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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 2 km in 1, 000 seconds. What is her speed? 3) How long would it take to run 100 metres if you run at 10 m/s? 4) Steve travels at 50 m/s for 20 s. How far does he go? 5) Susan drives her car at 85 mph (about 40 m/s). How long does it take her to drive 20 km? 6) Convert 450 m/s into km/hr.
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
Vectors • 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 20 m/s Velocity is “speed in a given direction”… This car is travelling at a velocity of 20 m/s east
Scalar vs. Vector Scalar has only magnitude…. . mass This car has a mass of 2000 kg Vector has magnitude and direction ……. . Weight This car has a Weight of 20000 N
Distance and Displacement Scalar- Distance travelled 200 m Vector. Displacement 120 m
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
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 m 6 m 4 m Vectors in opposite directions: 6 m s-1 10 m s-1 = 4 m s-1 6 N 9 N = 3 N
The Parallelogram 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
Vector Addition Speed in still air 120 m/s Res ulta nt Wind 50 m/s R 2 = 1202 + 502 = 14400 + 2500 = 16900 Tan = 50/120 R = 130 m/s = 22. 60
Problem: Resultant of 2 Vectors 2004 HL Section B Q 5 (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 a 5 N b θ 12 N 13 12 d N 5 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 v • 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 Resolving a Vector Into Perpendicular Components x
Practical Applications • Here we see a table being pulled by a force of 50 N at a 30º angle to the horizontal y=25 N 50 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
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 Proof: θ x=v Cos θ x y
Problem: Calculating the magnitude of perpendicular components 2002 HL Sample Paper Section B Q 5 (a) A force of 15 N acts on a box as shown. What is the horizontal component of the force? 15 N 12. 99 N Vertical Component Solution: 60º Horizontal Component 7. 5 N
H/W - 2003 HL Section B Q 6 • 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 8 N 56. 2 1 10º parallel to ramp: 80º Component of weight perpendicular to ramp: 10º 900 N 886. 33 N
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: v • 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/s 2) time taken (in s) A T 1) A cyclist accelerates from 0 to 10 m/s in 5 seconds. What is her acceleration? 2) A ball is dropped and accelerates downwards at a rate of 10 m/s 2 for 12 seconds. How much will the ball’s velocity increase by? 3) A car accelerates from 10 to 20 m/s with an acceleration of 2 m/s 2. How long did this take? 4) A rocket accelerates from 1, 000 m/s to 5, 000 m/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 30 Shallow line 50 40 3) = 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 50 s? 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. 5 x 10 x 20=100 Velocity m/s 40 20 0. 5 x 10 x 40=200 0 0. 5 x 20 x 60 =600 40 x 20=800 10 20 30 40 Total Distance Traveled =200+100+800+600=1700 m 50 T/s
Motion Formula v = u + at A car starts from rest and accelerates for 12 s at 2 ms-2. Find the final velocity. U=0 a=2 and t = 12 find v=? Using V = U + at = 0 + 2 x 12 = 24 m/s v 2 = u 2 + 2 as A car traveling at 30 m/s takes 200 m to stop what is it’s deceleration? U=30 s=200 and v = 0 find a=? Using V 2 = U 2 + 2 as 0 = 900 + 2 a (200) a = -900/400=-2. 25 ms-2
Motion Formula S = ut + 0. 5 at 2 A train accelerates from rest at 10 ms-2 for 12 s find the distance it has traveled. Using S = ut + 0. 5 at 2 = 0 x 12 +0. 5 x 10 x 144 =720 m
Proofs If you want to start with a any formula you can get any other try to prove
Velocity and Acceleration t 1 Light beam Air track Dual timer t 2 l Photogate Pulley Card s Slotted weights
H/W • LC H 2008 • Q 1
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.
1) What is friction? 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=2 kg Mass of canon=150 kg ub=400 m/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. 3 m/s
Momentum 10 m/s V=? m/s 3 kg 2 kg 6 kg 3 kg 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 3 kg x 10 m/s = 3 kg x (-2 m/s) + 6 kg x v 6 v = 30 + 6 V = 6 m/s
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM t 1 l t 2 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 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.
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.
H/W • LC Ord • 2007 Q 1
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 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 µ F t 1 Light beam Air track Dual timer t 2 l Photogate Pulley Card s Slotted weights
TO SHOW THAT a µ F t 1 Dual timer Photog ate Light beam t 1 time for card to pass first photo-gate
TO SHOW THAT a µ F t 1 t 2 Dual timer Photog ate Light beam t 2 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 t 1 and t 2. 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.
F/N 1/. t 1/s t 2/s V/m/s U/m/s A/m/s 2 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/s 2) F M A
1) A force of 1000 N is applied to push a mass of 500 kg. How quickly does it accelerate? 2) A force of 3000 N acts on a car to make it accelerate by 1. 5 m/s 2. How heavy is the car? 3) A car accelerates at a rate of 5 m/s 2. If it ‘s mass is 500 kg how much driving force is the engine applying? 4) A force of 10 N is applied by a boy while lifting a 20 kg mass. How much does it accelerate by? Using F=ma 10=20 xa a=0. 5 m/s 2 Using F=ma 1000=500 xa a=2 m/s 2 Using F=ma 3000=mx 1. 5 m=2000 kg Using F=ma F=5 x 500 F=2500 N
Net Force creates Acceleration Fnet=200 N F=-100 N F=-200 N F=200 N Fnet=100 N F=200 N Fnet=0 N F=-200 N Fnet=-200 N
Net Force creates Acceleration 800 kg F=-100 N F=200 N As net force causes acceleration F=m. a. Fnet=100 N = 800 kg. a a=100/800 = 0. 125 m/s 2
Acceleration gives Net Force 900 kg Friction=? Feng=5000 N As net force causes acceleration F=m. aa=3 m/s 2 Fnet = 900 kg. 3 m/s 2 Fnet= 2700 N So Friction = Feng – 2700 N Friction=2300 N
A car of mass 500 kg has an engine that produces 3 k. N 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 20 m/s when the engine cuts out?
H/W • LC H • 2007 Q 12 (a)
Archimedes Principle • A body in a fluid experiences an up-thrust equal to the weight of liquid displaced. 12 N 20 N 8 N
Floatation • A floating body displaces its own weight in water.
Floatation • A floating body displaces its own weight in water. = 10000 t
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… On n Moo he t Time New, lower terminal velocity reached Diver hits the ground
Weight vs. Mass Earth’s Gravitational Field Strength is 9. 8 m/s 2. In other words, a 1 kg mass is pulled downwards by a force of 9. 8 N. W Weight = Mass x acceleration due to gravity (in N) (in kg) (in m/s 2) M g 1) What is the weight on Earth of a book with mass 2 kg? 2) What is the weight on Earth of an apple with mass 100 g? 3) Dave weighs 700 N. What is his mass? 4) On the moon the gravitational field strength is 1. 6 N/kg. What will Dave weigh if he stands on the moon?
Weight vs. Mass 900 kg • 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 0 N
Homework • LC Higher Level • 2007 Q 1
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. 81 m/s 2 • That means every falling body gets 9. 81 m/s faster every second T=1 s v=0 m/s v=9. 81 m/s T=2 s v=19. 62 m/s T=3 s v=29. 43 m/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. Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
Finding Drag The sky diver accelerates at 2 m/s 2 what is his drag? Force due to gravity=80. g =80. (9. 8)=784 N Net Force=m. a=80. 2=160 N Drag=784 -160=624 N Drag 80 kg
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 m 1 m 2 d 2 Where F = Gravitational Force m 1. m 2 = 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 m 1 m 2 d 2 Example What is the force on a man of mass 100 kg standing on the surface of Mars mass=6. 6 x 1023 kg and radius=3. 4 x 106 m G=6. 67 x 10 -11 Nm 2 kg-2 F = G m 1 m 2 d 2 = 6. 67 x 10 -11 x 6. 6 x 1023 x 100 (3. 4 x 106)2 = 380 N
• 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 2 d above surface is 3 d 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 • Q 6
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=0 N Length=5 cm Ext. =0 cm 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Force=6 N Length=8 cm Ext. =3 cm 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Force=12 N Length=11 cm Ext. =6 cm 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Force =24 N Length =17 cm Ext. =12 cm
Hookes Law Example Force =Constant (k) x Extension Example a/. A mass of 3 kg causes an extension of 0. 3 m what is the spring constant? 3 x 9. 8 = k x 0. 3 K=98 N/m B/. What is the extension if 40 N 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 Q 6
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 mv 2 in m/s
Some example questions… A 70 kg boy is running at about 10 m/s. What is his kinetic energy? Using KE=½mv 2=0. 5 x 70 x 10=3500 J A braking force of 1000 N is applied by a driver to stop his car. The car covered 50 m before it stopped. How much work did the brakes do? Work Done = force x distance = 1000 x 50 = 50000 J What is the kinetic energy of a 100 g tennis ball being thrown at a speed of 5 m/s? Using KE=½mv 2=0. 5 x 0. 1 x 5 x 5=1. 25 J A crane is lifting a 50 kg load up into the air with a constant speed. If the load is raised by 200 m how much work has the crane done? Work Done = force x distance = 50 x 9. 81 x 200 = 98100 J KE = ½ mv 2
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 mv 2 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 mv 2 H=10 cm gh = ½ 2 v v 2 = 2 gh v 2 = 2(9. 8)0. 1 v = 1. 4 m/s
Power • The rate at which work is done • POWER = Work Done time taken Example A jet takes 2 mins to climb to 4000 m. If the jet has mass 200 tonnes find the work done and the power? Work Done = Force x Distance = 200 x 1000 x 9. 81 x 4000 =7. 8 x 109 Joules Power = Work Done / Time = 7. 8 x 109 Joules / 120 = 6. 5 x 107 Watts
H/W • LC H 2013 • Q 12(a)
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/m 2) = Force (in N) Area (in m 2) P A
Some example questions… 1) A circus elephant weighs 10, 000 N and can stand on one foot. This foot has an area of 50 cm 2. How much pressure does he exert on the floor (in Pa)? 2) A 50 kg woman copies the elephant by standing on the heel of one of her high-heeled shoes. This heel has an area of 1 cm 2. How much pressure does she exert on the floor? Pressure=Force/area = 500 N/ 0. 0001 m 2 = 5000000 Pa Extension task: Atmospheric pressure is roughly equivalent to 1 kg pressing on every square centimetre on our body. What does this equate to in units called Pascals? (1 Pascal = 1 N/m 2)
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 200 km 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. 4 m high. • 1 Atmosphere is 760 mm 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. P IV I = P F V F 1) A gas has a volume of 3 m 3 at a pressure of 20 N/m 2. What will the pressure be if the volume is reduced to 1. 5 m 3? 2) A gas increases in volume from 10 m 3 to 50 m 3. If the initial pressure was 10, 000 N/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 3 m 3. What was the initial volume? 4) The pressure of a gas changes from 100 N/m 2 to 20 N/m 2. What is the ratio for volume change?
Pressure and Volume in gases Pressure Volume Pressure x volume 20 10 200 1 200 4 50 200
Boyles Law Pressure is inversely proportional to volume
VERIFICATION OF BOYLE’S LAW Volume scale Bicycle pump Valve Tube with volume of air trapped by oil Reservoir of oil Pressure gauge
ØUsing 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. ØRead the volume V of the air column from the scale. ØTake the corresponding pressure reading from the gauge and record the pressure P of the trapped air. ØReduce 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. ØRecord the corresponding values for the volume V and pressure P. ØRepeat steps two to five to get at least six pairs of readings.
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 1000 N? Pressure in Slave = 1000/10=100 Pa Pressure in Master = Force/1 = 100 Pa Force in the master only 100 N 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 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 × 103 kg m– 3)
H/W • LC H • 2006 Q 12(a)
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 = 10 N x 5 m = 50 Nm Perpendicular distance=5 m FORCE =10 N
More than two forces ? N 15 N 10 15 N 50 5 N 60 70 10 N ? 90 5 N • 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 15 N 10 15 N 50 5 N 60 70 ? 10 N • First law coplanar forces • Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N 90 5 N
20 N 15 N A 10 15 N 32. 5 50 5 N 60 70 10 N ? 90 5 N • Second law coplanar forces • Take moments about A Clockwise Moments = Anticlockwise Moments 10 x 15 + 50 x 5 + 70 x 10 + 90 x 5 = 60 x 15 + dx 20 150 + 250 + 700 + 450 = 900 + dx 20 1550 -900 = dx 20 so d=32. 5 cm
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES (2) Support Newton balance Paperclips w 1 w 2 w 3 w 4
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 H • 2007 Q 2
Couples of Forces • Two equal forces that cause a solid to rotate around an axis • Moment = Force x Distance • Moment = 5 Nx 0. 06 m • Moment = 0. 3 Nm
Motion in a circle Velocity always at 90 o 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 4 s what is it’s angular velocity?
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 6 m
Centripetal Acceleration a = r 2 Always towards the centre So the acceleration in the previous example a= 6 ( /2)2 =14. 8 m/s 2
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
Satellites balance forces • Balance of Gravity and Centripetal • ((GMm)/d 2)=mv 2/d Gravity F=-Gm. M/r 2
Eq uat Th e Fo e rce s l ance C ss of Ma llite sate d rio =Pe for T e Tim bit) ( Or 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 nce ista V=D ime t
Geostationary or Clarke Orbit Same period (For earth 24 hrs) and angular velocity as the planet surface so stays above same spot. What is it’s height above the earth?
In a test we do it like this
Example of Orbits What is the parking orbit height above Saturn if it is 200000 km in radius. It rotates every 4 days and has mass 2 x 1031 Kg. The Universal gravitational Constant is 6. 7 x 10 -11 Using T 2=4 2 d 3 /GM (4 x 24 x 60)2=4 2 d 3 /(2 x 1031)(6. 7 x 10 -11) d 3 = 1 x 1030 d = 1 x 1010 m Height =h= d - r =1 x 1010 m - 2 x 108 m = 9. 8 x 109 m h d r
Classwork 2015 H q 6
Simple Harmonic Motion • Is a vibration where the acceleration is proportional to the displacement a -s in book a = ω2 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
Calculations in ω •
Classwork • 2007 H Q 6
Pendulum Split cork l Bob Timer 20: 30 • If we displace the bob by a small angle it vibrates with SHM
T 2 l
H/W • LC H • 2008 Q 1