Lecture 9: Equation of Motion

Section 13.1 Newton’s Laws of Motion Newton’s Laws of Motion (1 of 2) 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 (2 of 2) 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 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 m1m2 r2 where F = force of attraction between the two bodies, G = universal constant of gravitation , m1, m2 = 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 Versus FPS System (1 of 2) 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 W (N)  m (kg) g (m/s 2 )  N  kg  m/s g  9.81 m/s2. 2 Units: SI System Versus FPS System (2 of 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 m (slugs)  W (lb)/g ( ft/s 2 32.2 )  slug  lb  s 2 /ft ft/s 2 . Section 13.2 Equation of Motion 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 free-body 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. Section 13.3 Equation of Motion for a System of Particles Equation of Motion for a 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 aG as the acceleration of the center of mass G of the particles, then m aG   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, not the result of an analytical proof. 2. Mass (a 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  mg. 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 (1 of 3) Given: A 25-kg block is subjected to the force F=100 N. The spring has a stiffness of k = 200 N/m and is unstretched when the block is at A. The contact surface is smooth. Find: Draw the free-body and kinetic diagrams of the block when s=0.4 m. Plan: 1. Define an inertial coordinate system. 2. Draw the block’s free-body diagram, showing all external forces. 3. Draw the block’s kinetic diagram, showing the inertial force vector in the proper direction. Example (2 of 3) 1. An inertial x-y frame can be defined as fixed to the ground. 2. Draw the free-body diagram of the block: The weight force (W) acts through the block’s center of mass. F is the applied load and Fs  200  is the spring force, where  is the spring deformation. When s = 0.4,   0.5  0.3  0.2 The normal force (N) is perpendicular to the surface. There is no friction force since the contact surface is smooth. Example (3 of 3) 3. Draw the kinetic diagram of the block. The block will be moved to the right. The acceleration can be directed to the right if the block is speeding up or to the left if it is slowing down. Section 13.4 Rectangular Coordinates Rectangular Coordinates The equation of motion, F  ma 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 three components in the Cartesian (rectangular) coordinate system as F  ma or  FX i   FY j   FZ k  m(a x i  a y j  a z k ) or, as scalar equations,  FX  ma X ,  FY  maY and  FZ  maZ Procedure For Analysis (1 of 3) • Free Body Diagram (is always critical!!) 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 kinetic diagram. Make sure any friction forces act opposite to the direction of motion! If the particle is connected to an elastic linear spring, a spring force equal to ‘k s’ should be included on the FBD. Procedure For Analysis (2 of 3) • 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 the best approach. A Cartesian vector formulation of the second law is  ma or  FX i   FY j   FZ k  m( a X i  aY j  a Z k ) F 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 (3 of 3) • Kinematics 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 kinematics 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 (1 of 3) Given: The motor winds in the cable with a constant acceleration such that the 20-kg crate moves a distance s = 6 m in 3 s, starting from rest. μk  0.3. Find: The tension developed in the cable. Plan: 1. Draw the free-body and kinetic diagrams of the crate. 2. Using a kinematic equation, determine the acceleration of the crate. 3. Apply the equation of motion to determine the cable tension. Example (2 of 3) Solution: 1. Draw the free-body and kinetic diagrams of the crate. Since the motion is up the incline, rotate the x-y axes so the x-axis aligns with the incline. Then, motion occurs only in the xdirection. There is a friction force acting between the surface and the crate. Why is it in the direction shown on the FBD? Example (3 of 3) 2. Using kinematic equation s  v0 t  ½ a t 2  6  (0) 3  ½ a (32 )  a  1.333 m/s 2 3. Apply the equations of motion  T  20 (981) (sin30)  0.3(169.9)  20 (1.333)  T  176 N