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DYNAMIC FORCES Equations of motion. 1, Acceleration (rate of change of Velocity) velocity is the vector version of speed, (speed in a straight line) v.

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Presentation on theme: "DYNAMIC FORCES Equations of motion. 1, Acceleration (rate of change of Velocity) velocity is the vector version of speed, (speed in a straight line) v."— Presentation transcript:

1 DYNAMIC FORCES Equations of motion

2 1, Acceleration (rate of change of Velocity) velocity is the vector version of speed, (speed in a straight line) v – u t = a u = initial velocity (in m/s) v= final velocity (in m/s) t = time (in seconds (s)) a = m/s 2 Can be rearranged to v = u + at

3 Sample question A car increases it’s velocity from 10 m/s to 20m/s in 5 seconds. What is it’s acceleration?

4 Sample question (answer) v – u t = a 20 – 10 5 = 2m/s 2 v = 20m/s u =10m/s t = 5 sec

5 Sample question A car moving at 10 m/s accelerates at 2m/s 2 for 4 seconds. What is it’s final velocity?

6 Sample question (answer) v = u + at v = x 4 = 18m/s Can be re-arranged to u = 10m/s a = 2m/s 2 t = 4 sec

7 Sample question A stone block with a mass of 800Kg is lifted from rest with a uniform acceleration by a crane such that it reached a velocity of 14m/s. after 10 seconds. Calculate the tension in the lifting cable

8 Sample question To do this calculation we need to use another equation Force = mass x acceleration F = ma Weight (force acting downwards) = mass x acceleration due to gravity Wt = mg Wt = m x 9.81 (acceleration due to gravity on earth)

9 Sample question (answer) v – u = a v = a 14 = 1.4m/s 2 t t 10 acceleration Force due to acceleration F = ma Tension Weight mg U = 0 from standing start First find the acceleration

10 Sample question (answer) F = ma F = 800 x 1.4 = 1120N Wt of box mg = 800 x 9.81 = 7848N (7.8kN) Tension in cable = = 8938N (8.9kN) acceleration Force due to acceleration F = ma Tension Weight mg

11 Sample question If the stone is lowered at the same rate of acceleration the tension is less than the weight of the box Tension in cable = = 6728N (6.7kN) Force due to acceleration F = ma Tension acceleration Weight mg

12 Lift You get the same effect when you travel in a lift. You feel heavier when the lift begins and accelerates upwards and lighter when the lift accelerates downwards

13 A man standing on weighing scales in a lift has a mass of 60 Kg. The lift accelerates uniformly at a rate of 2m/s 2 Calculate the reading on the scales during the period of acceleration.

14 acceleration Force due to acceleration F = ma Weight mg

15 F = ma F = 60 x 2 = 120N Wt of man mg = 60 x 9.81 = 588.6N Force acting on scale = = 708.6N The reading on the scale is ÷ 9.81 = 72.23Kg

16 2, Displacement (s) is the vector version of distance (distance in a straight line) Displacement = average velocity x time s = (v + u) x t 2

17 Sample question A car moving at 10 m/s increases it’s velocity to 20m/s in 4 seconds. How far will it have travelled during this time?

18 Sample question (answer) s = (v + u) x t 2 v =20m/s u = 10m/s t = 4 sec s = ( ) x 4 2 S = 60m

19 Sample question How long will it take for an athlete to accelerate from rest to 4 m/s over 8m ?

20 Sample question (answer) s = (v + u) x t 2 u =0 so s = v x t 2 t = 2 s v t = 16 = 4 seconds 4

21 3, Displacement related to acceleration s = ut + at 2 2 With zero acceleration (constant velocity) a = 0 s = ut From a standing start ut = 0 s = at 2 2

22 Example A car accelerates uniformly from rest and after 12 seconds has covered 40m. What are its acceleration and its final velocity ?

23 Example 2s = at 2 2s = a t 2 80 = a = a 144 a = 0.56m/s 2

24 Finding v a = v – u t From a standing start u = 0 a = v t v = a x t v = 0.56 x 12 = 6.7 m/s

25 Or alternatively U = 0 from standing start s = (v + u) x t 2 s = v x t 2 v = 2s t v = 80 = 6.7m/s 12

26 4, final velocity, initial velocity acceleration and displacement v 2 = u as

27 Example If a car travelling at 10m/s accelerates at a constant rate of 2m/s 2, what is it’s final velocity after it has travelled 10m?

28 Sample Question (answer) v 2 = u as V 2 = (2 x 2 x 10) = = 140m/s 2

29 D’Alembert’s Principle Acceleration (a) Applied Force (F) Inertia Force (Fi) F + Fi = 0 so F = -Fi F = ma so ma = - Fi or Fi = - ma When an object is accelerating the applied force making it accelerate has to overcome the inertia. This is the force which resists the acceleration (or deceleration) and is equal and opposite to the applied force. This means that the total force acting on the body is zero mass

30 D’Alembert’s Principle Acceleration Friction Force 20N Applied Force 200N Spring Force 150N Mass 20Kg

31 Fr (Friction) 20N W (weight) Mg = 20 x 9.81 =196.2N F (applied) 200N Fs (spring) 150N Free body diagram

32 D’Alembert’s Principle Downward forces are minus and upward forces are positive. From the diagram -200N – 196.2N +20N +150N = N (which is downward as to be expected)

33 Conservation of energy (Gravitational potential energy) 10m 5Kg GPE = mass x gravity x height (mgh) = 5 x 9.81x joules Work done (in lifting the mass) = force x distance x 10 = 490.5joules (F = mg) Work done = energy gained

34 Kinetic energy 10m 5Kg GPE = mgh = joules KE (at the bottom) = 490.5J = mv 2 2 (v = velocity) Neglecting friction, all the GPE at the top of the slope converts to kinetic energy at the bottom

35 Kinetic energy 10m 5Kg GPE = mgh = joules The velocity at the bottom v =√(2Ke÷m) = √ (2 x ÷ 5) √196.2 = 14m/s Neglecting friction, all the GPE at the top of the slope converts to kinetic energy at the bottom

36 Work and energy Force distance Work done (in joules) = Force x distance moved (in direction of force) Work done = energy used

37 Example A car of mass 800Kg is stood on a uniform 1 in 10 slope when it’s handbrake is suddenly released and it runs 30 metres to the bottom of the slope against a uniform frictional force of 50N. What is the car’s velocity at the bottom of the slope?

38 Example First find the angle of the slope 1 10 Tan of the angle = opposite/adjacent = 1/10= 0.1 Tan = 5.7 o

39 Example Then find the height of the slope sin 5.7 o = opposite/hypotenuse Opposite = hypotenuse x sine5.7 o 30 x sine 5.7 o =2.98m ht 30m 5.7 o

40 Example 1 (conservation of energy method) Find the gravitational potential energy of the car at the top of the slope 800 x 9.81 x 2.98 = 23.4kN ht 30m 5.7 o

41 Example 1 (conservation of energy method) Work done against friction = force x distance = 50N x 30m = 1.5kj ht 30m 5.7 o

42 Example 1 (conservation of energy method) Kinetic energy of the car at the bottom of the slope mv 2 /2 = = 21.9kj v= √(2 x Ke/m) v= √(2 x 21900/800) v= √54.75 v = 7.4 m/s ht 30m 5.7 o

43 Example 2 (Resolving forces method) Work out the forces involved Weight of car (force acting vertically downward) = mass x gravity = 800x N ht 30m 5.7 o

44 Example 2 (Resolving forces method) Work out the forces involved 5.7 o Fn Fs wt Wt = weight of car (7848N) Fn is the normal reaction force of the slope on the car Fs is the force on the car down the slope

45 Example 2 (Resolving forces method) Work out the forces involved Sine 5.7 o = Fs ÷ wt Fs = Wt x sin 5.7o Fs = 7848 sin 5.7o Fs = 779.5N (Force down the slope) 5.7 o Fn Fs wt

46 Example 2 (Resolving forces method) Work out the forces involved Resultant force down the slope = Fs – friction force – 50 =749.5N 5.7 o Fn Fs wt

47 Example 2 (Resolving forces method) Acceleration down the slope a= F/m 749.5/800 = 0.94m/s o Fn Fs wt

48 Example 2 (Resolving forces method) Velocity at the bottom of the slope v 2 = u 2 + 2as (u 2 = 0) so v 2 = 2as v 2 = 2 x 0.94 x 30 v =√56.4 v = 7.5m/s 5.7 o Fn Fs wt

49 Linear Momentum and Collisions Conservation of Energy

50 Momentum = mass x velocity The total momentum remains the same before and after a collision Momentum is a vector quantity

51 Linear Momentum and Collisions A railway coach of mass 25t is moving along a level track 36km/hr when it collides with and couples up to another coach of mass 20t moving in the same direction at 6km/hr. Both of the coaches continue in the same direction after coupling. What is the combined velocity of the two coaches?

52 Linear Momentum and Collisions Let the mass of the first coach be M1 and the mass of the second coach be M2 and the velocity of the first coach be V1 and the velocity of the second be V2

53 Linear Momentum and Collisions Before coupling the momentum of the first coach is 25 x 36 = 900tkm/hr and the momentum of the second is 20 x 6 =120 tkm/hr Which is a total of 1020tkm/hr

54 Linear Momentum and Collisions After the coupling the momentum of both is the same as before the coupling which is 1020 tkm/hr And the combined mass is 45t

55 Linear Momentum and Collisions Velocity after coupling is momentum divided by mass 1020÷ km/hr

56 Example A hammer of mass 200Kg falls 5m on to a pile of mass 300Kg and drives it 100mm into the ground a) Calculate the loss of energy on the impact. b) Calculate the work done by the resistance of the ground. c) calculate the average resistance to penetration.

57 Example Before falling the GPE of the hammer is 200kg x 9.81 x 5m = 9810j Kinetic energy of the hammer just before impact = 9810j (0.5 x m x v 2 ) Velocity of the hammer just before impact v 2 = (9810)/0.5 x m (9810)/0.5 X 200 =98.1 v = √98.1 = 9.9m/s

58 Example Momentum of hammer just before impact = mass x velocity = 200kg x 9.9m/s = 1980 kgm/s Momentum before collision = momentum after collision mass of hammer plus pile after collision = 200kg + 300kg = 500kg thus 500kg x velocity after collision = 1980 kgm/s Velocity after collision = 1980 ÷ 500 = 3.96m/s

59 Example Kinetic energy after collision (0.5 x mass x v 2 ) = 0.5 x 500 x = 3920j Loss of energy = 9810 – 3920 = 5890j To find deceleration of pile hammer v 2 = u 2 +2as (v 2 =0) u 2 = - 2as

60 Example a = - u 2 /2s / m/s 2 The minus sign means deceleration, Resistive force of the ground = mass x deceleration 500 x kN

61 Example Work done by ground = Force x distance 39.2kN x 0.1 m 3.92kj (this agrees with the fact that fact that the kinetic energy of the hammer and pile after impact was 3.92kj and zero when the pile stopped moving in the ground


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