Topics in textbook n Actual Work n Definition

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

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

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

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

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

B 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. A (spring stretched length is 0. 5 m) Pos B Horizontal Force P: constant Pos A Weight W: constant. “Active Force” Spring Force Fs. : varying Force Normal Force NB : constant “Inactive Force”

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

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

15

18

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

M 3/107) Calculate the work done on 10 -kg object with the constant Force ( F= 8 N ) during the curve path AB. x-y y x mg F=8 N (const) N Does not do the work If F is not constant, how to calculate it? Ans

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

M 3/107) Calculate the work done by F during the curve path AB. n-t Fn doest not effect works! s engine thrust a

M 3/107) Calculate the work done by F during the curve path AB. r-q reference point central force 23

M 3/121) 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. mg general position F Work-Energy Eq. 0 N Does not do the work r-q coordinate reference point 0 25

Work on frictionless connected particles 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. ) q internal force R and –R will have the same displacement. q So, the sum of these works are zero. 33

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) initial state Final state N causes no work! System selection is not so good (you have to calculate Tension T for its work) 34

M 3/131) The ball is released from position A with a velocity of 3 m/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. T does no work system mg 35

Power path q Power is defined as time rate of work A (scalar quantity) q 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) q Unit of power: Watt (W) = J/s = N-m/s 37

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

A car has a mass of 2 Mg and an engine efficiency of e = 0. 65. The car uniformly accelerates at 5 m/s 2, starting from rest. During that constant acceleration, the wind outside creates a drag resistance on the car of FD = 1. 2 v 2 N, where v is the velocity in m/s. Find the engine output input when t=4 s. a x Constant acceleration: 39

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. v=? F = 30 N (const) 50 N s = 10 m (start from rest) 2 F= 2(30) B 50 N Energy Approach

Work and Energy kinetic energy at B path B kinetic energy at A A summation of all forces We found that …. Work from all other forces (not spring & gravitation) Work from Gravity Force Gravitational Potential Energy Work from spring Elastic Potential Energy 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 42

Work of Gravity Force >0 Only depends on position at final state (2) Only depends on position at initial state (2) any path 1 energy level (higher) W=mg 2 energy level (lower) 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 h Think in Term of “Potential Energy” Fixed reference line Work done by Gravity Force: from position 1 to position 2 Work = “Energy in Transfer” (for convenience) point function Potential Energy - Energy from gravity field Energy “Emission”: from position 1 to position 2

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

Work of Spring Force 1 L Only depends on position at final state (2) Only depends on position at initial state (1) natural length (unstretched length) 2 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) Work done by Spring Force: from position 1 to position 2 x : distance , stretched or compressed from natural length point function Energy Emission: from position 1 to position 2

Work-Energy Equation FBD ** (Use Energy Concept) FBD Not Recommended Method in this course N (Conser vative F orce) Think of Energy N Work-Energy Equation (1 st Form) Virtual work by nonconservative forces. Energy Concept (2 nd Form)

Work-Energy Equation (1 st Form) FBD Work-Energy Equation (Conser vative F orce) Think of Energy N (2 nd Form)

M 3/173) 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 At position B 49

Advantage q Integral Equation (not instantaneous equation like 2 nd Law) q Scalar equation. (easy to handle with 1 unknown) q Get change in velocity directly. (No need to find acceleration first) q Handle with only active forces. q it can be applied to system of particles with frictionless and non-deformable links We will see this later, when applying at system of particles 57

Work on frictionless connected particles A B C O 58

initial state 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) Final state T T F F Object A sys tem Object B F T is internal force (excluding from Work Calculation) We have no interest in T, thus object separation (separating object A and B) is not good in this problem. 60

64

M 3/158) 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 (assume) up Position B: block B moves down as 1 meter datum unsolvable

M 3/158) 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 (assume) up Position B: block B moves down as 1 meter datum

H 14/16) 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. 20 N System: block A + block B + cord+ 2 Pulleys Position A: at rest Position B: block B moves down as d meter 50 N 67

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

Power path q Power is defined as time rate of work A (scalar quantity) q 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) q Unit of power: Watt (W) = J/s = N-m/s 70

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

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. v=? No! F = 30 N (const) 50 N 2 F= 2(30) Energy Approach s = 10 m (start from rest) F = 30 N (const) 50 N

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

Recommended Problem M 3/155 M 3/144 M 3/166 M 3/168 M 3/160 H 14/93 , H 14/96 75

77