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 ScalarsA scalar quantity is a quantity that has magnitude only and has no direction in spaceExamples of Scalar Quantities:LengthAreaVolumeTimeMass
5 VectorsA vector quantity is a quantity that has both magnitude and a direction in spaceExamples of Vector Quantities:DisplacementVelocityAccelerationForce
6 Speed vs. Velocity Speed is simply how fast you are travelling… This car is travelling at a speed of 20m/sVelocity 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. VectorScalar has only magnitude….. massThis car has a mass of 2000kgVector has magnitude and direction …….. WeightThis car has a Weight of 20000N
8 Distance and Displacement Scalar- Distance travelled 200mVector- Displacement 120m
9 Vector Diagrams Vector diagrams are shown using an arrow The length of the arrow represents its magnitudeThe direction of the arrow shows its direction
10 Resultant of Two Vectors The resultant is the sum or the combined effect of two vector quantitiesVectors in the same direction:6 N 4 N = 10 N6 m= 10 m4 mVectors in opposite directions:6 m s m s-1 = 4 m s-16 N 9 N = 3 N
11 The Parallelogram Law The Triangle Law When two vectors are joined tail to tailComplete the parallelogramThe resultant is found by drawing the diagonalThe Triangle LawWhen two vectors are joined head to tailDraw the resultant vector by completing the triangle
12 Vector Addition Resultant Speed in still air 120m/s Wind 50m/s = 16900R = 130m/sTan = 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 forceThe magnitude of the resultant is found using Pythagoras’ Theorem on the triangle abc12 Nadθ5 N13 N5bc12Resultant 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 resultantWe usually resolve a vector into components that are perpendicular to each othervyxHere 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 horizontal50 Ny=25 N30ºx=43.3 NWhen 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 θy = v Sin θvy=v Sin θyProof:θx=v Cos θx
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 N15 NComponentVertical60ºHorizontalComponent7.5 N
18 H/W HL Section B Q6A 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 weightparallel to ramp:N10º80º10ºComponent of weight perpendicular to ramp:N900 N
19 SummaryIf 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 θy= v Sin θvy=v Sin θyθx=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?
22 Constant Acceleration Velocity-time graphs80604020Upwards line =Constant Acceleration4) Downward line =DecelerationVelocitym/s3) Shallow line =Less Acceleration2) Horizontal line =Constant VelocityT/s
23 80604020Velocitym/sT/sHow 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 80604020Velocitym/sT/sThe area under the graph is the distance travelled by the object
25 Total Distance Traveled 806040200.5x10x20=100Velocitym/s0.5x20x60=60040x20=8000.5x10x40=200T/sTotal Distance Traveled= =1700m
26 A car traveling at 30m/s takes 200m to stop what is it’s deceleration? Motion FormulaA car starts from rest and accelerates for 12s at 2ms-2. Find the final velocity.v = u + atU=0 a=2 and t = 12 find v=?Using V = U + at = 0 + 2x12 = 24m/sv2 = u2 + 2asA 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 + 2as0 = 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
34 Friction is the force that opposes motion The unit is called the Newton (N)Lubricationreduces frictionFriction 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 frictionWithout 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 ShootMass Canon x Velocity Canon =Mass of Ball x Velocity of Ball2 x 400150 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 conservedMomentum Before = Momentum AfterMass Moving x velocity before = Mass moving x velocity after3kg x 10m/s = 3kg x (-2m/s) + 6kg x v6v =V = 6m/s
41 VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM Dual timert1t2PhotogateLight beamCardlVehicle 2Air trackVehicle 1Velcro 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.
47 Newton’s Laws1 /. Every body stays in it’s state of rest or constant motion until an outside force acts on it2/. The rate of change of momentum is proportional to the applied force and in the direction of the applied force.F=ma3/. 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 2force Rate of change of MomentumForcem.aOr Force=k.m.a where k=constantAs 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 CardlPulleyLight beamSlotted weightsAir tracks
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 ProcedureSet 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
55 Balanced and unbalanced forces ReactionConsider 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 ReactionWhat 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.
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)FAM
63 Using F=ma1000=500xaa=2m/s2A 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=ma3000=mx1.5m=2000kgUsing F=maF=5x500F=2500NUsing F=ma10=20xaa=0.5m/s2
64 Net Force creates Acceleration Fnet=200NF=-100NF=200NFnet=100NF=-200NF=200NFnet=0NF=-200NFnet=-200N
66 Net Force creates Acceleration 800kgF=200NFnet=100NF=-100NAs net force causes acceleration F=m.a100N = 800kg.aa=100/800 = 0.125m/s2
67 Acceleration gives Net Force 900kgFeng=5000Na=3m/s2Friction=?As net force causes acceleration F=m.aFnet = 900kg. 3m/s2Fnet= 2700NSo Friction = Feng – 2700NFriction=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 PrincipleA body in a fluid experiences an up-thrust equal to the weight of liquid displaced.12N20N8N
71 FloatationA 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 Velocity4) 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 downVelocitySpeed increases…Terminal velocity reached…On the MoonDiver hits the groundNew, lower terminal velocity reachedTime
77 Weight vs. MassEarth’s Gravitational Field Strength is 9.8m/s2. In other words, a 1kg mass is pulled downwards by a force of 9.8N.WgMWeight = 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 900kg900kgMass is the amount of matter in usSame on Earth and Space9000 N0 NWeight is the pull of gravity on usDifferent on Earth and Space
81 Gravity all bodies have gravity we feel it only from planet sized objects T=0 v=0m/sAcceleration due to gravity is 9.81m/s2That means every falling body gets 9.81m/s faster every secondT=1s v=9.81m/sT=2s v=19.62m/sT=3s v=29.43m/s
84 MEASUREMENT OF g h Electromagnet Electronic timer Switch Ball bearing MEASUREMENT OF gElectromagnetElectronic timerSwitchBall bearinghTrapdoor
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 trapdoorMeasure 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 DragDragThe sky diver accelerates at 2m/s2 what is his drag?Force due to gravity=80.g=80.(9.8)=784 NNet Force=m.a=80.2=160NDrag= =624N80kg
89 Launching a satelliteThe cannon ball is constantly falling towards the earth but earth curve is same as it’s pathThe 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 positiveCalled an inverse square lawF m1m2d2WhereF = Gravitational Forcem1.m2 = Product of massesd = 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 constantG The UNIVERSAL GRAVITATIONAL CONSTANTExample What is the force on a man of mass 100kg standing on the surface of Mars.Mars mass=6.6x1023 kg and radius=3.4x106mG=6.67x10-11 Nm2kg-2F = G m1 m2d2= 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−2Distance from the centre of the earth to the centre of the moon = 3.84 × 108 mAssume 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?
95 More force means more Extension - they are proportional Hookes LawForce1234567891011121314151617181920ExtensionMore force means more Extension - they are proportional
96 Hookes Law Calculation 1234567891011121314151617181920123456789101112131415161718192012345678910111213141516171819201234567891011121314151617181920Force=24NLength=17cmExt.=12cmForce=0NLength=5cmExt.=0cmForce=6NLength=8cmExt.=3cmForce=12NLength=11cmExt.=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 0.3K=98N/mB/. What is the extension if 40N is put on the same spring?Force = Spring Constant x Extension40 = 98 x sS = 40/98 = 0.41 m
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 Movedin J in N in mWDF
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 squaredin J in kg in m/sKE = 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=3500JA 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 = 50000JWhat is the kinetic energy of a 100g tennis ball being thrown at a speed of 5m/s?Using KE=½mv2=0.5x0.1x5x5=1.25JA 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 = 98100JKE = 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 upThe formula is for high objects:Potential energy = mass x g x heightPE = 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 PEPE at top=KE at bottomAt the bottom the bob has no PE only KEPE = mghhKE = 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 bottommgh = ½ mv2mgh = ½ mv2H=10cmgh = ½ v2v2 = 2ghv2 = 2(9.8)0.1v = 1.4m/s
108 Power The rate at which work is done POWER = Work Done time taken ExampleA 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 JoulesPower = Work Done / Time = 7 x 109 Joules / 120= 5.83 x 107 Watts
110 Pressure F A P Pressure depends on two things: How much force is applied, andHow big (or small) the area on which this force is applied is.Pressure can be calculated using the equation:FAPPressure (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 = PaExtension 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 FluidsPressure increases with depth
113 As the frog goes deeper there is a greater weight of water above it. Pressure and DepthAs 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 calledATMOSPHERIC PRESSUREHeavy!
115 Proving Atmospheric Pressure Very full glass of water
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 pressureWe can measure the altitude using a barometer with a different scale.
120 Aneroid BarometerWorks 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 = PFVFA 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
124 Boyles LawPressure is inversely proportional to volume
125 VERIFICATION OF BOYLE’S LAW 1. . VolumescaleTube with volume of air trapped by oilBicycle pumpReservoir of oilValvePressure 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.
131 Hydraulic systemsPressure is constant throughout this liquid
132 Hydraulic systemsBasically, 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=100PaPressure in Master = Force/1 = 100PaForce 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)
135 Center of GravityThings stay standing (STABLE) because their Center of Gravity acts through their base.The perpendicular from the COG passes inside the support
136 Unstable EquilibriumThings fall over because the center of gravity is outside the base
137 Moments (Also called TORQUE) =Force x Perpendicular distance FulcrumPerpendicular distanceFORCE
138 Moments =Force x Perpendicular distance = 10N x 5m = 50NmPerpendicular distance=5mFORCE =10N
139 More than two forces15N?N5090607010?15N5N10N5NFirst 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 50906070?1015N5N10N5NFirst law coplanar forcesForces Up = Forces Down15 + x =x = 20 N
141 Second law coplanar forces Take moments about A 32.550607090?10A15N5N10N5NSecond law coplanar forcesTake moments about AClockwise Moments = Anticlockwise Moments10x x5 + 70x x5 = 60x15 + dx20= dx20= dx so d=32.5cm
142 INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES(2)SupportNewton balanceNewton balancePaperclipsw2w4w1w3
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 InternetOk 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 outNotice 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/WLC Ord2003 Q12(a)Last h/w before xmas (honest)
148 Couples of ForcesTwo equal forces that cause a solid to rotate around an axisMoment = Force x DistanceMoment = 5Nx0.06mMoment = 0.3 Nm
149 Motion in a circleVelocity always at 90o to the force or acceleration
150 Circular Motion Angular Velocity =θ/t Units of Radians per second Angle timeA 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 accelerationIf the radius is 6m
152 Centripetal Acceleration a = r 2Always towards the centreSo 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 2Tension in string of weight spun around headForce on tyres (Or camel) as we go around corner
158 Example of OrbitsWhat 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 6.7x10-11Using T2=42 d3 /GM(4x24x60x60)2=42 d3 /(2x1031)(6.7x10-11)d3 = 1x1030d = 1x1010mHeight =h= d - r =1x1010m - 2x108m= 9.8x109mhdr
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 displacementa -sFurther from centre =more acceleration
161 So motion under Hookes law is SHM Hooke’s Law as SHMForce ExtensionF -sm.a -sIf mass is constanta -sSo motion under Hookes law is SHM