<p>MR. SURRETTE VAN NUYS HIGH SCHOOL</p><p>CHAPTER 3: DYNAMICS NEWTON’S LAWS CLASS NOTES</p><p>DYNAMICS Kinematics uses the definitions of displacement, velocity, and acceleration to describe motion. Dynamics investigates motion in terms of the forces that cause it. The three laws of motion were formulated by Isaac Newton three centuries ago.</p><p>NEWTON’S FIRST LAW Newton’s first law is the Law of Inertia: “an object at rest will remain at rest and an object in motion will remain in motion with a constant velocity unless acted on by a net external force.” (force is later defined in Newton’s Second Law).</p><p>MASS AND INERTIA Mass is the same as inertia. The greater an object’s mass, the harder it is to change its motion. For example, interstate trucks take much longer to speed up and slow down than motorcycles.</p><p>MASS VERSUS WEIGHT Weight is a mathematical relationship: w = mg</p><p>MASS VERSUS WEIGHT For example, to calculate the weight of a 10 kg object on earth: (1) w = mg (2) w = (10 kg)(9.8 m/s2) (3) w = 98 N (The unit “newton” is defined in Newton’s Second Law). </p><p>NEWTON’S SECOND LAW Newton’s second law, the Law of Acceleration, states: “the acceleration of an object is directly proportional to the resultant force acting on it and inversely proportional to its mass.” </p><p>NEWTON’S SECOND LAW Newton’s second law is written mathematically as: F = ma Force [N] = (Mass [kg])(Acceleration[m/s2]. (One newton is about 0.25 english pounds.)</p><p>NET FORCE Net force is the sum of all forces acting on an object. Net force is written as: F or FNET (The symbol  means, “the sum of”.)</p><p>1 | P a g e PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL</p><p>EQUILIBRIUM Equilibrium exists when the net force acting on an object is zero. An object in equilibrium has zero acceleration. Zero acceleration is when the velocity is constant or zero.</p><p>FRICTION Friction is a force that opposes motion. It is caused by the interaction of two different mediums. It is always in the opposite direction of motion. Mathematically, it is written as: f = n (Friction) = (Coefficient of Friction) x (Normal Force)</p><p>NORMAL FORCE The normal force is a passive response to an object's weight. Objects resting on a surface exert their weight on top of the surface. As a result, the molecules of the surface "lock up" and support the object. The greater an object's weight, the greater the normal force.</p><p>NEWTON’S THIRD LAW Newton’s third law, the Law of Action-Reaction, states: “when two bodies interact, the force which “A” exerts on body “B” (the action force) is equal in magnitude and opposite in direction to the force which body “B” exerts on body “A” (the reaction force).”</p><p>NEWTON’S THIRD LAW A consequence of the third law is that forces occur in pairs. Remember that the action force and the reaction force act on different objects.</p><p>NEWTON’S THIRD LAW Graphically, this can be represented as:</p><p>Note: Action and Reaction forces occur at the same time.</p><p>2 | P a g e PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL</p><p>CHAPTER 3: DYNAMICS FREE-BODY DIAGRAMS CLASS NOTES</p><p>FREE-BODY DIAGRAMS Free-body diagrams are an important step in the application of Newton’s laws. Diagrams should include a vector labeled to identify each force acting on the bodies. When a system has more than one body, construct a free-body diagram for each body. Free body diagrams use the equation F = ma and apply this equation in both the x (Fx) and y (Fy) directions. </p><p>STATIC EQUILIBRIUM A book resting on a table is in static equilibrium because it is not moving. The weight of the book in the downwards direction is equal and opposite to the normal force exerted by the table.</p><p>Free-Body Diagram</p><p>Example 1. There are six books in a stack, each with a weight of 3 N. The coefficient of friction between each pair of books is 0.1. What is the horizontal force needed to start sliding the top five books off the bottom one? 1A.</p><p>(1) Fx: FNET = ma à FNET = m(0) à F – f = 0 (there is no acceleration here, that is what the phrase “start sliding” means) (2) F = f</p><p>(3) Fy: FNET = ma à FNET = m(0) à n – w = 0 (4) n = w </p><p>3 | P a g e PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL</p><p>1a. (continued...) (5) Determine friction: f = n (6) f = (0.1)(15 N) (7) f = 1.5 N (8) F = 1.5 N</p><p>Example 2. As a 3.0 kg bucket is being lowered into a 10 m deep well, starting from the top, the tension in the rope is 9.8 N. What is the time to reach the bottom? 2A.</p><p>(T = 9.8 N, m = 3.0 kg, d = 10 m)</p><p>(1) Fx: 0</p><p>(2) Fy: FNET = ma à FNET = m(- a) à T – w = - ma (3) - ma = T – w (4) ma = - (T – w) (5) ma = w – T (6) a = (w – T) / m Determine weight of bucket: (7) w = mg (8) w = (3.0 kg)(9.8 m/s2) (9) w = 29.4 N Calculate the acceleration: (10) a = (29.4 N – 9.8 N) / 3.0 kg (Note: remember N = kg.m/s2) (11) a = 6.53 m/s2 Determine the time: 2 (12) d = vot + ½ at (13) d = 0 + ½ at2 (14) ½ at2 = d (15) 2[ ½ at2 = d] (16) at2 = 2d (17) t = (2d / a)1/2 (18) t = (2(10m) / 6.53 m/s2)1/2 (19) t = 1.7 s</p><p>4 | P a g e PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL</p><p>Example 3. A woman at an airport is pulling a 5 kg suitcase with wheels at constant speed by pulling on a strap at an angle  above the horizontal. She pulls on the strap with a 30 N force, and the frictional force between the suitcase wheels and the floor is 24 N. What is the angle? 3A.</p><p>(T = 30 N, f = 24 N, m = 5 kg)</p><p>(1) Fx: FNET = ma à FNET = m(0) à T cos  – f = 0 (when a problem says “constant speed”, that means no acceleration. Use 0 instead of “ma” for these questions). (2) T cos f</p><p>(3) Fy: FNET = ma à FNET = m(0) </p><p>(4) Fy: T sin  + n – w = m(0) = 0 (5) T sin n = w</p><p>(6) T cos  = f (from Fx) (7) cos  = f / T (8) cos  = 24 N / 30 N (9) cos  = 0.8 (10)  = arc cos (0.8) (11)  = 37o</p><p>Example 4. A 15 kg block rests on a level frictionless surface and is attached by a light string to a 5 kg hanging mass where the string passes over a massless frictionless pulley. The tension in the string is 49 N. If g = 9.8 m/s2, what is the acceleration of the system when released? 4A.</p><p>5 | P a g e PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL</p><p>4A. (continued…) (1) FOR MASS 1: Fx: FNET = ma à T = m1a Fy: FNET = ma à n – w1 = 0 à n = w1 (2) FOR MASS 2:</p><p>Fx: 0 Fy: FNET = ma à T – w2 = - m2a (3) T = m1a (4) a = (T / m1) (5) a = (49 N / 15 kg) (6) a = 3.27 m/s2</p><p>6 | P a g e PHYSICS</p>