Presentation is loading. Please wait.

Presentation is loading. Please wait.

Forces By Neil Bronks Force causes a body to change velocity…….. accelerate The unit is called the Newton (N)

Similar presentations

Presentation on theme: "Forces By Neil Bronks Force causes a body to change velocity…….. accelerate The unit is called the Newton (N)"— Presentation transcript:


2 Forces By Neil Bronks

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

4 Distance, Speed and Time Speed = distance (in metres) time (in seconds) D TS 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.

5 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

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

7 Speed vs. Velocity Speed is simply how fast you are travelling… Velocity is “speed in a given direction”… This car is travelling at a speed of 20m/s This car is travelling at a velocity of 20m/s east

8 Scalar vs. Vector Scalar has only magnitude….. mass Vector has magnitude and direction …….. Weight This car has a mass of 2000kg This car has a Weight of 20000N

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

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

11 Vectors in opposite directions: 6 m s -1 10 m s -1 =4 m s -1 6 N9 N=3 N Vectors in the same direction: 6 N4 N=10 N 6 m =10 m 4 m TThe resultant is the sum or the combined effect of two vector quantities

12  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

13 Vector Addition Speed in still air 120m/s Wind 50m/s Resultant R 2 = 120 2 + 50 2 = 14400 + 2500 = 16900 R = 130m/s  Tan  = 50/120  = 22.6 0

14 Solution: CComplete the parallelogram (rectangle) θ TThe diagonal of the parallelogram ac represents the resultant force 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? 5 N 12 N 5 12 a bc d  The magnitude of the resultant is found using Pythagoras’ Theorem on the triangle abc RResultant displacement is 13 N 67 º with the 5 N force 13 N

15 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 y v x HHere a vector v is resolved into an x component and a y component

16 Here we see a table being pulled by a force of 50 N at a 30 º angle to the horizontal 50 N y=25 N x=43.3 N 30 º 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

17 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 x=v Cos θ θ y  Proof: x

18 60 º 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 Horizontal Component Solution: 15 N 7.5 N 12.99 N

19 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 10 º 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. 10 º 80 º 900 N Complete the parallelogram. Component of weight parallel to ramp: Component of weight perpendicular to ramp: 156.28 N 886.33 N H/W - 2003 HL Section B Q6

20 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 x=v Cosθ θ y

21 Acceleration V-U TA Acceleration = change in velocity (in m/s) (in m/s 2 ) time taken (in s) 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/s 2 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/s 2. How long did this take? 4)A rocket accelerates from 1,000m/s to 5,000m/s in 2 seconds. What is its acceleration?

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

23 Velocity-time graphs 80 60 40 20 0 10 20 30 4050 Velocity m/s T/s 1)Upwards line = Constant Acceleration 2) Horizontal line = Constant Velocity 3) Shallow line = Less Acceleration 4) Downward line = Deceleration

24 80 60 40 20 0 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? 10 20 30 4050 Velocity m/s T/s

25 80 60 40 20 0 The area under the graph is the distance travelled by the object 10 20 30 4050 Velocity m/s T/s

26 80 60 40 20 0 10 20 30 4050 Velocity m/s T/s 40x20=800 Total Distance Traveled =200+100+800+600=1700m 0.5x10x40=200 0.5x10x20=100 0.5x20x60 =600

27 Motion Formula v = u + at A car starts from rest and accelerates for 12s at 2ms -2. Find the final velocity. U=30 s=200 and v = 0 find a=? v2 v2 = u2 u2 + 2as A car traveling at 30m/s takes 200m to stop what is it’s deceleration? Using V 2 = U 2 + 2as 0 = 900 + 2a (200) a = -900/400=-2.25ms -2 Using V = U + at = 0 + 2x12 = 24m/s U=0 a=2 and t = 12 find v=?

28 Motion Formula S = ut + 0.5at 2 A train accelerates from rest at 10ms -2 for 12s find the distance it has traveled. Using S = ut + 0.5at 2 = 0x12 +0.5x10x144 =720m




32 Velocity and Acceleration Card l t1t1 t2t2 s Dual timer Light beam Photogate Pulley Air track Slotted weights


34 H/W LC Ord 2008 Q 1

35 Friction is the force that opposes motion The unit is called the Newton (N) Friction is the force between two bodies in contact. Lubrication reduces friction

36 Lubrication reduces friction and separates the two bodies

37 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.

38 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?

39 Recoil Momentum of the Shoot Momentum of Recoil = Mass of Ball x Velocity of Ball Mass Canon x Velocity Canon = V= 800/150 = 2 x 400 150 x U c = u b =400m/s m=2kg Mass of canon=150kg 5.3m/s

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

41 Internet Calculations

42 VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM Velcro pad Dual timer Photogate Air track t1t1 t2t2 Light beam Card l Vehicle 1 Vehicle 2

43 1.Set up apparatus as in the diagram. Level the air-track. To see if the track is level carry out these tests: 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 m 1 and m 2 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 t 1 and t 2 for the card through the two beams.

44 Calculate the velocity before the collision, and after the collision, momentum before the collision=momentum after the collision, m 1 u = (m 1 + m 2 ) v. Repeat several times, with different velocities and different masses.



47 H/W LC Ord 2007 Q1

48 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

49 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

50 TO SHOW THAT a  F Card l t1t1 t2t2 s Dual timer Light beam Photogate Pulley Air track Slotted weights

51 TO SHOW THAT a  F t1t1 Dual timer Light beam Photo gate t 1 time for card to pass first photo-gate

52 TO SHOW THAT a  F t1t1 Dual timer Light beam Photo gate t 2 time for card to pass second photo-gate t2t2

53 Procedure Make sure the card cuts both light beams as it passes along the track. 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. Level the air track. Set the weight F at 1 N. Release the vehicle. Note the times t 1 and t 2. repeat 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.

54 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 F/Nt1/st2/sV/m/sU/m.sA/m/s 2

55 Newton’s Laws on the Internet

56 Balanced and unbalanced forces Consider a camel standing on a road. What forces are acting on it? Weight Reaction These two forces would be equal – we say that they are BALANCED. The camel doesn’t move anywhere.

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

58 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

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

60 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 Engine force Friction Gravity Reaction

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

63 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/s 2 ) F AM

64 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/s 2. How heavy is the car? 3)A car accelerates at a rate of 5m/s 2. 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 1000=500xa a=2m/s 2 Using F=ma 3000=mx1.5 m=2000kg Using F=ma F=5x500 F=2500N Using F=ma 10=20xa a=0.5m/s 2

65 Net Force creates Acceleration F net =200N F=200N F net =100N F net =0N F net =-200N F=-200N F=-100N F=-200N F=200N

66 H/W LC Ord 2004 Q6

67 Net Force creates Acceleration F=200N F net =100N F=-100N 800kg As net force causes acceleration F=m.a 100N = 800kg.a a=100/800 = 0.125m/s 2

68 Acceleration gives Net Force F eng =5000N a=3m/s 2 Friction=? 900kg As net force causes acceleration F=m.a F net = 900kg. 3m/s 2 F net = 2700N So Friction = F eng – 2700N Friction=2300N

69 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/s 2 ? If the engine stops how long before the car stops if it is travelling at 20m/s when the engine cuts out?

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

71 Internet Diagram

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

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

74 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.specific gravityrelative densityliquids

75 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

76 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 ________ _______.

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

78 Weight vs. Mass Earth’s Gravitational Field Strength is 9.8m/s 2. In other words, a 1kg mass is pulled downwards by a force of 9.8N. W gM Weight = Mass x acceleration due to gravity (in N) (in kg) (in m/s 2 ) 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?

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

80 Homework LC Ordinary Level 2002 Q6

81 Galileo’s Falling Balls

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

83 Internet Even proved it in real life

84 All bodies fall at the same rate

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

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

87 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

88 Finding Drag The sky diver accelerates at 2m/s 2 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 80kg Drag

89 Newton’s Cannon

90 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.

91 Newton's Law of Gravitation This force is always positiveThis force is always positive Called an inverse square lawCalled an inverse square law F  m 1 m 2 d 2 Where F = Gravitational Force m 1. m 2 = Product of masses d = Distance between their center of gravity

92 Gravity Calculations 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=6.6x10 23 kg and radius=3.4x10 6 m G=6.67x10 -11 Nm 2 kg -2 F = G m 1 m 2 d 2 = 6.67x10 -11 x 6.6x10 23 x100 (3.4x10 6 ) 2 = 380N F = G m 1 m 2 d 2

93 2010 Question 6 [Higher Level] (Radius of the earth = 6.36 × 10 6 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 × 10 8 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

94 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 H/W LC Ord 2008 Q 6

96 Hookes Law 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Force Extension More force means more Extension - they are proportional

97 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=0N Length=5cm Ext.=0cm Force=6N Length=8cm Ext.=3cm Force=12N Length=11cm Ext.=6cm Force =24N Length =17cm Ext. =12cm


99 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

100 Homework LC Ord 2003 Q6

101 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 DF

102 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 in m/s KE = ½ mv 2

103 Some example questions… A 70kg boy is running at about 10m/s. What is his kinetic energy? Using KE=½mv 2 =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=½mv 2 =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 mv 2

104 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

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

106 Consider an oscillating pendulum

107 PE = mgh KE = ½ mv 2 At the top of the oscillation the pendulum bob stops. All it’s energy is PE At the bottom the bob has no PE only KE PE at top=KE at bottom h

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

109 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 10 9 Joules Power = Work Done / Time = 7 x 10 9 Joules / 120 = 5.83 x 10 7 Watts

110 H/W LC Ord 2007 Q 6

111 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: Pressure (in N/m 2 ) = Force (in N) Area (in m 2 ) F AP

112 Some example questions… 1)A circus elephant weighs 10,000N and can stand on one foot. This foot has an area of 50cm 2. 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 1cm 2. How much pressure does she exert on the floor? Pressure=Force/area = 500N/ 0.0001m 2 = 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/m 2 )

113 Pressure – in Fluids Pressure increases with depth

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

115 Atmospheric Pressure The earth is covered with layer of Gas.The earth is covered with layer of Gas. We are at the bottom of a gas ocean 200km deep.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.The effect of this huge column of gas is 1 Tonne of weight on our shoulders. This is calledThis is called ATMOSPHERIC PRESSUREATMOSPHERIC PRESSURE Heavy!

116 Proving Atmospheric Pressure Very full glass of water

117 Proving Atmospheric Pressure

118 Now the atmospheric Pressure holds the card in place

119 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.

120 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.

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

122 Pressure and Volume in gases This can be expressed using the equation: Initial Pressure x Initial Volume = Final Press. x Final Vol. P I V I = P F V F 1)A gas has a volume of 3m 3 at a pressure of 20N/m 2. What will the pressure be if the volume is reduced to 1.5m 3 ? 2)A gas increases in volume from 10m 3 to 50m 3. If the initial pressure was 10,000N/m 2 what is the new pressure? 3)A gas decreases in pressure from 100,000 Pascals to 50,000 Pascals. The final volume was 3m 3. What was the initial volume? 4)The pressure of a gas changes from 100N/m 2 to 20N/m 2. What is the ratio for volume change?

123 Pressure and Volume in gases PressureVolumePressure x volume 2010200 2001200 450200

124 Internet Demo

125 Boyles Law Pressure is inversely proportional to volume

126 VERIFICATION OF BOYLE’S LAW 1.. Bicycle pump Reservoir of oil Pressure gauge Tube with volume of air trapped by oil Volume scale Valve

127  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.

128 P



131 Hydraulic systems

132 Pressure is constant throughout this liquid

133 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. 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 Magic!

134 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 × 10 5 Pa. The temperature of the lake is 4 o C. 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 × 10 3 kg m –3 )

135 H/W LC Ord 2005 Q6

136 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

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

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

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

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

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

142 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 10 60 90 70 50 ? 15N 5N 10N 5N 20N A 32.5

143 INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES (2) Newton balance Support Paperclips w1w1 w2w2 w3w3 w4w4

144 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

145 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.

146 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 outhere Notice the units of torque are included as we should.

147 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?

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

149 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

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

151 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t

152 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

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

154 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

155 Centripetal Acceleration

156 Satellites balance forces Balance of Gravity and Centripetal ((GMm)/d 2 )=mv 2 /d Gravity F=-GmM/r 2

157 Period of Orbit ((GMm)/d 2 )=mv 2 /d (GM)/d=v 2 (GM)/d=(2  d/T) 2 T 2 =4  2 d 3 /GM Equate The Forces V=Distance time T=Period (Time for Orbit) Cancel Mass of satellite

158 In a test we do it like this

159 Example of Orbits Using T 2 =4  2 d 3 /GM (4x24x60x60) 2 =4  2 d 3 /(2x10 31 )(6.7x10 -11 ) d 3 = 1x10 30 d = 1x10 10 m Height =h= d - r =1x10 10 m - 2x10 8 m = 9.8x10 9 m What is the parking orbit height above Saturn if it is 200000km in radius. It rotates every 4 days and has mass 2x10 31 Kg. The Universal gravitational Constant is 6.7x10 -11 d r h

160 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?

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

162 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

163 H/W LC Ord 2006 Q6

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

165 l T2T2

166 Time to go over h/w LC Ord 2008 Q1 LC Ord 2007 Q1 LC Ord2004 Q6 LC Ord2002 Q6 LC Ord 2008Q6 LC Ord2003 Q6 LC Ord 2007Q6 LC Ord2005 Q6 LC Ord2003 Q12(a) LC Ord2006 Q6

Download ppt "Forces By Neil Bronks Force causes a body to change velocity…….. accelerate The unit is called the Newton (N)"

Similar presentations

Ads by Google