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Transcript
Newton’s Laws http://mathsforeurope.digibel.be/images Tuesday, September 18, 2007 Re-introduction to Newton’s 3 Laws What is Force? A push or pull on an object. Unbalanced forces cause an object to accelerate… To speed up To slow down To change direction Force is a vector! Types of Forces Contact forces: involve contact between bodies. Field forces: act without necessity of contact. Normal, Friction Gravity, Electromagnetic, Strong, Weak Question: Is there really any such thing as a contact force? Forces and Equilibrium If the net force on a body is zero, it is in equilibrium. An object in equilibrium may be moving relative to us (dynamic equilibrium). An object in equilibrium may appear to be at rest (static equilibrium). Galileo’s Thought Experiment © The Physics Classroom, Tom Henderson 1996-2007 Galileo’s Thought Experiment This thought experiment lead to Newton’s First Law. © The Physics Classroom, Tom Henderson 1996-2007 Newton’s First Law The Law of Inertia. A body in motion stays in motion in a straight line unless acted upon by an external force. This law is commonly applied to the horizontal component of velocity, which is assumed not to change during the flight of a projectile. Newton’s Second Law A body accelerates when acted upon by a net external force The acceleration is proportional to the net (or resultant) force and is in the direction which the net force acts. This law is commonly applied to the vertical component of velocity. SF = ma Newton’s Third Law For every action there exists an equal and opposite reaction. If A exerts a force F on B, then B exerts a force of -F on A. Commonly Confused Terms Inertia: or the resistance of an object to being accelerated Mass: the same thing as inertia (to a physicist). Weight: gravitational attraction inertia = mass weight Sample Problem – 1st Law • Two forces, F1 = (4i – 6j + k) N and F2 = (i – 2j - 8k) N, act upon a body of mass 3.0 kg as it is moving at constant speed. What do you know must be true? Sample Problem – 2nd Law • Two forces, F1 = (4i – 6j + k) N and F2 = (i – 2j - 8k) N, act upon a body of mass 3.0 kg. No other forces act upon the body at this time. What do you know must be true? Sample Problem – 3rd Law • A tug-of-war team ties a rope to a tree and pulls hard horizontally to create a tension of 30,000 N in the rope. Suppose the team pulls equally hard when, instead of a tree, the other end of the rope is being pulled by another tug-of-war team such that no movement occurs. What is the tension in the rope in the second case? Working a 2nd Law Problem Working Newton’s 2nd Law Problems is best accomplished in a systematic fashion. The more complicated the problem, the more important it is to have a general procedure to follow in working it. 2nd Law Procedure 1. 2. 3. 4. 5. 6. 7. Identify the body to be analyzed. Select a reference frame, stationary or moving, but not accelerating Draw a force or free body diagram. Set up ΣF = ma equations for each dimension. Use kinematics or calculus where necessary to obtain acceleration. Substitute known quantities. Calculate the unknown quantities. Wednesday, September 19, 2007 Types of Forces Commonly Found in Newton’s Law Problems Sample Problem A 5.00-g bullet leaves the muzzle of a rifle with a speed of 320 m/s. The bullet is accelerated by expanding gases while it travels down the 0.820 m long barrel. Assuming constant acceleration and negligible friction, what is the force on the bullet? Sample Problem A 3.00 kg mass undergoes an acceleration given by a = (2.50i + 4.10j) m/s2. Find the resultant force F and its magnitude. Normal force The force that keeps one object from invading another object. Our weight is the force of attraction of our body for the center of the planet. We don’t fall to the center of the planet. The normal force keeps us up. Normal Force on Flat surface Normal Force on Flat surface N mg N = mg for objects resting on horizontal surfaces. Normal Force on Ramp Normal Force on Ramp N = mgcos N mgsin mg The normal force is perpendicular to angled ramps as well. It’s usually equal to the component of weight perpendicular to the surface. mgcos Normal Force on Ramp What will acceleration be in this situation? N SF= ma mgsin = ma mgsin gsin = a N = mgcos mg mgcos Normal Force on Ramp N = mgcos N How could you keep the block from accelerating? T mgsin mg mgcos Tension A pulling force. Generally exists in a rope, string, or cable. Arises at the molecular level, when a rope, string, or cable resists being pulled apart. Tension (static 2D) The horizontal and vertical components of the tension are equal to zero if the system is not accelerating. 30o 45o 1 2 3 15 kg Tension (static 2D) The horizontal and vertical components of the tension are equal to zero if the system is not accelerating. 30o T3 mg 45o 1 2 3 15 kg SFx = T1 0 T2 SFy = T3 0 Tension (elevator) What about when an elevator is accelerating upward? M Tension (elevator) T M Mg What about when an elevator is accelerating upward? Tension (elevator) What about when the elevator is moving at constant velocity between floors? M Tension (elevator) T M Mg What about when the elevator is moving at constant velocity between floors? Tension (elevator) What about when the elevator is slowing at the top floor? M Tension (elevator) T M Mg What about when the elevator is slowing at the top floor? Tension (elevator) What about if the elevator cable breaks? M Tension (elevator) What about if the elevator cable breaks? M Mg Pulley problems Magic pulleys simply bend the coordinate system. m1 m2 Pulley problems Magic pulleys simply bend the coordinate system. N T T m1g -x m1 SF = ma m2g = (m1+m2)a m2g m2 x Pulley problems All problems should be started from a force diagram. m2 Pulley problems Tension is determined by examining one SF = (m1+m2)a block or the other T T N m1g m2 m2g – T+T – m1gsin = (m1+m2)a m2g Pulley problems Tension is determined by examining one block or the other T T N m1g m2 SF2 = m2a m2g - T = m2a SF1 = m1a T-m1gsin = m1a m2g Atwood machine m2 m1 A device for measuring g. If m1 and m2 are nearly the same, slows down freefall such that acceleration can be measured. Then, g can be measured. Atwood machine T m1 m1g m2 T A device for measuring g. If m1 and m2 are nearly the same, slows down freefall such that acceleration can be measured. Then, g can be measured. m2g SF = ma m2g-m1g = (m2+m1)a Thursday, September 20, 2007 Post-Test Review Friday, September 21, 2007 Workday Exam Corrections Homework Review Atwood Machine Mini-lab Draw diagram of apparatus in lab book. Record all data. Calculate g. Easy Problem How fast will the block be sliding at the bottom of the frictionless ramp? 5.0 kg 20o L = 12 m Easy Problem How high up the frictionless ramp will the block slide? v = 12.0 m/s 5.0 kg 20o Tuesday, September 25, 2007 Friction Moderate Problem Describe acceleration of the 5 kg block. Table and pulley are magic and frictionless. 1.0 kg 20o Friction Friction opposes a sliding motion. Static friction exists before sliding occurs (fs sN). Kinetic friction exists after sliding occurs f k = k N Friction on flat surfaces y y x Draw a free body diagram for a braking car. x Draw a free body diagram for a car accelerating from rest. But we don’t want cars to skid! Why don’t we? Let’s use DataStudio to see if we can detect the difference in magnitude between static and kinetic friction. Wednesday, September 26, 2007 Friction Lab Thursday, September 27, 2007 Friction on a ramp Sliding down Sliding up Friction is always parallel to surfaces…. A 1.00 kg book is held against a wall by pressing it against the wall with a force of 50.00 N. What must be the minimum coefficient of friction between the book and the wall, such that the book does not slide down the wall? f F N W (0.20) Problem #1 Assume a coefficient of static friction of 1.0 between tires and road. What is the minimum length of time it would take to accelerate a car from 0 to 60 mph? Problem #2 Assume a coefficient of static friction of 1.0 between tires and road and a coefficient of kinetic friction of 0.80 between tires and road. How far would a car travel after the driver applies the brakes if it skids to a stop? Centripetal Force Inwardly directed force which causes a body to turn; perpendicular to velocity. Centripetal force always arises from other forces, and is not a unique kind of force. Sources include gravity, friction, tension, electromagnetic, normal. ΣF = ma a = v2/r ΣF = m v2/r Highway Curves z R Friction turns the vehicle r Normal force turns the vehicle Sample problem • Find the minimum safe turning radius for a car traveling at 60 mph on a flat roadway, assuming a coefficient of static friction of 0.70. Sample problem • Derive the expression for the period best banking angle of a roadway given the radius of curvature and the likely speed of the vehicles. Friday, September 28, 2007 More on Circular Motion Conical Pendulum z T = 2p L cos g L r T For conical pendulums, centripetal force is mg provided by a component o the tension. Sample problem • Derive the expression for the period of a conical pendulum with respect to the string length and radius of rotation. Non-uniform Circular Motion Consider circular motion in which either speed of the rotating object is changing, or the forces on the rotating object are changing. If the speed changes, there is a tangential as well as a centripetal component to the force. In some cases, the magnitude of the centripetal force changes as the circular motion occurs. Sample problem You swing a 0.25-kg rock in a vertical circle on a 0.80 m long rope at 2.0 Hz. What is the tension in the rope a) at the top and b) at the bottom of your swing? Monday, October 1 Time-dependent forces Sample problem A 40.0 kg child sits in a swing supported by 3.00 m long chains. If the tension in each chain at the lowest point is 350 N, find a) the child’s speed at the lowest point and b) the force exerted by the seat on the child at the lowest point. Sample problem A 900-kg automobile is traveling along a hilly road. If it is to remain with its wheels on the road, what is the maximum speed it can have as it tops a hill with a radius of curvature of 20.0 m? Non-constant Forces • • • Forces can vary with time. Forces can vary with velocity. Forces can vary with position. Calculus Concepts for Forces that Vary with Time • Differentiation • • • the tangent (or slope) of a function position -> velocity -> acceleration Integration • • the area under a curve acceleration -> velocity -> position Evaluating Integrals If a(t) = tn then tn dt = tn+1 / (n+1) + C t t tn dt 0 = tn+1 / (n+1) 0 Sample problem Consider a force that is a function of time: F(t) = (3.0 t – 0.5 t2)N If this force acts upon a 0.2 kg particle at rest for 3.0 seconds, what is the resulting velocity and position of the particle? Sample problem Consider a force that is a function of time: F(t) = (16 t2 – 8 t + 4)N If this force acts upon a 4 kg particle at rest for 1.0 seconds, what is the resulting change in velocity of the particle? Drag Forces Drag forces slow an object down as it passes through a fluid. act in opposite direction to velocity. are functions of velocity. impose terminal velocity. Drag as a Function of Velocity fD = bv + cv2 b and c depend upon shape and size of object properties of fluid b is important at low velocity c is important at high velocity Drag Force in Free Fall fD fD fD mg mg mg mg when fD equals mg, terminal velocity has been reached Drag Force in Free Fall FD = bv + c v2 FD for fast moving objects for slow moving objects FD = c v2 FD = b v c = 1/2 D r A Where D = drag coefficient r = density of fluid A = cross-sectional area mg Sample Problem: Slow moving objects FD = bv mg Show that vT = mg/b Sample Problem: Slow Moving Objects Show that v(t) = (mg/b)(1 – e FD = bv mg –bt/m) Sample Problem: Fast moving objects FD = 1/2DrAv2 mg Show that vT = (2mg / DrA)1/2 Sample Problem: Fast moving objects Derive an expression for the velocity of the object as a function of time FD = 1/2DrAv2 mg