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1 Topics in textbook Actual Work Definition of Work Work on Particle Potential Energy Definition of Energy Potential Energy Kinetics Energy Definition.

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Presentation on theme: "1 Topics in textbook Actual Work Definition of Work Work on Particle Potential Energy Definition of Energy Potential Energy Kinetics Energy Definition."— Presentation transcript:

1 1 Topics in textbook Actual Work Definition of Work Work on Particle Potential Energy Definition of Energy Potential Energy Kinetics Energy Definition of Energs Work-Energy Equation

2 2 Three approaches for solving dynamics 1) Direct Method 2) Work and Energy Kinematics Eq: 3) Impulse and Momentum linear impulse of Force i work (and potential energy) of Force i along the path Newton’s 2 nd Law From 2 nd Law (kinetics Eq) From 2 nd Law A path

3 3 Work and Energy Work of Force i along the path From 2 nd Law Newton’s 2 nd Law Principle of work and Energy A path Work of a force during small displacement kinetic energy Usually convenient when F = F(s), and you want to find velocity at final state (without finding acc. first). change of kinetic energy kinetic energy at A kinetic energy at B B

4 4 Work & Energy Work - “something” successful done by force. - it depends on - “Indicator” has sign  Energy - Capacity/Potential to do the work. ( the work is not yet done) The objective of force is to create the movement. ( “indicator” of successiveness of force ) [Unit: Newton- Meter (J: joule) ] and

5 5 Work done on Particle P Work by a force path Work done over particle A sum of works done by all forces over the particle A Since, the total work done on object is …… (inactive force)

6 6  is positive when Note on work path  Active force is the force that does the work has the same direction.  Reactive force = constrain force that does not do the work  Unit of work is N-m or Joule (J).

7 The 10-kg block rest on a smooth incline. If the spring is originally stretched 0.5 m, determine the total work done by all forces acting on the block when a horizontal force P = 400 N pushes the block up the plane s = 2 m. The block is not tipping. Horizontal Force P: constant Spring Force F s.: varying Force Weight W: constant. Normal Force N B : constant A (spring stretched length is 0.5m) B “Inactive Force” “Active Force” Pos A Pos B

8 8 The 10-kg block rest on a smooth incline. If the spring is originally stretched 0.5 m, determine the total work done by all forces acting on the block when a horizontal force P = 400 N pushes the block up the plane s = 2 m. scalar calculation (aware of +/- ) P = 400 N constant along the path Spring = k(0.5+x) x=0 to 0.5 P =400N Does not do the work N mg Spring mg = (10)(9.81) constant along the path Normal Force This force does no work since it is always perpendicular to the displacement.

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10 Work done on Particle P path Work done over particle A sum of works done by all forces over the particle A Pos A Pos B t Work done on particle P during path A->B, is to increase kinetic energy of particle

11 11 Work and Energy Work of Force i along the path From 2 nd Law Newton’s 2 nd Law Principle of work and Energy A path Work of a force during small displacement kinetic energy Usually convenient when F = F(s), and you want to find velocity at final state (without finding acc. first). change of kinetic energy kinetic energy at A kinetic energy at B B

12 12 Kinetics Energy Principle of work and Energy A path  T is the work done on a particle to accelerate it from rest to the velocity v  Unit of T is N-m or Joule (J)  No need to find acceleration first  it can be applied to system of particles with frictionless and non-deformable links Advantage  Get change in velocity directly from active forces.  Scalar equation. (1 unknown)  Integral Equation (not instantaneous eq like 2 nd Law)

13 Motivation : Work-Energy Equation Solve for every instant t (Newtonian Equation is instantaneous Equation) We have to write FBD of general position of time t New way: Use Work-Energy Equation initial final

14 14 SP3/3 The block A is released from rest, and pulls the 200- kg log up. Determine the velocity of the block A when it hits the ground constant at any time t 200(g) T N 2T T Static? The pulley C is moving T A 125(g )

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16 16 Three approaches for solving dynamics Direct Method Work and Energy Kinematics: Impulse and Momentum linear impulse of Force i work (and potential energy) of Force i along the path Newton’s 2 nd Law From 2 nd Law

17 17 Advantage of Work-Energy Eq. Principle of work and Energy A path Usually convenient when F = F(s) Work-Energy Eq.  No need to find acceleration first  Get change in velocity directly from active forces.  can be applied to system of particle using frictionless and non-deformable links

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19 19 How to calculate Work path displacement x force component in the direction of displacement In general In xyz-coord In nt-coord displacement in the direction of force x force component In r  -coord scalar (be careful Of +/-)

20 F=8N (const) Does not do the work M3/107) Calculate the work done on 10-kg object with the constant Force ( F= 8N ) during the curve path AB. N mg x y x-y Ans If F is not constant, how to calculate it?

21 If F is not constant 21 More general case or x y

22 M3/107) Calculate the work done by F during the curve path AB. s engine thrust n-t Fn doest not effect works! 

23 23 M3/107) Calculate the work done by F during the curve path AB. central force r-  reference point r- 

24 24 M3/107) Calculate the work done by F during the curve path AB. central force s r-  reference point engine thrust N n-t co-ord

25 25 M3/121) The 0.2-kg slider moves freely along the fixed curved rod from A to B in the vertical plane under the action of the constant 5-N tension in the cord. If the slider is released from rest at A, calculate its velocity v as it reaches B. Work-Energy Eq. r-  coordinate reference point Does not do the work N mg F 0 general position 0

26 26 Hibbeler Ex Given: x 1 = 0.5 m P = 400 N s = 2 m Find: Total work done by all forces Plan: 1) Draw a FBD 2) Calculate for work done by each forces (Horizontal P, Spring force F s, Weight W, and Normal force N). Then sum all works together.

27 27 Hibbeler Ex (cont.)

28 28 Hibbeler Ex (cont.)

29 29 Hibbeler Ex (cont.)

30 30 Work and Curvilinear Motion

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32 32  No need to find acceleration first  it can be applied to system of particles using frictionless and non-deformable links Advantage of Work-Energy Eq.  Get change in velocity directly from active forces.  Work-Energy Eq. is a scalar equation.

33 33 Work on frictionless connected particles  internal force R and –R will have the same displacement.  So, the sum of these works are zero. Only the external forces are needed to calculate the total work on a system of particles. (If frictions exist, the sum of action and reaction of the friction may not be zero.)

34 34 The system starts from rest at Configuration 1. Find the velocity of A at configuration where d = 0.5 m. F is 20 N (constant) System selection is not so good (you have to calculate Tension T for its work) initial state Final state N causes no work!

35 35 M3/131) The ball is released from position A with a velocity of 3m/s and swings in a vertical plane. At the bottom position, the cord strikes the fixed bar at B, and the ball continues to swing in the dashed arc. Calcuate the velocity v of the ball as it passes position C. Work-Energy Eq. system r-  coordinate reference point T mg system does no work

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37 37 Power  Power is defined as time rate of work  For a machine, power tells how much work it can do in a period of time. (small machine can deliver lots of energy given enough time) A path  Unit of power: Watt (W) = J/s = N-m/s (scalar quantity)

38 38 Mechanical Efficiency  Mechanical Efficiency  Since machine consists of moving parts which may have frictions, so extra energy or power is needed to overcome the frictions. If energy applied to the machine occurs during the same time interval at which it is removed.

39 39 A car has a mass of 2 Mg and an engine efficiency of  = The car uniformly accelerates at 5 m/s2, starting from rest. During that constant acceleration, the wind outside creates a drag resistance on the car of F D = 1.2v 2 N, where v is the velocity in m/s. Find the engine output input when t=4 s. x Constant acceleration: a

40 50N A 50-N load (B) is lifted up by the motor from rest until the distance is 10 m. The motor M has an efficiency of 0.76 and exerts a constant force of 30 N. Find the power supplied to the motor at that instant. Neglect the mass of the pulleys and cable. s = 10 m (start from rest) F = 30 N (const) v=? 2F= 2(30) 50 N B Energy Approach

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42 42 Work and Energy A path kinetic energy kinetic energy at A kinetic energy at B B summation of all forces Work from spring Work from Gravity Force Work from all other forces ( not spring & gravitation) We found that …. It is much easier to solve dynamic problem, if we think the work done by spring and gravity force in the form of Potential Energy Elastic Potential Energy Gravitational Potential Energy

43 Energy “Emission”: from position 1 to position 2 Work of Gravity Force 43 Work done over Object Work by couple is - positive if M has the same sense as d  - negative if M has the inverse sense as d  W=mg Fixed reference line h 2 1 Only depends on position at final state (2) Only depends on position at initial state (2) Work done by W, only depends on the initial state position and final state position only, i.e., it does not depends on actual path any path Think in Term of “Potential Energy ” (for convenience) Potential Energy - Energy from gravity field Work done by Gravity Force: from position 1 to position 2 > 0 energy level (lower) energy level (higher) Work = “Energy in Transfer” point function

44 44 when change in g is significant Define as negative of work done from the position to  the potential energy at r is  from

45 Work of Spring Force L 1 2 Only depends on position at final state (2) Only depends on position at initial state (1) Work done by Spring, depends only on the initial state and final state only, i.e., it does not depends on actual path any path Think in Term of “ Energy ” (for convenience) Energy Emission: from position 1 to position 2 Work done by Gravity Force: from position 1 to position 2 (direction may be opposite) x : distance, stretched or compressed from natural length Energy Emission: from position 1 to position 2 Work done by Spring Force: from position 1 to position 2 natural length (unstretched length) point function

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47 47 Work-Energy Equation initial location final location FBD Virtual work by non- conservative forces. Energy Concept (Conservative Force) Think of Energy Work-Energy Equation FBD ** (Use Energy Concept) N (1 st Form) (2 nd Form) Not Recommended Method in this course N

48 The virtual work done by external active forces (other than gravitational and spring forces accounted for in the potential energy terms) on an ideal mechanical system in equilibrium, equals the corresponding change in the total elastic and gravitational potential energy of the system for any and all virtual displacements consistent with the constraints. Work-Energy Equation (1 st Form) (2 nd Form) FBD (Conservative Force) Think of Energy N

49 49 M3/173) The 0.6-kg slider is released from rest at A and slides down under the influence of its own weight and of the spring of k = 120 N/m. Determine the speed of the slider and the normal force at point B. The unstrecthed length of the spring is 200 mm. gravitational Potential datum gravitational potential datum At position B

50 50

51 51 Hibbeler Ex (cont.)

52 52 H14/92) The 1-kg collar has a speed of 1.5 m/s at A. the attached spring has an unstretched length of 0.6 m and a stiffness of k = 150 N/m. When it reaches point B, determine its velocity, the rate of decrease in its speed, and the normal force of the collar to the rod. N mg F F=kx N

53 53 CFO system M3/172) The cars of an amusement-park ride have a speed v=90 km/h at the lowest part of the track. Determine their speed v at the highest part of the track. Neglect energy loss due to friction. gravitational potential datum

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55 55 Hibbeler Ex (cont.) Datum a n = v 2 / ,  = 20 m

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57 57  Get change in velocity directly. (No need to find acceleration first)  it can be applied to system of particles with frictionless and non-deformable links Advantage  Handle with only active forces.  Scalar equation. (easy to handle with1 unknown)  Integral Equation (not instantaneous equation like 2 nd Law) We will see this later, when applying at system of particles

58 58 Work on frictionless connected particles  internal force R and –R will have the same displacement.  So, the sum of these works are zero. Only the external forces are needed to calculate the total work on a system of particles. (If frictions exist, the sum of action and reaction of the friction may not be zero.) A B C O

59 59 Work on frictionless connected particles  internal force R and –R will have the same displacement.  So, the sum of these works are zero. Only the external forces are needed to calculate the total work on a system of particles. (If frictions exist, the sum of action and reaction of the friction may not be zero.)

60 60 The system starts from rest at Configuration 1. Find the velocity of A at configuration where d = 0.5 m. F is 20 N (constant) We have no interest in T, thus object separation (separating object A and B) is not good in this problem. initial stateFinal state F T T Object A F T T Object B system F T is internal force (excluding from Work Calculation)

61 61 Constrain Force (reactive force) causes no work because the movement is not possible. Work on Connected Rigid Bodies A B C O Internal forces (action/reaction pair) are cancelled with each other. friction-less system AFD (Active Force Diagram ) Active Force: Forces that do work. AFD

62 62 Constrain Force (reactive force) causes no work because the movement is not possible. Work on Connected Rigid Bodies A B C O Internal forces (action/reaction pair) are cancelled with each other. friction-less system AFD (Active Force Diagram ) Active Force: Forces that do work. AF D

63 63 Constrain Force (reactive force) causes no work because the movement is not possible. Work on Connected Rigid Bodies A B C O Internal forces (action/reaction pair) are cancelled with each other. friction-less system AFD (Active Force Diagram ) Active Force: Forces that do work. AFD

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65 datum 65 M3/158) If the system is released from rest, determine the speeds of both masses after B have move 1 m. Neglect friction and the masses of pulleys. System: block A + block B + cord+ 2 Pulleys Position A: at rest Position B: block B moves down as 1 meter (assume) unsolvable up

66 datum 66 M3/158) If the system is released from rest, determine the speeds of both masses after B have move 1 m. Neglect friction and the masses of pulleys. System: block A + block B + cord+ 2 Pulleys Position A: at rest Position B: block B moves down as 1 meter (assume) up

67 67 20 N 50 N H14/16) Block A rest on a surface which has friction. Determine the distance d cylinder B must move down so that A has a speed of starting from rest. System: block A + block B + cord+ 2 Pulleys Position A: at rest Position B: block B moves down as d meter

68 68 3/168) The system is released from rest with  =180, where the uncompressed spring of stiffness k= 900 N/m is just touch the underside of 4-kg collar. Determine the angle  corresponding to the maximum spring compression. O2-1 O2-2 O1 System: O1+O2+O3+4 rods Position A: at rest with  =180 Position B: maximum compression r L r datum

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70 70 Power  Power is defined as time rate of work  For a machine, power tells how much work it can do in a period of time. (small machine can deliver lots of energy given enough time) A path  Unit of power: Watt (W) = J/s = N-m/s (scalar quantity)

71 71 Mechanical Efficiency  Mechanical Efficiency  Since machine consists of moving parts which may have frictions, so extra energy or power is needed to overcome the frictions. If energy applied to the machine occurs during the same time interval at which it is removed.

72 50N A 50-N load (B) is lifted up by the motor from rest until the distance is 10 m. The motor M has an efficiency of 0.76 and exerts a constant force of 30 N. Find the power supplied to the motor at that instant. Neglect the mass of the pulleys and cable. 2F= 2(30) 50 N Energy Approach No! v=? F = 30 N (const) s = 10 m (start from rest) F = 30 N (const) 50 N

73 73 A car has a mass of 2 Mg and an engine efficiency of  = The car uniformly accelerates at 5 m/s2, starting from rest. During that constant acceleration, the wind outside creates a drag resistance on the car of F D = 1.2v 2 N, where v is the velocity in m/s. Find the engine output input when t=4 s. x Constant acceleration: a

74 74 Summary  Make sure you write FBD (no FBD, no score)  Equation itself is not hard to solve, but calculating work may be more difficult than you thought.   Scalar Equation (Only 1 unknown) or

75 75 Recommended Problem M3/144 M3/155 M3/160 M3/166 M3/168 H14/93, H 14/96

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