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Forces By Neil Bronks.

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1 Forces By Neil Bronks

2 Force causes a body to change velocity…….. accelerate
The unit is called the Newton (N)

3 Distance, Speed and Time
Speed = distance (in metres) time (in seconds) Dave walks 200 metres in 40 seconds. What is his speed? Laura covers 2km in 1,000 seconds. What is her speed? How long would it take to run 100 metres if you run at 10m/s? Steve travels at 50m/s for 20s. How far does he go? Susan drives her car at 85mph (about 40m/s). How long does it take her to drive 20km? Convert 450m/s into km/hr.

4 Scalars 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

5 Vectors A vector quantity is a quantity that has both magnitude and a direction in space Examples of Vector Quantities: Displacement Velocity Acceleration Force

6 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

7 This car has a Weight of 20000N
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

8 Distance and Displacement
Scalar- Distance travelled 200m Vector- Displacement 120m

9 Vector Diagrams Vector diagrams are shown using an arrow
The length of the arrow represents its magnitude The direction of the arrow shows its direction

10 Resultant of Two Vectors
The resultant is the sum or the combined effect of two vector quantities Vectors in the same direction: 6 N 4 N = 10 N 6 m = 10 m 4 m Vectors in opposite directions: 6 m s m s-1 = 4 m s-1 6 N 9 N = 3 N

11 The Parallelogram Law The Triangle Law
When two vectors are joined tail to tail Complete the parallelogram The resultant is found by drawing the diagonal The Triangle Law When two vectors are joined head to tail Draw the resultant vector by completing the triangle

12 Vector Addition Resultant Speed in still air 120m/s  Wind 50m/s
= 16900 R = 130m/s Tan  = 50/120  = 22.60

13 Problem: Resultant of 2 Vectors
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 12 N a d θ 5 N 13 N 5 b c 12 Resultant displacement is 13 N 67º with the 5 N force

14 Resolving a Vector Into Perpendicular Components
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 v y x Here a vector v is resolved into an x component and a y component

15 Practical Applications
Here we see a table being pulled by a force of 50 N at a 30º angle to the horizontal 50 N 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

16 Calculating the Magnitude of the Perpendicular Components
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 v y=v Sin θ y Proof: θ x=v Cos θ x

17 Problem: Calculating the magnitude of perpendicular components
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? Solution: 12.99 N 15 N Component Vertical 60º Horizontal Component 7.5 N

18 H/W 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 parallel to ramp: N 10º 80º 10º Component of weight perpendicular to ramp: N 900 N

19 Summary 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 v y=v Sin θ y θ x=v Cosθ

20 Acceleration V-U T A Acceleration = change in velocity (in m/s)
(in m/s2) time taken (in s) A cyclist accelerates from 0 to 10m/s in 5 seconds. What is her acceleration? 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? A car accelerates from 10 to 20m/s with an acceleration of 2m/s2. How long did this take? A rocket accelerates from 1,000m/s to 5,000m/s in 2 seconds. What is its acceleration?

21 1/.Constant Acceleration
Velocity-Time Graphs V V t t 1/.Constant Acceleration 2/.Constant Velocity V t 3/.Deceleration

22 Constant Acceleration
Velocity-time graphs 80 60 40 20 Upwards line = Constant Acceleration 4) Downward line = Deceleration Velocity m/s 3) Shallow line = Less Acceleration 2) Horizontal line = Constant Velocity T/s

23 80 60 40 20 Velocity m/s T/s How fast was the object going after 10 seconds? What is the acceleration from 20 to 30 seconds? What was the deceleration from 30 to 50s? How far did the object travel altogether?

24 80 60 40 20 Velocity m/s T/s The area under the graph is the distance travelled by the object

25 Total Distance Traveled
80 60 40 20 0.5x10x20=100 Velocity m/s 0.5x20x60=600 40x20=800 0.5x10x40=200 T/s Total Distance Traveled = =1700m

26 A car traveling at 30m/s takes 200m to stop what is it’s deceleration?
Motion Formula A car starts from rest and accelerates for 12s at 2ms-2. Find the final velocity. v = u + at 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 = a (200) a = -900/400=-2.25ms-2

27 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 = 0x x10x144 =720m




31 Velocity and Acceleration
Dual timer t1 t2 Photogate Card l Pulley Light beam Slotted weights Air track s


33 H/W LC Ord 2008 Q 1

34 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.

35 Lubrication reduces friction and separates the two bodies

36 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.

37 Friction What is friction? Give 3 examples where it is annoying:
Give 3 examples where it is useful: What effect does friction have on the surfaces?

38 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 2 x 400 150 x Uc = V= 800/150 = 5.3m/s

39 Momentum 10m/s V=? m/s 2kg 3kg 3kg 6kg 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 = V = 6m/s

40 Internet Calculations

Dual timer t1 t2 Photogate Light beam Card l Vehicle 2 Air track Vehicle 1 Velcro pad

42 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.   

43 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.



46 H/W LC Ord 2007 Q1

47 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

48 As this is the basic constant so we say k=1 and Force=m.a
Newton 2 force Rate of change of Momentum 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

49 TO SHOW THAT a µ F Dual timer Photogate l Pulley Light beam
Card l Pulley Light beam Slotted weights Air track s

50 TO SHOW THAT a µ F Dual timer Photogate Light beam t1
t1 time for card to pass first photo-gate

51 TO SHOW THAT a µ F Dual timer Photogate Light beam t1 t2
t2 time for card to pass second photo-gate

52 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.

53 F/N t1/s t2/s V/m/s U/m.s A/m/s2
1/. 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

54 Newton’s Laws on the Internet

55 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

56 Balanced and unbalanced forces
Reaction What would happen if we took the road away? Weight

57 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

58 Balanced and unbalanced forces
1) This animal is either ________ or moving with _____ _____… 2) This animal is getting _________… 3) This animal is getting _______…. 4) This animal is…

59 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.

60 Reaction Friction Engine force Gravity

61 Force and acceleration
If the forces acting on an object are unbalanced then the object will accelerate, like these wrestlers:

62 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 A M

63 Using F=ma 1000=500xa a=2m/s2 A force of 1000N is applied to push a mass of 500kg. How quickly does it accelerate? A force of 3000N acts on a car to make it accelerate by 1.5m/s2. How heavy is the car? A car accelerates at a rate of 5m/s2. If it ‘s mass is 500kg how much driving force is the engine applying? A force of 10N is applied by a boy while lifting a 20kg mass. How much does it accelerate by? Using F=ma 3000=mx1.5 m=2000kg Using F=ma F=5x500 F=2500N Using F=ma 10=20xa a=0.5m/s2

64 Net Force creates Acceleration
Fnet=200N F=-100N F=200N Fnet=100N F=-200N F=200N Fnet=0N F=-200N Fnet=-200N

65 H/W LC Ord 2004 Q6

66 Net Force creates Acceleration
800kg F=200N Fnet=100N F=-100N As net force causes acceleration F=m.a 100N = 800kg.a a=100/800 = 0.125m/s2

67 Acceleration gives Net Force
900kg Feng=5000N a=3m/s2 Friction=? As net force causes acceleration F=m.a Fnet = 900kg. 3m/s2 Fnet= 2700N So Friction = Feng – 2700N Friction=2300N

68 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?

69 Archimedes Principle A body in a fluid experiences an up-thrust equal to the weight of liquid displaced. 12N 20N 8N

70 Internet Diagram

71 Floatation A floating body displaces its own weight in water.

72 = Floatation A floating body displaces its own weight in water. 10000t

73 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.

74 Terminal Velocity Consider a skydiver:
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

75 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 ________ _______.

76 Velocity-time graph for terminal velocity…
Parachute opens – diver slows down Velocity Speed increases… Terminal velocity reached… On the Moon Diver hits the ground New, lower terminal velocity reached Time

77 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 g M Weight = Mass x acceleration due to gravity (in N) (in kg) (in m/s2) What is the weight on Earth of a book with mass 2kg? What is the weight on Earth of an apple with mass 100g? Dave weighs 700N. What is his mass? On the moon the gravitational field strength is 1.6N/kg. What will Dave weigh if he stands on the moon?

78 Weight vs. Mass Mass is the amount of matter in us
900kg 900kg Mass is the amount of matter in us Same on Earth and Space 9000 N 0 N Weight is the pull of gravity on us Different on Earth and Space

79 Homework LC Ordinary Level 2002 Q6

80 Galileo’s Falling Balls

81 Gravity all bodies have gravity we feel it only from planet sized objects
T=0 v=0m/s Acceleration due to gravity is 9.81m/s2 That means every falling body gets 9.81m/s faster every second T=1s v=9.81m/s T=2s v=19.62m/s T=3s v=29.43m/s

82 Internet Even proved it in real life

83 All bodies fall at the same rate

84 MEASUREMENT OF g h Electromagnet Electronic timer Switch Ball bearing
MEASUREMENT OF g Electromagnet Electronic timer Switch Ball bearing h Trapdoor

85 When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens

86 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. Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release

87 The sky diver accelerates at 2m/s2 what is his drag?
Finding Drag 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= =624N 80kg

88 Newton’s Cannon

89 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. Discuss what happens when they throw a ball- it drops to the Earth. If an object is projected at VERY high speed then it goes further round the Earth. An object must be moving at high speed to orbit the Earth. Talk about satellites and the fact that the Moon is a satellite of the Earth.

90 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

91 Gravity Calculations F = G m1 m2 d2 F = G m1 m2 d2
To make an equation we add a constant G The UNIVERSAL GRAVITATIONAL CONSTANT 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

92 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

93 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?

94 H/W LC Ord 2008 Q 6

95 More force means more Extension - they are proportional
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

96 Hookes Law Calculation
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 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 Force=0N Length=5cm Ext.=0cm Force=6N Length=8cm Ext.=3cm Force=12N Length=11cm Ext.=6cm


98 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

99 Homework LC Ord 2003 Q6

100 Work done = Force x Distance Moved
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 D F

101 Kinetic energy = ½ x mass x velocity squared
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 in m/s KE = ½ mv2

102 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

103 Potential energy = mass x g x height
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

104 Work Done = Energy Converted
Work Done raising an object = PE Stored

105 Consider an oscillating pendulum Consider an oscillating pendulum

106 At the bottom the bob has no PE only KE
At the top of the oscillation the pendulum bob stops. All it’s energy is PE PE at top=KE at bottom At the bottom the bob has no PE only KE PE = mgh h KE = ½ mv2

107 mgh = ½ mv2 mgh = ½ mv2 gh = ½ v2 H=10cm v2 = 2gh v2 = 2(9.8)0.1
PE at top=KE at bottom mgh = ½ mv2 mgh = ½ mv2 H=10cm gh = ½ v2 v2 = 2gh v2 = 2(9.8)0.1 v = 1.4m/s

108 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 x 109 Joules Power = Work Done / Time = 7 x 109 Joules / 120 = 5.83 x 107 Watts

109 H/W LC Ord 2007 Q 6

110 Pressure F A P Pressure depends on two things:
How much force is applied, and How big (or small) the area on which this force is applied is. Pressure can be calculated using the equation: F A P Pressure (in N/m2) = Force (in N) Area (in m2)

111 Some example questions…
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)? 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/ m2 = 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)

112 Pressure – in Fluids Pressure increases with depth

113 As the frog goes deeper there is a greater weight of water above it.
Pressure and Depth As the frog goes deeper there is a greater weight of water above it.

114 Atmospheric Pressure 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!

115 Proving Atmospheric Pressure
Very full glass of water

116 Proving Atmospheric Pressure

117 Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place

118 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.

119 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.

120 Aneroid Barometer Works on changes in size of small can.(Get it!)

121 Pressure and Volume in gases
This can be expressed using the equation: Initial Pressure x Initial Volume = Final Press. x Final Vol. PIVI = PFVF 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? A gas increases in volume from 10m3 to 50m3. If the initial pressure was 10,000N/m2 what is the new pressure? A gas decreases in pressure from 100,000 Pascals to 50,000 Pascals. The final volume was 3m3. What was the initial volume? The pressure of a gas changes from 100N/m2 to 20N/m2. What is the ratio for volume change?

122 Pressure and Volume in gases
Pressure x volume

123 Internet Demo

124 Boyles Law Pressure is inversely proportional to volume

1.  .    Volume scale Tube with volume of air trapped by oil Bicycle pump Reservoir of oil Valve Pressure gauge

126 Using the pump, increase the pressure on the air in the tube
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.

127 P



130 Hydraulic systems

131 Hydraulic systems Pressure is constant throughout this liquid

132 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! 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

133 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)

134 H/W LC Ord 2005 Q6

135 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

136 Unstable Equilibrium Things fall over because the center of gravity is outside the base

137 Moments (Also called TORQUE) =Force x Perpendicular distance
Fulcrum Perpendicular distance FORCE

138 Moments =Force x Perpendicular distance
= 10N x 5m = 50Nm Perpendicular distance=5m FORCE =10N

139 More than two forces 15N ?N 50 90 60 70 10 ? 15N 5N 10N 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)

140 First law coplanar forces Forces Up = Forces Down
50 90 60 70 ? 10 15N 5N 10N 5N First law coplanar forces Forces Up = Forces Down 15 + x = x = 20 N

141 Second law coplanar forces Take moments about A
32.5 50 60 70 90 ? 10 A 15N 5N 10N 5N Second law coplanar forces Take moments about A Clockwise Moments = Anticlockwise Moments 10x x5 + 70x x5 = 60x15 + dx20 = dx20 = dx so d=32.5cm

EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES (2) Support Newton balance Newton balance Paperclips w2 w4 w1 w3

143 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

144 For each situation (1). Forces up = Forces down i. e
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.

145 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.

146 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?

147 H/W LC Ord 2003 Q12(a) Last h/w before xmas (honest)

148 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

149 Motion in a circle Velocity always at 90o to the force or acceleration

150 Circular Motion Angular Velocity =θ/t Units of Radians per second
Angle  time A particle goes round a circle in 4s what is it’s angular velocity? t

151 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

152 Centripetal Acceleration
a = r 2 Always towards the centre So the acceleration in the previous example a= 6 (/2)2 =14.8m/s2

153 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

154 Centripetal Acceleration

155 Satellites balance forces
Balance of Gravity and Centripetal ((GMm)/d2)=mv2/d Gravity F=-GmM/r2

156 Period of Orbit ((GMm)/d2)=mv2/d (GM)/d=v2 (GM)/d=(2d/T)2
Equate The Forces ((GMm)/d2)=mv2/d (GM)/d=v2 (GM)/d=(2d/T)2 T2=42 d3 /GM Cancel Mass of satellite V=Distance time T=Period (Time for Orbit)

157 In a test we do it like this

158 Example of Orbits What is the parking orbit height above Saturn if it is km 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

159 Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot. What is it’s height above the earth?

160 Simple Harmonic Motion
Is a vibration where the acceleration is proportional to the displacement a  -s Further from centre =more acceleration

161 So motion under Hookes law is SHM
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

162 H/W LC Ord 2006 Q6

163 Pendulum If we displace the bob by a small angle it vibrates with SHM
Split cork If we displace the bob by a small angle it vibrates with SHM l Timer Bob 20:30

164 T2 l

165 Time to go over h/w LC Ord 2008 Q1 LC Ord 2007 Q1 LC Ord 2004 Q6
LC Ord Q12(a) LC Ord Q6

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