NEWTONS LAWS OF MOTION EQUATION OF MOTION Sections

NEWTON’S LAWS OF MOTION (EQUATION OF MOTION) (Sections 13. 1 -13. 3) Today’s Objectives: Students will be able to: a) Write the equation of motion In-Class Activities: for an accelerating body. • Check homework, if any b) Draw the free-body and • Reading quiz kinetic diagrams for an • Applications accelerating body. • Newton’s laws of motion • Newton’s law of gravitational attraction • Equation of motion for a particle or system of particles • Concept quiz • Group problem solving • Attention quiz

APPLICATIONS 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?

APPLICATIONS (continued) 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?

NEWTON’S LAWS OF MOTION The motion of a particle is governed by Newton’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.

NEWTON’S LAWS OF MOTION (continued) The first and third laws were used in developing the concepts of statics. Newton’s second law forms the basis of the study of dynamics. Mathematically, Newton’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. Newton’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).

NEWTON’S LAW OF GRAVITATIONAL ATTRACTION Any two particles or bodies have a mutually attractive gravitational force acting between them. Newton 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.

MASS AND 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 Newton’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.

UNITS: 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 (N). On the earth’s surface, g can be taken as g = 9. 81 m/s 2. W (N) = m (kg) g (m/s 2) => N = 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

EQUATION OF MOTION The motion of a particle is governed by Newton’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 = FR = ma where FR 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. Next, draw the kinetic diagram, showing the inertial force ma acting in the same direction as the resultant force FR.

INERTIAL FRAME OF REFERENCE This equation of motion is only valid if the acceleration is measured in a Newtonian 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.

SYSTEM OF PARTICLES 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 = mi 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 = miai. 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 POINTS 1) Newton’s second law is a “Law of Nature”--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!

PROCEDURE FOR THE APPLICATION OF THE EQUATION OF MOTION 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 mk. Find: Draw the free-body and kinetic diagrams of the crate. Plan: 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.

EXAMPLE (continued) Solution: 1) An inertial x-y frame can be defined as fixed to the ground. 2) Draw the free-body diagram of the crate: y W = mg The weight force (W) acts through the crate’s center of mass. T is the tension x force in the cable. The normal force (N) is perpendicular to the surface. The friction force (F = u. KN) acts in a direction F = u KN opposite to the motion of the crate. N 3) Draw the kinetic diagram of the crate: T 30° 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 SOLVING Given: Each block has a mass m. The coefficient of kinetic friction at all surfaces of contact is m. A horizontal force P is applied to the bottom block. Find: Draw the free-body and kinetic diagrams of each block. Plan: 1) 2) 3)

GROUP PROBLEM SOLVING (continued) Solution: 1) An inertial x-y frame can be defined as 2) Draw the ________ of each block: Block B: y x 3) Draw the kinetic diagram of each block: Block B: Block A:

EQUATIONS OF MOTION: RECTANGULAR COORDINATES (Section 13. 4) Today’s Objectives: Students will be able to apply Newton’s second law to determine forces and accelerations for particles in rectilinear motion. In-Class Activities: • Check homework, if any • Reading quiz • Applications • Equations of motion using rectangular (Cartesian) Coordinates • Concept quiz • Group problem solving • Attention Quiz

APPLICATIONS 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?

APPLICATIONS (continued) 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 vo 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?

EQUATION OF MOTION 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 Fx i + Fy j + Fz k = m(ax i + ay j + az k) or, as scalar equations, Fx = max , Fy = may , and Fz = maz.

PROCEDURE FOR ANALYSIS • Free Body Diagram 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.

PROCEDURE FOR ANALYSIS (continued) • 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 Fx i + Fy j + Fz k = m(ax i + ay j + az k) Three scalar equations can be written from this vector equation. You may only need two equations if the motion is in 2 -D.

PROCEDURE FOR ANALYSIS (continued) • Kinematics The second law only provides solutions 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!

EXAMPLE Given: WA = 10 lb WB = 20 lb vo. A = 2 ft/s mk = 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 (continued) Solution: 2 T Free-body and kinetic diagrams of B: = WB Apply the equation of motion to B: m. Ba. B + ¯ Fy = ma y WB - 2 T = m. B a. B 20 - 2 T = 20 a. B 32. 2 (1)

EXAMPLE (continued) Free-body and kinetic diagrams of A: y WA x T N = m. Aa. A F = m k. N Apply the equations of motion to A: + F = ma ¬ + Fy = ma y = 0 x x N = WA = 10 lb F = m N = 2 lb k F - T = m. A a. A 10 2 - T = 32 2 a. A. (2)

EXAMPLE (continued) Now consider the kinematics. Constraint equation: s. A Datums A s. B B 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) (Notice a. A is considered positive to the left and a. B is positive downward. )

EXAMPLE (continued) Now combine equations (1), (2), and (3). T = 22 = 7. 33 lb 3 ft 2 ® ft 2 a. A = -17. 16 s = 17. 16 s Now use the kinematic equation: v. A 2 2 v = o. A + 2 a. A ( s. A - so. A ) v. A 2 = 2 2 + 2 (17. 16 )(4 ) ft v. A = 11. 9 s ®

GROUP PROBLEM SOLVING Given: The 400 kg mine car is hoisted up the incline. The force in the cable is F = (3200 t 2) N. The car has an initial velocity of vi = 2 m/s at t = 0. Find: Plan: The velocity when t = 2 s.

GROUP PROBLEM SOLVING (continued) Solution: 1) Draw the free-body and kinetic diagrams of the mine car:

GROUP PROBLEM SOLVING (continued) 2) Apply the equation of motion in the x-direction: 3) Use kinematics to determine the velocity:

EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES (Section 13. 5) Today’s Objectives: Students will be able to apply the equation of motion using normal and tangential coordinates. In-Class Activities: • Check homework, if any • Reading quiz • Applications • Equation of motion in n-t coordinates • Concept quiz • Group problem solving • Attention quiz

APPLICATIONS Race tracks are often banked in the turns to reduce the frictional forces required to keep the cars from sliding at high speeds. If the car’s maximum velocity and a minimum coefficient of friction between the tires and track are specified, how can we determine the minimum banking angle (q) required to prevent the car from sliding?

APPLICATIONS (continued) Satellites are held in orbit around the earth by using the earth’s gravitational pull as the centripetal force – the force acting to change the direction of the satellite’s velocity. Knowing the radius of orbit of the satellite, how can we determine the required speed of the satellite to maintain this orbit?

NORMAL & TANGENTIAL COORDINATES When a particle moves along a curved path, it may be more convenient to write the equation of motion in terms of normal and tangential coordinates. The normal direction (n) always points toward the path’s center of curvature. In a circle, the center of curvature is the center of the circle. The tangential direction (t) is tangent to the path, usually set as positive in the direction of motion of the particle.

EQUATIONS OF MOTION Since the equation of motion is a vector equation , F = ma, it may be written in terms of the n & t coordinates as Ftut + Fnun = mat + man Here Ft & Fn are the sums of the force components acting in the t & n directions, respectively. This vector equation will be satisfied provided the individual components on each side of the equation are equal, resulting in the two scalar equations: Ft = mat and Fn = man. Since there is no motion in the binormal (b) direction, we can also write Fb = 0.

NORMAL AND TANGENTIAL ACCERLERATIONS The tangential acceleration, at = dv/dt, represents the time rate of change in the magnitude of the velocity. Depending on the direction of Ft, the particle’s speed will either be increasing or decreasing. The normal acceleration, an = v 2/r, represents the time rate of change in the direction of the velocity vector. Remember, an always acts toward the path’s center of curvature. Thus, Fn will always be directed toward the center of the path. Recall, if the path of motion is defined as y = f(x), the radius of curvature at any point can be obtained from r = dy 2 3/2 [1 + ( ) ] dx d 2 y dx 2

SOLVING PROBLEMS WITH n-t COORDINATES • Use n-t coordinates when a particle is moving along a known, curved path. • Establish the n-t coordinate system on the particle. • Draw free-body and kinetic diagrams of the particle. The normal acceleration (an) always acts “inward” (the positive ndirection). The tangential acceleration (at) may act in either the positive or negative t direction. • Apply the equations of motion in scalar form and solve. • It may be necessary to employ the kinematic relations: at = dv/dt = v dv/ds an = v 2/r

EXAMPLE Given: At the instant q = 60°, the boy’s center of mass G is momentarily at rest. The boy has a weight of 60 lb. Neglect his size and the mass of the seat and cords. Find: The boy’s speed and the tension in each of the two supporting cords of the swing when q = 90°. Plan: 1) Since the problem involves a curved path and finding the force perpendicular to the path, use n-t coordinates. Draw the boy’s free-body and kinetic diagrams. 2) Apply the equation of motion in the n-t directions. 3) Use kinematics to relate the boy’s acceleration to his speed.

EXAMPLE (continued) Solution: 1) The n-t coordinate system can be established on the boy at some arbitrary angle q. Approximating the boy and seat together as a particle, the free-body and kinetic diagrams can be drawn. Free-body diagram Kinetic diagram n n 2 T q = man mat W t t T = tension in each cord W = weight of the boy

EXAMPLE (continued) 2) Apply the equations of motion in the n-t directions. (a) Fn = man => 2 T – W sin q = man Using an = v 2/r = v 2/10, w = 60 lb, and m = w/g = (60/32. 2), we get: 2 T – 60 sin q = (60/32. 2)(v 2/10) (1) (b) Ft = mat => W cos q = mat => 60 cos q = (60/32. 2) at Solving for at: at = 32. 2 cos q (2) Note that there are 2 equations and 3 unknowns (T, v, at). One more equation is needed.

EXAMPLE (continued) 3) Apply kinematics to relate at and v. v dv = at ds where ds = r dq = 10 dq => v dv = 32. 2 cosq ds = 32. 2 cosq (10 dq ) v => ò v dv = 0 => v 2 2 90 ò 32. 2 cosq dq 60 90 = 32. 2 sinq => v = 9. 29 ft/s 60 This v is the speed of the boy at q = 90. This value can be substituted into equation (1) to solve for T. 2 T – 60 sin(90°) = (60/32. 2)(9. 29)2/10 T = 38. 0 lb (the tension in each cord)

GROUP PROBLEM SOLVING Given: A 200 kg snowmobile with rider is traveling down the hill. When it is at point A, it is traveling at 4 m/s and increasing its speed at 2 m/s 2. Find: The resultant normal force and resultant frictional force exerted on the tracks at point A. Plan: 1) 2) 3)

GROUP PROBLEM SOLVING (continued) Solution: 1) The n-t coordinate system is established on the snowmobile at point A.

GROUP PROBLEM SOLVING (continued) 2) Apply the equations of motion in the n-t directions:

GROUP PROBLEM SOLVING (continued) 3) Determine r Determine q

EQUATIONS OF MOTION: CYLINDRICAL COORDINATES (Section 13. 6) Today’s Objectives: In-Class Activities: Students will be able to • Check homework, if any analyze the kinetics of a • Reading quiz particle using cylindrical coordinates. • Applications • Equations of motion using cylindrical coordinates • Angle between radial and tangential directions • Concept quiz • Group problem solving • Attention quiz

APPLICATIONS The forces acting on the 100 -lb boy can be analyzed using the cylindrical coordinate system. If the boy slides down at a constant speed of 2 m/s, can we find the frictional force acting on him?

APPLICATIONS (continued) When an airplane executes the vertical loop shown above, the centrifugal force causes the normal force (apparent weight) on the pilot to be smaller than her actual weight. If the pilot experiences weightlessness at A, what is the airplane’s velocity at A?

EQUATIONS OF MOTION: CYLINDRICAL COORDINATES This approach to solving problems has some external similarity to the normal & tangential method just studied. However, the path may be more complex or the problem may have other attributes that make it desirable to use cylindrical coordinates. Equilibrium equations or “Equations of Motion” in cylindrical coordinates (using r, q , and z coordinates) may be expressed in scalar form as: . . . 2 Fr = mar = m(r – rq ). . Fq = maq = m(rq – 2 rq). . Fz = maz = mz

EQUATIONS OF MOTION (continued) If the particle is constrained to move only in the r – q plane (i. e. , the z coordinate is constant), then only the first two equations are used (as shown below). The coordinate system in such a case becomes a polar coordinate system. In this case, the path is only a function of q. Fr = mar = m(r – Fq = maq = m(rq – 2 rq) rq 2) . . . . Note that a fixed coordinate system is used, not a “bodycentered” system as used in the n – t approach.

TANGENTIAL AND NORMAL FORCES If a force P causes the particle to move along a path defined by r = f (q ), the normal force N exerted by the path on the particle is always perpendicular to the path’s tangent. The frictional force F always acts along the tangent in the opposite direction of motion. The directions of N and F can be specified relative to the radial coordinate by using angle y.

DETERMINATION OF ANGLE y The angle y, defined as the angle between the extended radial line and the tangent to the curve, can be required to solve some problems. It can be determined from the following relationship. tan y = r dq dr = r dr dq If y is positive, it is measured counterclockwise from the radial line to the tangent. If it is negative, it is measured clockwise.

EXAMPLE Given: The ball (P) is guided along the vertical circular path. . W = 0. 5 lb, q 2 = 0. 4 rad/s, q = 0. 8 rad/s , rc = 0. 4 ft Find: Force of the arm OA on the ball when q = 30. Plan: Draw a FBD. Then develop the kinematic equations and finally solve the kinetics problem using cylindrical coordinates. Solution: Notice that r = rc cosq, therefore: . . r = -2 rc sinq q. . . 2 r = -2 rc cosq q – 2 rc sinq q

EXAMPLE (continued) Free Body Diagram: Establish the r , q inertial coordinate system and draw the particle’s free body diagram. Notice that the radial acceleration is negative. q mg q r t y q maq q = Ns q NOA q n q = 30° mar

EXAMPLE (continued) Kinematics: at q = 30 r = 2(0. 4) cos(30 ) = 0. 693 ft. r = -2(0. 4) sin(30 )(0. 4) = -0. 16 ft/s. . r = -2(0. 4) cos(30 )(0. 4)2 – 2(0. 4) sin(30 )(0. 8) = -0. 431 ft/s 2 Acceleration components are. . . 2 ar = r – rq = -0. 431 – (0. 693)(0. 4)2 = -0. 542 ft/s 2. . aq = rq + 2 rq = (0. 693)(0. 8) + 2(-0. 16)(0. 4) = 0. 426 ft/s 2 tan y = r/(dr/dq) where dr/dq = -2 rc sinq tan y = (2 rc cosq)/(-2 rc sinq) = -1/tanq y = 120

EXAMPLE (continued) Kinetics: Fr = mar 0. 5 Ns cos(30 ) – 0. 5 sin(30 ) = (-0. 542) 32. 2 Ns = 0. 279 lb Fq = maq NOA + 0. 279 sin(30 ) – 0. 5 cos(30 ) = NOA = 0. 3 lb 0. 5 (0. 426) 32. 2

GROUP PROBLEM SOLVING Given: A plane flies in a vertical loop as shown. v. A = 80 ft/s (constant) W = 130 lb Find: Normal force on the pilot at A. Plan: Solution: Kinematics:

GROUP PROBLEM SOLVING (continued) Therefore v A = Since r = 600 ft · 80 = 0. 133 rad at A, q = s 600 Since v. A is constant, aq = = - 42. 67 ft ·· r= · 2 ·· = ar r rq = = - 53. 33 ft s 2

GROUP PROBLEM SOLVING (continued) Free Body Diagram & Kinetic Diagram Kinetics: Fr = mar