Unformatted text preview: Lecture 22
• THE WORK OF A FORCE • PRINCIPLE OF WORK AND ENERGY • PRINCIPLE OF WORK AND ENERGY FOR A SYSTEM OF PARTICLES • POWER AND EFFICIENCY
Ehab Zalok THE WORK OF A FORCE, PRINCIPLE OF WORK AND ENERGY, & PRINCIPLE OF WORK AND ENERGY FOR A SYSTEM OF PARTICLES
Objectives: Students will be able to: 1. Calculate the work of a force. 2. Apply the principle of work and energy to a particle or system of particles. 2 Dynamics: Lecture Notes for Sections 14.1-14.4 1 APPLICATIONS A roller coaster makes use of gravitational forces to assist the cars in reaching high speeds in the “valleys” of the track. How can we design the track to control the forces experienced by the passengers? e.g.: • The height, h, • The radius of curvature, ρ
3 APPLICATIONS (continued) Crash barrels are often used along roadways for crash protection. The barrels absorb the car’s kinetic energy by deforming. If we know the typical velocity of an oncoming car and the amount of energy that can be absorbed by each barrel, How can we design a crash cushion?
4 Dynamics: Lecture Notes for Sections 14.1-14.4 2 WORK AND ENERGY Another equation for working kinetics problems involving particles can be derived by integrating the equation of motion (F = ma) with respect to displacement. By substituting at = v (dv/ds) into Ft = mat, the result is integrated to yield an equation known as the principle of work and energy. This principle is useful for solving problems that involve force, velocity, and displacement. It can also be used to explore the concept of power. To use this principle, we must first understand how to calculate the work of a force.
5 WORK OF A FORCE (Section 14.1) A force does work on a particle when the particle undergoes a displacement along the line of action of the force. Work is defined as the product of force and displacement components acting in the same direction. So, if the angle between the force and displacement vector is θ, the increment of work dU done by the force is dU = F cos θ ds By using the definition of the dot product and integrating, the total work can be U = 1-2 written as: r2 r1 ∫ F • dr
6 Dynamics: Lecture Notes for Sections 14.1-14.4 3 WORK OF A FORCE (continued) If F is a function of position (a common case) this becomes
s2 U1-2 = ∫ F cos θ ds
s1 If both F and θ are constant (F = Fc), this equation further simplifies to U1-2 = Fc cos θ (s2 - s1) • • • Work is positive if the force and the movement are in the same direction. If they are opposing, then the work is negative. If the force and the displacement directions are perpendicular, the work is zero. Recall cos 90 = 0
7 WORK OF A WEIGHT The work done by the gravitational force acting on a particle (or weight of an object) can be calculated by using
y2 U1-2 = y1 ∫ - W dy = - W (y2 - y1) = - W Δy The work of a weight is the product of the magnitude of the particle’s weight and its vertical displacement. If Δy is upward, the work is negative since the weight force always acts downward (Recall cos 180 = -1). 8 Dynamics: Lecture Notes for Sections 14.1-14.4 4 PRINCIPLE OF WORK AND ENERGY (Section 14.2 & Section 14.3)
By integrating the equation of motion, ∑ Ft = mat = mv(dv/ds), the principle of work and energy can be written as ∑ U1-2 = 0.5m(v2)2 – 0.5m(v1)2 or T1 + ∑ U1-2 = T2 ∑U1-2 is the work done by all the forces acting on the particle as it moves from point 1 to point 2. Work can be either a positive or negative scalar. T1 and T2 are the kinetic energies of the particle at the initial and final position, respectively. Thus, T1 = 0.5 m (v1)2 and T2 = 0.5 m (v2)2. The kinetic energy is always a positive scalar (velocity is squared!). So, the particle’s initial kinetic energy plus the work done by all the forces acting on the particle as it moves from its initial to final position is equal to the particle’s final kinetic energy. 9 PRINCIPLE OF WORK AND ENERGY (continued) Note that the principle of work and energy (T1 + ∑ U1-2 = T2) is not a vector equation! Each term results in a scalar value. Both kinetic energy and work have the same units, that of energy! In the SI system, the unit for energy is called a joule (J), where 1 J = 1 N·m. In the FPS system, units are ft·lb. The principle of work and energy cannot be used, in general, to determine forces directed normal to the path, since these forces do no work (Recall cos 90 = 0). The principle of work and energy can also be applied to a system of particles by summing the kinetic energies of all particles in the system and the work due to all forces acting on the system.
10 Dynamics: Lecture Notes for Sections 14.1-14.4 5 PROBLEM SOLVING Given: Block A has a weight of 60 lb and block B has a weight of 10 lb. The coefficient of kinetic friction between block A and the incline is µk = 0.2. Neglect the mass of the cord and pulleys. Find: The speed of block A after it moves 3 ft down the plane, starting from rest. Plan: 1) Define the kinematic relationships between the blocks. 2) Draw the FBD of each block. 3) Apply the principle of work and energy to the system of blocks. 11 Solution: PROBLEM SOLVING (continued) 1) The kinematic relationships can be determined by defining position coordinates sA and sB, and then differentiating. sA sB Since the cable length is constant: 2sA + sB = l 2ΔsA + ΔsB = 0 ΔsA = 3ft => ΔsB = -6 ft and 2vA + vB = 0 => vB = -2vA Note that, by this definition of sA and sB, positive motion for each block is defined as downwards. 12 Dynamics: Lecture Notes for Sections 14.1-14.4 6 PROBLEM SOLVING (continued) 2) Draw the FBD of each block. WA y 2T x A 5 4 µNA B T 3 NA WB Sum forces in the y-direction for block A (note that there is no motion in this direction): ∑Fy = 0: NA –(4/5)WA = 0 => NA = (4/5)WA
13 PROBLEM SOLVING (continued) 3) Apply the principle of work and energy to the system (the blocks start from rest).
∑T1 + ∑U1-2 = ∑T2 (0.5mA(vA1)2 +0.5mB(vB1)2) + ((3/5)WA – 2T – µNA)ΔsA + (WB – T)ΔsB = (0.5mA(vA2)2 + 0.5mB(vB2)2) vA1 = vB1 = 0, ΔsA = 3ft, ΔsB = -6 ft, vB = -2vA, NA = (4/5)WA => 0 + 0 + (3/5)(60)(3) – 2T(3) – (0.2)(0.8)(60)(3) + (10)(-6) – T(-6) = 0.5(60/32.2)(vA2)2 + 0.5(10/32.2)(-2vA2)2 => vA2 = 3.52 ft/s Note that the work due to the cable tension force on each block cancels out.
14 Dynamics: Lecture Notes for Sections 14.1-14.4 7 POWER AND EFFICIENCY
Objectives: Students will be able to: 1. Determine the power generated by a machine, engine, or motor. 2. Calculate the mechanical efficiency of a machine. 15 APPLICATIONS (continued) The speed at which a vehicle can climb a hill depends in part on the power output of the engine and the angle of inclination of the hill. For a given angle, how can we determine the speed of this jeep, knowing the power transmitted by the engine to the wheels? 16 Dynamics: Lecture Notes for Sections 14.1-14.4 8 POWER AND EFFICIENCY (Section 14.4) Power is defined as the amount of work performed per unit time If a machine or engine performs a certain amount of work, dU, within a given time interval, dt, the power generated can be calculated as P = dU/dt Since the work can be expressed as dU = F • dr, the power can be written P = dU/dt = (F • dr)/dt = F • (dr/dt) = F • v Thus, power is a scalar defined as the product of the force and velocity components acting in the same direction. 17 POWER Using scalar notation, power can be written P = F • v = F v cos θ where θ is the angle between the force and velocity vectors. So if the velocity of a body acted on by a force F is known, the power can be determined by calculating the dot product or by multiplying force and velocity components. The unit of power in the SI system is the watt (W) where 1 W = 1 J/s = 1 (N ·m)/s . In the “Foot-Pound-Second” FPS system, power is usually expressed in units of horsepower (hp) where 1 hp = 550 (ft · lb)/s = 746 W .
18 Dynamics: Lecture Notes for Sections 14.1-14.4 9 EFFICIENCY The mechanical efficiency of a machine is the ratio of the useful power produced (output power) to the power supplied to the machine (input power): ε = (power output)/(power input) If energy input and removal occur at the same time, efficiency may also be expressed in terms of the ratio of output energy to input energy or ε = (energy output)/(energy input) Machines will always have frictional forces. Since frictional forces dissipate energy, additional power will be required to overcome these forces. Consequently, the efficiency of a machine is always less than 1.
19 PROCEDURE FOR ANALYSIS • Find the resultant external force acting on the body causing its motion. It may be necessary to draw a free-body diagram. • Determine the velocity of the point on the body at which the force is applied. Energy methods or the equation of motion and appropriate kinematic relations, may be necessary. • Multiply the force magnitude by the component of velocity acting in the direction of F to determine the power supplied to the body (P = F v cos θ). • In some cases, power may be found by calculating the work done per unit of time (P = dU/dt). • If the mechanical efficiency of a machine is known, either the power input or output can be determined.
20 Dynamics: Lecture Notes for Sections 14.1-14.4 10 EXAMPLE Given: A sports car has a mass of 2 Mg and an engine efficiency of ε = 0.65. Moving forward, the wind creates a drag resistance on the car of FD = 1.2v2 N, where v is the velocity in m/s. The car accelerates at 5 m/s2, starting from rest. Find: The engine’s input power when t = 4 s. Plan: 1) Draw a free body diagram of the car. 2) Apply the equation of motion and kinematic equations to find the car’s velocity at t = 4 s. 3) Determine the power required for this motion. 4) Use the engine’s efficiency to determine input power. 21 Solution: 1) Draw the FBD of the car. EXAMPLE (continued) The drag force and weight are known forces. The normal force Nc and frictional force Fc represent the resultant forces of all four wheels. The frictional force between the wheels and road pushes the car forward. 2) The equation of motion can be applied in the x-direction, with ax = 5 m/s2: + ∑Fx = max => Fc – 1.2v2 = (2000)(5) => Fc = (10,000 + 1.2v2) N
22 Dynamics: Lecture Notes for Sections 14.1-14.4 11 EXAMPLE (continued) 3) The constant acceleration equations can be used to determine the car’s velocity. vx = vxo + axt = 0 + (5)(4) = 20 m/s 4) The power output of the car is calculated by multiplying the driving (frictional) force and the car’s velocity: Po = (Fc)(vx ) = [10,000 + (1.2)(20)2](20) = 209.6 kW 5) The power developed by the engine (prior to its frictional losses) is obtained using the efficiency equation. Pi = Po/ε = 209.6/0.65 = 322 kW
23 PROBLEM SOLVING
Given: A 50-lb load (B) is hoisted by the pulley system and motor M. The motor has an efficiency of 0.76 and exerts a constant force of 30 lb on the cable. Neglect the mass of the pulleys and cable. Find: The power supplied to the motor when the load has been hoisted 10 ft. The block started from rest. Plan: 1) Relate the cable and block velocities by defining position coordinates. Draw a FBD of the block. 2) Use the equation of motion or energy methods to determine the block’s velocity at 10 feet. 3) Calculate the power supplied by the motor and to the motor.
24 Dynamics: Lecture Notes for Sections 14.1-14.4 12 PROBLEM SOLVING (continued) Solution: 1) Define position coordinates to relate velocities. sm Here sm is defined to a point on the cable. Also sB is sB defined only to the lower pulley, since the block moves with the pulley. From kinematics, sm + 2sB = l => vm + 2vB = 0 => vm = -2vB Draw the FBD of the block: 2T Since the pulley has no mass, a force balance requires that the tension in the B lower cable is twice the tension in the WB = 50 lb upper cable.
25 PROBLEM SOLVING (continued) 2) The velocity of the block can be obtained by applying the principle of work and energy to the block (recall that the block starts from rest). + T1 + ∑U1-2 = T2 0.5m(v1)2 + [2T(s) – wB(s)] = 0.5m (v2)2 0 + [2(30)(10) - (50)(10)] = 0.5(50/32.2)(v2)2 => v2 = vB = 11.35 ft/s Since this velocity is upwards, it is a negative velocity in terms of the kinematic equation coordinates. The velocity of the cable coming into the motor (vm) is calculated from the kinematic equation. vm = - 2vB = - (2)(-11.35) = 22.70 ft/s
26 Dynamics: Lecture Notes for Sections 14.1-14.4 13 PROBLEM SOLVING (continued) 3) The power supplied by the motor is the product of the force applied to the cable and the velocity of the cable: Po = F • v = (30)(22.70) = 681 (ft ·lb)/s The power supplied to the motor is determined using the motor’s efficiency and the basic efficiency equation. Pi = Po/ε = 681/0.76 = 896 (ft ·lb)/s Converting to horsepower Pi = 896/550 = 1.63 hp 27 A guy walks into work, and both of his ears are all bandaged up. The boss says, “What happened to your ears?” He says, “Yesterday I was ironing a shirt when the phone rang and (hold iron to ear) shhh! I accidentally answered the iron.” The boss says, “Well, that explains one ear, but what happened to your other ear?” He says, “Well, jeez, I had to call the doctor!” 28 Dynamics: Lecture Notes for Sections 14.1-14.4 14 ...
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