Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chapter 5 Applying Newton’s Laws PowerPoint® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Copyright © 2012 Pearson Education Inc. Congratulations! Copyright © 2012 Pearson Education Inc. Congratulations! Copyright © 2012 Pearson Education Inc. Congratulations! Copyright © 2012 Pearson Education Inc. Congratulations! Copyright © 2012 Pearson Education Inc. Congratulations! On the first try…. Thursday! • Ahmed, Christian, Nick, Fern • Troy, Ali, Russell, Moses, Andrew • Dolly, Samuel, Ray, Hye Friday! Hairu, Guohao, Dan, Quinlan On the second try…. • • • • • Robert, Saminder, Jaikar, Kelsi Joshua, Rex, Steven, Shean, Hesham Jordan, Camille, Andrew, Bryan David, Omar, Sugei, Louis, Eduardo Roger, Sue, Becky, Adolfo, Jasmine Copyright © 2012 Pearson Education Inc. Goals for Chapter 5 • Use Newton’s 1st law for bodies in equilibrium (statics) • Use Newton’s 2nd law for accelerating bodies (dynamics) • Study types of friction & fluid resistance • Solve circular motion problems Copyright © 2012 Pearson Education Inc. Using Newton’s First Law when forces are in equilibrium • A body is in equilibrium when it is at rest or moving with constant velocity in an inertial frame of reference. • Follow Problem-Solving Strategy 5.1. Copyright © 2012 Pearson Education Inc. Copyright © 2012 Pearson Education Inc. Using Newton’s First Law when forces are in equilibrium • VISUALIZE (create a coordinate system; decide what is happening?) • SKETCH FREE-BODY DIAGRAM •Isolate one point/body/object •Show all forces in that coordinate system ON that body (not acting by that body!) • LABEL all forces clearly, consistently •Normal forces from surfaces •Friction forces from surfaces •Tension Forces from ropes •Contact forces from other objects •Weight from gravity Copyright © 2012 Pearson Education Inc. Using Newton’s First Law when forces are in equilibrium • BREAK ALL applied forces into components based on your coordinate system. • Apply Newton’s Laws to like components ONLY •SFx = max; SFy = may Copyright © 2012 Pearson Education Inc. One-dimensional equilibrium: Tension in a massless rope • A gymnast hangs from the end of a massless rope. • Example 5.1 (mg = 50 kg; what is weight & force on rope?) Copyright © 2012 Pearson Education Inc. One-dimensional equilibrium: Tension in a massless rope • A gymnast hangs from the end of a massless rope. • Example (mg = 50 kg; what is weight & force on rope?) Copyright © 2012 Pearson Education Inc. One-dimensional equilibrium: Tension in a massless rope • A gymnast hangs from the end of a massless rope. • Example (mg = 50 kg; what is weight & force on rope?) Tension of Rope on Gymnast Copyright © 2012 Pearson Education Inc. One-dimensional equilibrium: Tension in a rope with mass • What is the tension in the previous example if the rope has mass? (Example 5.2 weight of rope = 120 N) Copyright © 2012 Pearson Education Inc. Two-dimensional equilibrium • A car engine hangs from several chains. • Weight of car engine = w; ignore chain weights Copyright © 2012 Pearson Education Inc. A car on an inclined plane • An car rests on a slanted ramp (car of weight w) Copyright © 2012 Pearson Education Inc. A car on an inclined plane • Coordinate system choice #1: – (y) parallel to slope & – (x) perpendicular to slope Copyright © 2012 Pearson Education Inc. A car on an inclined plane • Coordinate system choice #1: – (y) parallel to slope & – (x) perpendicular to slope N (in y) T (in x) W (in x & y) Copyright © 2012 Pearson Education Inc. A car on an inclined plane • Coordinate system choice #2: – (y) perpendicular to ground – (x) parallel to ground N in both x & y! y T (in x & y!) W (in y only) x Copyright © 2012 Pearson Education Inc. A car on an inclined plane • Coordinate system choice #2: – (y) perpendicular to ground – (x) parallel to ground N T y Ny Tx Tx Nx W x Copyright © 2012 Pearson Education Inc. Example 5.5: Bodies connected by a cable and pulley • Cart connected to bucket by cable passing over pulley. •Initially, assume pulley is massless and frictionless! • Pulleys REDIRECT force – they don’t amplify or reduce. Tension in rope pulls upwards along slope Copyright © 2012 Pearson Education Inc. SAME tension in rope pulls upwards on bucket Example 5.5: Bodies connected by a cable and pulley • Draw separate free-body diagrams for the bucket and the cart. Copyright © 2012 Pearson Education Inc. A note on free-body diagrams • Only the force of gravity acts on the falling apple. • ma does not belong in a free-body diagram! It is the SUM of all the forces you find! Copyright © 2012 Pearson Education Inc. Ex 5.6 Straight-line motion with constant force • Wind exerts a constant horizontal force on the boat. • 4.0 s after release, v = 6.0 m/s; mass = 200 kg. Copyright © 2012 Pearson Education Inc. Ex 5.6 Straight-line motion with constant force • Wind exerts a constant horizontal force on the boat. • 4.0 s after release, v = 6.0 m/s; mass = 200 kg. • Find W, the force of the wind! Copyright © 2012 Pearson Education Inc. Example 5.7: Straight-line motion with friction • For the ice boat in the previous example, a constant horizontal friction force of 100 N opposes its motion • What constant force needed by wind to create the same acceleration (a = + 1.5 m/s/s)? Copyright © 2012 Pearson Education Inc. Example 5.7: Straight-line motion with friction • For the ice boat in the previous example, a constant horizontal friction force now opposes its motion (100N); what constant force needed by wind to create the same acceleration (a = + 1.5 m/s/s)? Copyright © 2012 Pearson Education Inc. Example 5.8: Tension in an elevator cable • Elevator (800 kg) is moving downward @ 10 m/s but slowing to a stop over 25.0 m. • What is the tension in the supporting cable? Copyright © 2012 Pearson Education Inc. Example 5.8: Tension in an elevator cable • Elevator (800 kg) is moving downward @ 10 m/s but slowing to a stop over 25.0 m. • What is the tension in the supporting cable? Copyright © 2012 Pearson Education Inc. Example 5.8: Tension in an elevator cable • Compare Tension to Weight while elevator slows? Copyright © 2012 Pearson Education Inc. Example 5.8: Tension in an elevator cable • Compare Tension to Weight while elevator slows? • What if elevator was accelerating upwards at same rate? Copyright © 2012 Pearson Education Inc. Example 5.8: Tension in an elevator cable • What if elevator was accelerating upwards at same rate? • Same Free Body Diagram! Same result! Copyright © 2012 Pearson Education Inc. Ex 5.9 Apparent weight in an accelerating elevator • A woman inside the elevator of the previous example is standing on a scale. How will the acceleration of the elevator affect the scale reading? Copyright © 2012 Pearson Education Inc. Ex 5.9 Apparent weight in an accelerating elevator • A woman inside the elevator of the previous example is standing on a scale. How will the acceleration of the elevator affect the scale reading? Copyright © 2012 Pearson Education Inc. Ex 5.9 Apparent weight in an accelerating elevator • What if she was accelerating downward, rather than slowing? Increasing speed down? Copyright © 2012 Pearson Education Inc. Acceleration down a hill • What is the acceleration of a toboggan sliding down a friction-free slope? Copyright © 2012 Pearson Education Inc. Acceleration down a hill • What is the acceleration of a toboggan sliding down a friction-free slope? Follow Example 5.10. Copyright © 2012 Pearson Education Inc. Two common free-body diagram errors • The normal force must be perpendicular to the surface. • There is no separate “ma force.” Copyright © 2012 Pearson Education Inc. Two bodies with the same acceleration • We can treat the milk carton and tray as separate bodies, or we can treat them as a single composite body. Copyright © 2012 Pearson Education Inc. Two bodies with the same acceleration • Push a 1.00 kg food tray with constant 9 N force, no friction. What is acceleration of tray and force of tray on the carton of milk? Copyright © 2012 Pearson Education Inc. Two bodies with the same magnitude of acceleration • The glider on the air track and the falling weight move in different directions, but their accelerations have the same magnitude and relative direction (both increasing, or both decreasing) Copyright © 2012 Pearson Education Inc. Two bodies with the same magnitude of acceleration • What is the tension T, and the acceleration a, of the system? Copyright © 2012 Pearson Education Inc. Frictional forces • When a body rests or slides on a surface, the friction force is parallel to the surface. • Friction between two surfaces arises from interactions between molecules on the surfaces. Copyright © 2012 Pearson Education Inc. Kinetic and static friction • Kinetic friction acts when a body slides over a surface. • The kinetic friction force is fk = µkn. • Static friction acts when there is no relative motion between bodies. • The static friction force can vary between zero and its maximum value: fs ≤ µsn. Copyright © 2012 Pearson Education Inc. Static friction followed by kinetic friction • Before the box slides, static friction acts. But once it starts to slide, kinetic friction acts. Copyright © 2012 Pearson Education Inc. Some approximate coefficients of friction Copyright © 2012 Pearson Education Inc. Friction in horizontal motion – example 5.13 • Move a 500-N crate across a floor with friction by pulling with a force of 230 N. • Initially, pull harder to get it going; later pull easier (at 200N once it is going). What are ms? mk? Copyright © 2012 Pearson Education Inc. Friction in horizontal motion • Before the crate moves, static friction acts on it. Copyright © 2012 Pearson Education Inc. Friction in horizontal motion • After it starts to move, kinetic friction acts. Copyright © 2012 Pearson Education Inc. Static friction can be less than the maximum • Static friction only has its maximum value just before the box “breaks loose” and starts to slide. Force Force builds in time to maximum value, then object starts moving & slipping Time Copyright © 2012 Pearson Education Inc. Pulling a crate at an angle • The angle of the pull affects the normal force, which in turn affects the friction force. • Follow Example 5.15. Copyright © 2012 Pearson Education Inc. Motion on a slope having friction – ex 5.16 • Consider a toboggan going down a slope at constant speed. What is m? • Now consider same toboggan on steeper hill, so it is now accelerating. What is a? Copyright © 2012 Pearson Education Inc. Fluid resistance and terminal speed • Fluid resistance on a body depends on the speed of the body. • Resistance can depend upon v or v2 and upon the shape moving through the fluid. • Fresistance = -kv or - Dv2 • These will result in different terminal speeds Copyright © 2012 Pearson Education Inc. Fluid resistance and terminal speed • A falling body reaches its terminal speed when resisting force equals weight of the body. • If F = -kv for a falling body, vterminal = mg/k • If F = - Dv2 for a falling body, vterminal = (mg/D)½ Copyright © 2012 Pearson Education Inc. Dynamics of circular motion • If something is in uniform circular motion, both its acceleration and net force on it are directed toward center of circle. • The net force on the particle is Fnet = mv2/R, always towards the center. Copyright © 2012 Pearson Education Inc. Using Newton’s First Law when forces are in equilibrium • IF you see uniform circular motion… r v Copyright © 2012 Pearson Education Inc. Using Newton’s First Law when forces are in equilibrium • IF you see uniform circular motion… THEN remember centripetal force is NOT another force – it is the SUM of one or more forces already present! r •SFx = mv2/R NOT •mv2/R + T – mg = ma Copyright © 2012 Pearson Education Inc. v Avoid using “centrifugal force” • Figure (a) shows the correct free-body diagram for a body in uniform circular motion. • Figure (b) shows a common error. • In an inertial frame of reference, there is no such thing as “centrifugal force.” Copyright © 2012 Pearson Education Inc. Force in uniform circular motion • A 25 kg sled on frictionless ice is kept in uniform circular motion by a 5.00 m rope at 5 rev/minute. What is the force? Copyright © 2012 Pearson Education Inc. Force in uniform circular motion • A 25 kg sled on frictionless ice is kept in uniform circular motion by a 5.00 m rope at 5 rev/minute. What is the force? Copyright © 2012 Pearson Education Inc. What if the string breaks? • If the string breaks, no net force acts on the ball, so it obeys Newton’s first law and moves in a straight line. Copyright © 2012 Pearson Education Inc. A conical pendulum – Ex 5.20 • A bob at the end of a wire moves in a horizontal circle with constant speed. Copyright © 2012 Pearson Education Inc. A car rounds a flat curve • A car rounds a flat unbanked curve. What is its maximum speed? Copyright © 2012 Pearson Education Inc. A car rounds a flat curve • A car rounds a flat unbanked curve. What is its maximum speed? Copyright © 2012 Pearson Education Inc. A car rounds a banked curve • At what angle should a curve be banked so a car can make the turn even with no friction? Copyright © 2012 Pearson Education Inc. A car rounds a banked curve • At what angle should a curve be banked so a car can make the turn even with no friction? Copyright © 2012 Pearson Education Inc. Uniform motion in a vertical circle – Ex. 5.23 • A person on a Ferris wheel moves in a vertical circle at constant speed. • What are forces on person at top and bottom? Copyright © 2012 Pearson Education Inc. Uniform motion in a vertical circle – Ex. 5.23 • A person on a Ferris wheel moves in a vertical circle. Copyright © 2012 Pearson Education Inc.