular
from the COG passes
inside the support
Unstable Equilibrium
• Things fall over
because the
center of gravity
is outside the
base
Moments
(Also called TORQUE)
=Force x Perpendicular distance
Fulcrum
Perpendicular
distance
FORCE
Moments
=Force x Perpendicular distance
= 10N x 5m = 50Nm
Perpendicular
distance=5m
FORCE
=10N
More than two forces
?N
15N
10
15N
50
5N
60
70
10N
?
90
5N
• First prove all coplanar forces on a body in
equilibrium add up to zero.
(Forces Up = Forces Down)
• Then take moments about one end.
(Clockwise moments=Anti-clockwise moments)
?N
15N
10
15N
50
5N
60
70
?
10N
• First law coplanar forces
• Forces Up = Forces Down
15 + x = 15 + 5 +10 + 5
x = 20 N
90
5N
20N
15N
A
10
15N
32.5
50
5N
60
70
10N
?
90
5N
• Second law coplanar forces
• Take moments about A
Clockwise Moments = Anticlockwise Moments
10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20
150 + 250 + 700 + 450 = 900 + dx20
1550-900 = dx20 so d=32.5cm
INVESTIGATION OF THE LAWS OF
EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Support
Newton
balance
Newton
balance
Paperclips
w1
w2
w3
w4
1. Use a balance to find the centre of gravity of
the metre stick and its weight.
2. The apparatus was set up as shown and a
equilibrium point found.
3.
Record the reading on each Newton balance.
4. Record the positions on the metre stick of
each weight, each Newton balance and the centre
of gravity of the metre stick
For each situation
(1) Forces up = Forces down
i.e. the sum of the readings on the balances
should be equal to the sum of the weights plus
the weight of the metre stick.
(2)The sum of the clockwise moments about
an axis through any of the chosen points
should be equal to the sum of the
anticlockwise moments about the same axis.
Internet
• Ok its not the most exciting thing doing all
the calculations so here Walter has done
them for us and we just play and see how
they are laid out
• Notice the units of torque are included as
we should.
• 2011 Question 6 (b) [Higher Level]
• State the conditions necessary for the equilibrium of
a body under a set of co-planar forces.
• Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium.
• The fulcrum of the see-saw is at its centre of gravity.
• A child of mass 30 kg sits 1.8 m to the left of the
fulcrum and another child of mass 40 kg sits 0.8 m
to the right of the fulcrum.
• Where should the third child of mass 45 kg sit, in
order to balance the see-saw?
H/W
• LC Ord
• 2003 Q12(a)
Couples of Forces
• Two equal forces that
cause a solid to rotate
around an axis
• Moment = Force x Distance
• Moment = 5Nx0.06m
• Moment = 0.3 Nm
Motion in a circle
Velocity always
at 90o to the
force or
acceleration
Circular Motion
• Angular Velocity
• =θ/t
• Units of Radians per
second
• Angle  time
t

A particle goes round a
circle in 4s what is it’s
angular velocity?
2 

 rads / second
4
2
Circular Motion
•
•
•
•
•
Linear Velocity(V)
m/s
V=  r
r=radius of motion
Always changing as
direction is always
changing this creates
acceleration
• If the radius is 6m


2
rads / second
v  r.  6.

2
 9.42m / s
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous
example a= 6 (/2)2
=14.8m/s2
Centripetal Force
If we have an acceleration we must have
a force.
Centripetal force f = ma = m r 2
Tension in string of weight spun around
head
Force on tyres (Or camel) as we go
around corner
Centripetal Acceleration
Satellites balance forces
• Balance of Gravity and Centripetal
• ((GMm)/d2)=mv2/d
Gravity F=-GmM/r2
Period of Orbit
((GMm)/d2)=mv2/d
(GM)/d=v2
(GM)/d=(2d/T)2
T2=42 d3 /GM
GMm mv

2
d
d
In a test we do
it like this
GMm mv2

d2
d
GMm mv

d
1
2
2
GM
 2d 
2
v 

d
 T 
GMm mv2

d2
d
4 d
T 
GM
2
2
3
GMm mv2

d2
d
2
Example of Orbits
What is the parking orbit height above Saturn if it
is 200000km in radius. It rotates every 4 days and
has mass 2x1031Kg. The Universal gravitational
Constant is 6.7x10-11
Using T2=42 d3 /GM
(4x24x60x60)2=42 d3 /(2x1031)(6.7x10-11)
d3 = 1x1030
d = 1x1010m
Height =h= d - r =1x1010m - 2x108m
= 9.8x109m
h
d
r
Geostationary or
Clarke Orbit
Same period (For earth 24hrs) and
angular velocity as the planet surface
so stays above same spot. What is
it’s height above the earth?
Classwork
2015 H q 6
Simple Harmonic Motion
• Is a vibration where the
acceleration is
proportional to the
displacement
a  -s
•Further from centre =more acceleration
Hooke’s Law as SHM
Force  Extension
F  -s
m.a  -s
If mass is constant
a  -s
So motion under
Hookes law is SHM
H/W
• LC Ord
• 2006 Q6
Pendulum
Split
cork
l
Bob
Timer
20:30
• If we displace
the bob by a
small angle it
vibrates with
SHM
T2
l
T  4
g
T 2 4 2


 slope
l
g
4 2
 g 
(slope)
2
l
2
Time to go over h/w
•
•
•
•
•
•
•
•
•
•
LC Ord
LC Ord
LC Ord
LC Ord
LC Ord
LC Ord
LC Ord
LC Ord
LC Ord
LC Ord
2008
2007
2004
2002
2008
2003
2007
2005
2003
2006
Q1
Q1
Q6
Q6
Q6
Q6
Q6
Q6
Q12(a)
Q6