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Forces

Forces. By Neil Bronks. The unit is called the Newton (N). Force causes a body to change velocity…….. accelerate. Speed = distance (in metres) time (in seconds). D. S. T. Distance, Speed and Time. Dave walks 200 metres in 40 seconds. What is his speed?

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Forces

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  1. Forces By Neil Bronks

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

  3. Speed = distance (in metres) time (in seconds) D S T Distance, Speed and Time • 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?

  4. Scalars • Examples of Scalar Quantities: • Length • Area • Volume • Time • Mass A scalar quantity is a quantity that has magnitude only and has no direction in space

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

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

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

  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 opposite directions: 6 m s-1 10 m s-1 = 4 m s-1 6 N 9 N = 3 N Vectors in the same direction: 6 N 4 N = 10 N 6 m = 10 m 4 m

  11. 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

  12. Vector Addition Speed in still air 120m/s  Wind 50m/s Resultant R2 = 1202 + 502 = 14400 + 2500 = 16900 R = 130m/s Tan  = 50/120  = 22.60

  13. Problem: Resultant of 2 Vectors 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 13 N 5 N 5 b c 12 • Resultant displacement is 13 N 67ºwith the 5 N 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?

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

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

  16. Calculating the Magnitude of the Perpendicular Components v y=v Sinθ y θ • Proof: x=v Cos θ x If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are: x = v Cosθ y = v Sinθ

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

  18. 2003 HL Section B Q6 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: 156.28 N 10º 80º 10º Component of weight perpendicular to ramp: 886.33 N 900 N 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).

  19. Summary v y=v Sinθ y θ x=v Cosθ 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

  20. V-U Acceleration = change in velocity (in m/s) (in m/s2) time taken (in s) A T Acceleration • 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. V V t t Velocity-Time Graphs V t 1/.Constant Acceleration 2/.Constant Velocity 3/.Deceleration

  22. 4) Downward line = Deceleration • Upwards line = • Constant Acceleration 2) Horizontal line = Constant Velocity 3) Shallow line = Less Acceleration Velocity-time graphs 80 60 40 20 0 Velocity m/s T/s 10 20 30 40 50

  23. 80 60 40 20 0 Velocity m/s T/s 10 20 30 40 50 • 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 0 Velocity m/s T/s 10 20 30 40 50 The area under the graph is the distance travelled by the object

  25. 80 60 40 20 0 0.5x10x20=100 Velocity m/s 0.5x20x60=600 40x20=800 0.5x10x40=200 T/s 10 20 30 40 50 Total Distance Traveled =200+100+800+600=1700m

  26. 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 = 900 + 2a (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 = 0x12 +0.5x10x144 =720m

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

  29. H/W • LC Ord 2008 • Q 1

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

  31. 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.

  32. 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?

  33. m=2kg ub=400m/s Momentum of the Shoot Momentum of Recoil = 2 x 400 150 x Uc = Recoil Mass of canon=150kg Mass Canon x Velocity Canon = Mass of Ball x Velocity of Ball 5.3m/s V= 800/150=

  34. 10m/s V=? m/s 2 m/s Momentum 6kg 3kg 3kg 2kg 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

  35. Internet Calculations

  36. VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM l Card Vehicle 2 Dual timer t1 t2 Photogate Light beam Air track Vehicle 1 Velcro pad

  37. 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.

  38. 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.

  39. H/W • LC Ord • 2007 Q1

  40. 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

  41. 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 andForce=m.a

  42. Dual timer Photogate l Pulley Light beam Card Slotted weights s Air track TO SHOW THAT aµ F t2 t1

  43. Dual timer Photogate Light beam TO SHOW THAT aµ F t1 t1 time for card to pass first photo-gate

  44. Dual timer Photogate Light beam TO SHOW THAT aµ F t1 t2 t2 time for card to pass second photo-gate

  45. 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.

  46. 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

  47. 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

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

  49. 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

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