NEWTON S LAWS OF MOTION, EQUATIONS OF MOTION, & EQUATIONS OF MOTION FOR A SYSTEM OF PARTICLES APPLICATIONS

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1 EWTO S LAWS OF MOTIO, EQUATIOS OF MOTIO, & EQUATIOS OF MOTIO FOR A SYSTEM OF PARTICLES Today s Objectives: Students will be able to: 1. Write the equation of motion for an accelerating body. 2. Draw the free-body and kinetic diagrams for an accelerating body. In-Class Activities: Check Homework Applications ewton s Laws of Motion ewton s Law of Gravitational Attraction Equation of Motion For A Particle or System of Particles Group Problem Solving APPLICATIOS The motion of an object depends on the forces acting on it. A parachutist relies on the atmospheric drag resistance force to limit his velocity. Knowing the drag force, how can we determine the acceleration or velocity of the parachutist at any point in time? 1

2 APPLICATIOS A freight elevator is lifted using a motor attached to a cable and pulley system as shown. How can we determine the tension force in the cable required to lift the elevator at a given acceleration? Is the tension force in the cable greater than the weight of the elevator and its load? EWTO S LAWS OF MOTIO (Section 13.1) The motion of a particle is governed by ewton s three laws of motion. First Law: A particle originally at rest, or moving in a straight line at constant velocity, will remain in this state if the resultant force acting on the particle is zero. Second Law: If the resultant force on the particle is not zero, the particle experiences an acceleration in the same direction as the resultant force. This acceleration has a magnitude proportional to the resultant force. Third Law: Mutual forces of action and reaction between two particles are equal, opposite, and collinear. 2

3 EWTO S LAWS OF MOTIO The first and third laws were used in developing the concepts of statics. ewton s second law forms the basis of the study of dynamics. Mathematically, ewton s second law of motion can be written F = ma where F is the resultant unbalanced force acting on the particle, and a is the acceleration of the particle. The positive scalar m is called the mass of the particle. ewton s second law cannot be used when the particle s speed approaches the speed of light, or if the size of the particle is extremely small (~ size of an atom). EWTO S LAW OF GRAVITATIOAL ATTRACTIO Any two particles or bodies have a mutually attractive gravitational force acting between them. ewton postulated the law governing this gravitational force as F = G(m 1 m 2 /r 2 ) where F = force of attraction between the two bodies, G = universal constant of gravitation, m 1, m 2 = mass of each body, and r = distance between centers of the two bodies. When near the surface of the earth, the only gravitational force having any sizable magnitude is that between the earth and the body. This force is called the weight of the body. 3

4 MASS AD WEIGHT It is important to understand the difference between the mass and weight of a body! Mass is an absolute property of a body. It is independent of the gravitational field in which it is measured. The mass provides a measure of the resistance of a body to a change in velocity, as defined by ewton s second law of motion (m = F/a). The weight of a body is not absolute, since it depends on the gravitational field in which it is measured. Weight is defined as W = mg where g is the acceleration due to gravity. UITS: SI SYSTEM VS. FPS SYSTEM SI system: In the SI system of units, mass is a base unit and weight is a derived unit. Typically, mass is specified in kilograms (kg), and weight is calculated from W = mg. If the gravitational acceleration (g) is specified in units of m/s 2, then the weight is expressed in newtons (). On the earth s surface, g can be taken as g = 9.81 m/s 2. W () = m (kg) g (m/s 2 ) => = kg m/s 2 FPS System: In the FPS system of units, weight is a base unit and mass is a derived unit. Weight is typically specified in pounds (lb), and mass is calculated from m = W/g. If g is specified in units of ft/s 2, then the mass is expressed in slugs. On the earth s surface, g is approximately 32.2 ft/s 2. m (slugs) = W (lb)/g (ft/s 2 ) => slug = lb s 2 /ft 4

5 EQUATIO OF MOTIO (Section 13.2) The motion of a particle is governed by ewton s second law, relating the unbalanced forces on a particle to its acceleration. If more than one force acts on the particle, the equation of motion can be written F = F R = ma where F R is the resultant force, which is a vector summation of all the forces. To illustrate the equation, consider a particle acted on by two forces. First, draw the particle s freebody diagram, showing all forces acting on the particle. ext, draw the kinetic diagram, showing the inertial force ma acting in the same direction as the resultant force F R. IERTIAL FRAME OF REFERECE This equation of motion is only valid if the acceleration is measured in a ewtonian or inertial frame of reference. What does this mean? For problems concerned with motions at or near the earth s surface, we typically assume our inertial frame to be fixed to the earth. We neglect any acceleration effects from the earth s rotation. For problems involving satellites or rockets, the inertial frame of reference is often fixed to the stars. 5

6 EQUATIO OF MOTIO FOR A SYSTEM OF PARTICLES (Section 13.3) The equation of motion can be extended to include systems of particles. This includes the motion of solids, liquids, or gas systems. As in statics, there are internal forces and external forces acting on the system. What is the difference between them? Using the definitions of m = m i as the total mass of all particles and a G as the acceleration of the center of mass G of the particles, then ma G = m i a i. The text shows the details, but for a system of particles: F = ma G where F is the sum of the external forces acting on the entire system. KEY POITS 1) ewton s second law is a Law of ature --experimentally proven and not the result of an analytical proof. 2) Mass (property of an object) is a measure of the resistance to a change in velocity of the object. 3) Weight (a force) depends on the local gravitational field. Calculating the weight of an object is an application of F = ma, i.e., W = m g. 4) Unbalanced forces cause the acceleration of objects. This condition is fundamental to all dynamics problems! 6

7 PROCEDURE FOR THE APPLICATIO OF THE EQUATIO OF MOTIO 1) Select a convenient inertial coordinate system. Rectangular, normal/tangential, or cylindrical coordinates may be used. 2) Draw a free-body diagram showing all external forces applied to the particle. Resolve forces into their appropriate components. 3) Draw the kinetic diagram, showing the particle s inertial force, ma. Resolve this vector into its appropriate components. 4) Apply the equations of motion in their scalar component form and solve these equations for the unknowns. 5) It may be necessary to apply the proper kinematic relations to generate additional equations. EXAMPLE Given: A crate of mass m is pulled by a cable attached to a truck. The coefficient of kinetic friction between the crate and road is μ k. Find: Plan: Draw the free-body and kinetic diagrams of the crate. 1) Define an inertial coordinate system. 2) Draw the crate s free-body diagram, showing all external forces applied to the crate in the proper directions. 3) Draw the crate s kinetic diagram, showing the inertial force vector ma in the proper direction. 7

8 EXAMPLE Solution: 1) An inertial x-y frame can be defined as fixed to the ground. y 2) Draw the free-body diagram of the crate: W = mg T The weight force (W) acts through the crate s center of mass. T is the tension x 30 force in the cable. The normal force () is perpendicular to the surface. The F = u friction force (F = u K ) acts in a direction K opposite to the motion of the crate. 3) Draw the kinetic diagram of the crate: ma The crate will be pulled to the right. The acceleration vector can be directed to the right if the truck is speeding up or to the left if it is slowing down. GROUP PROBLEM SOLVIG Given: Each block has a mass m. The coefficient of kinetic friction at all surfaces of contact is μ. A horizontal force P is applied to the bottom block. Find: Plan: Draw the free-body and kinetic diagrams of each block. 1) Define an inertial coordinate system. 2) Draw the free-body diagrams for each block, showing all external forces. 3) Draw the kinetic diagrams for each block, showing the inertial forces. 8

9 GROUP PROBLEM SOLVIG Solution: 1) An inertial x-y frame can be defined as fixed to the ground. 2) Draw the free-body diagram of each block: Block B: Block A: y W B = mg W A = mg B x x F fb = μ B T P y F fb = μ B B F fa = μ A The friction forces oppose the motion of each block relative to the surfaces on which they slide. 3) Draw the kinetic diagram of each block: Block B: Block A: ma B = 0 ma A A EQUATIOS OF MOTIO: RECTAGULAR COORDIATES Today s Objectives: Students will be able to: 1. Apply ewton s second law to determine forces and accelerations for particles in rectilinear motion. In-Class Activities: Applications Equations of Motion Using Rectangular (Cartesian) Coordinates Group Problem Solving 9

10 APPLICATIOS If a man is pushing a 100 lb crate, how large a force F must he exert to start moving the crate? What would you have to know before you could calculate the answer? APPLICATIOS Objects that move in any fluid have a drag force acting on them. This drag force is a function of velocity. If the ship has an initial velocity v o and the magnitude of the opposing drag force at any instant is half the velocity, how long it would take for the ship to come to a stop if its engines stop? 10

11 RECTAGULAR COORDIATES (Section 13.4) The equation of motion, F = m a, is best used when the problem requires finding forces (especially forces perpendicular to the path), accelerations, velocities or mass. Remember, unbalanced forces cause acceleration! Three scalar equations can be written from this vector equation. The equation of motion, being a vector equation, may be expressed in terms of its three components in the Cartesian (rectangular) coordinate system as F = ma or F x i + F y j + F z k = m(a x i + a y j + a z k) or, as scalar equations, F x = ma x, F y = ma y, and F z = ma z. Free Body Diagram PROCEDURE FOR AALYSIS Establish your coordinate system and draw the particle s free body diagram showing only external forces. These external forces usually include the weight, normal forces, friction forces, and applied forces. Show the ma vector (sometimes called the inertial force) on a separate diagram. Make sure any friction forces act opposite to the direction of motion! If the particle is connected to an elastic spring, a spring force equal to ks should be included on the FBD. 11

12 PROCEDURE FOR AALYSIS Equations of Motion If the forces can be resolved directly from the free-body diagram (often the case in 2-D problems), use the scalar form of the equation of motion. In more complex cases (usually 3-D), a Cartesian vector is written for every force and a vector analysis is often best. A Cartesian vector formulation of the second law is F = ma or F x i + F y j + F z k = m(a x i + a y j + a z k) Three scalar equations can be written from this vector equation. You may only need two equations if the motion is in 2-D. Kinematics PROCEDURE FOR AALYSIS The second law only provides solutions for forces and accelerations. If velocity or position have to be found, kinematics equations are used once the acceleration is found from the equation of motion. Any of the tools learned in Chapter 12 may be needed to solve a problem. Make sure you use consistent positive coordinate directions as used in the equation of motion part of the problem! 12

13 EXAMPLE Given: W A = 10 lb W B = 20 lb v oa = 2 ft/s μ k = 0.2 Find: v A when A has moved 4 feet. Plan: Since both forces and velocity are involved, this problem requires both the equation of motion and kinematics. First, draw free body diagrams of A and B. Apply the equation of motion. Using dependent motion equations, derive a relationship between a A and a B and use with the equation of motion formulas. EXAMPLE Solution: 2T Free-body and kinetic diagrams of B: = W B m B a B Apply the equation of motion to B: + B F = y ma W 2 T = y m 20 2T = a B (1) B a B 13

14 y EXAMPLE Free-body and kinetic diagrams of A: x W A T = m A a A + F y = may = 0 = W = 10 lb F Apply the equations of motion to A: = μ A = k 2 F = m k lb + F x = ma x F T = m A a A 10 2 T = Aa (2) EXAMPLE ow consider the kinematics. s A Datums A B s B Constraint equation: s A + 2 s B = constant or v A + 2 v B = 0 Therefore a A + 2 a B = 0 a A = -2 a B (3) (otice a A is considered positive to the left and a B is positive downward.) 14

15 EXAMPLE ow combine equations (1), (2), and (3). a A T = 22 = lb 3 ft ft s = s = 2 ow use the kinematic equation: 2 A v = v 2 oa Av 2 = a ( s A A s oa 2(17.16)(4) ft v A = s ) GROUP PROBLEM SOLVIG Given: The 400 kg mine car is hoisted up the incline. The force in the cable is F = (3200t 2 ). The car has an initial velocity of v i = 2 m/s at t = 0. Find: The velocity when t = 2 s. Plan: Draw the free-body diagram of the car and apply the equation of motion to determine the acceleration. Apply kinematics relations to determine the velocity. 15

16 Solution: GROUP PROBLEM SOLVIG 1) Draw the free-body and kinetic diagrams of the mine car: y x θ W = mg F = ma Since the motion is up the incline, rotate the x-y axes. θ = tan -1 (815) = Motion occurs only in the x-direction. GROUP PROBLEM SOLVIG 2) Apply the equation of motion in the x-direction: + F x = ma x => F mg(sinθ) = ma x => 3200t 2 (400)(9.81)(sin ) = 400a => a = (8t ) m/s 2 3) Use kinematics to determine the velocity: a = dv/dt => dv = a dt v dv = (8t ) dt, v 1 = 2 m/s, t = 2 s v 1 t 0 2 v 2 = (8/3t t) = => v = 14.1 m/s 0 16

17 READIG QUIZ 1. In dynamics, the friction force acting on a moving object is always A) in the direction of its motion. B) a kinetic friction. C) a static friction. D) zero. 2. If a particle is connected to a spring, the elastic spring force is expressed by F = ks. The s in this equation is the A) spring constant. B) un-deformed length of the spring. C) difference between deformed length and un-deformed length. D) deformed length of the spring. 17

18 ATTETIO QUIZ 1. Determine the tension in the cable when the 400 kg box is moving upward with a 4 m/s 2 acceleration. A) 2265 B) 3365 C) 5524 D) T a = 4 m/s 2 2. A 10 lb particle has forces of F 1 = (3i + 5j) lb and F 2 = (-7i + 9j) lb acting on it. Determine the acceleration of the particle. A) (-0.4 i j) ft/s 2 B) (-4 i + 14 j) ft/s 2 C) (-12.9 i + 45 j) ft/s 2 D) (13 i + 4 j) ft/s 2 18

19 READIG QUIZ 1. ewton s second law can be written in mathematical form as ΣF = ma. Within the summation of forces ΣF, are(is) not included. A) external forces B) weight C) internal forces D) All of the above. 2. The equation of motion for a system of n-particles can be written as ΣF i = Σ m i a i = ma G, where a G indicates. A) summation of each particle s acceleration B) acceleration of the center of mass of the system C) acceleration of the largest particle D) one of the above. COCEPT QUIZ 1. The block (mass = m) is moving upward with a speed v. Draw the FBD if the kinetic friction coefficient is μ k. mg A) μ k B) mg C) μ k mg D) one of the above. mg μ k v 19

20 COCEPT QUIZ 2. Packaging for oranges is tested using a machine that exerts a y = 20 m/s 2 and a x = 3 m/s 2, simultaneously. Select the correct FBD and kinetic diagram for this condition. y W ma y A) B) = ma x R x = ma x R y R x W R y x C) D) ma y W ma y = = ma x R y R y ATTETIO QUIZ 1. Internal forces are not included in an equation of motion analysis because the internal forces are. A) equal to zero B) equal and opposite and do not affect the calculations C) negligibly small D) not important 2. A 10 lb block is initially moving down a ramp with a velocity of v. The force F is applied to bring the block to rest. Select the correct FBD. 10 F A) B) C) μ k F μ k F μ k v F 20

21 COCEPT QUIZ 1. If the cable has a tension of 3, determine the acceleration of block B. A) 4.26 m/s 2 B) 4.26 m/s 2 C) 8.31 m/s 2 D) 8.31 m/s 2 10 kg μ k =0.4 4 kg 2. Determine the acceleration of the block. A) 2.20 m/s 2 B) 3.17 m/s 2 30 C) 11.0 m/s 2 D) 4.26 m/s 2 5 kg 60 21