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Transcript
9/12/2016 Chapter 4 Test Monday 9/19 Newton’s Laws The Study of Dynamics 1 9/12/2016 Isaac Newton • Arguably the greatest physical genius ever. • Came up with 3 Laws of Motion to explain the observations and analyses of Galileo and Johannes Kepler. • Invented Calculus. • Published his Laws in 1687 in the book Mathematical Principles of Natural Philosophy. What is Force? •A force is a push or pull on an object. •Forces cause an object to accelerate… •To speed up •To slow down •To change direction Contact forces arise from physical contact . Action-at-a-distance forces do not require contact and include gravity and electrical forces. 2 9/12/2016 The net force on an object is the vector sum of all forces acting on that object. The SI unit of force is the Newton (N). kg m N 2 s Arrows are used to represent forces. The length of the arrow is proportional to the magnitude of the force. 15 N 5N 3 9/12/2016 Arrows are used to represent forces. The length of the arrow is proportional to the magnitude of the force. Individual Forces 4N Net Force 10 N 6N Net Force Individual Forces 5N 37o 3N 4N 4 9/12/2016 Newton’s First Law •The Law of Inertia. •A body in motion stays in motion at constant velocity and a body at rest stays at rest 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. The First Law is Counterintuitive contrary to intuition or to common-sense expectation (but often nevertheless true). Aristotle firmly believed this. 5 9/12/2016 When a train suddenly starts, the passengers tend to fall backwards. This is because the lower part of the body which is in contact with the train begins to move while the upper part of the body tends to maintain its position of rest. As result, the upper part tends to fall backwards http://physics.tutorvista.com/motion/newton-s-first-law-of-motion.html A physics book is sitting at rest on a table top. Are there any forces acting on the book? 6 9/12/2016 A physics book is sitting at rest on a table top. Are there any forces acting on the book? A force diagram illustrating no net force 7 9/12/2016 A force diagram illustrating no net force Another example illustrating no net force 8 9/12/2016 Newton’s First Law •The Law of Inertia. •A body in motion stays in motion at constant velocity and a body at rest stays at rest unless acted upon by an external force. •If there is no NET force acting on an object there will be no acceleration. Newton’s Second Law •A body accelerates when acted upon by a net external force. •The acceleration is proportional to the net force and is in the direction which the net force acts. •This law is commonly applied to the vertical component of velocity. 9 9/12/2016 Newton’s Second Law •∑F = ma •where ∑F is the net force measured in Newtons (N) •m is mass (kg) •a is acceleration (m/s2) A free-body-diagram is a diagram that represents the object and the forces that act on it. Ftable Fg 10 9/12/2016 Working a Newton’s 2nd Law Problem F 395 275 560 110 N F ma 395 N 110 N (1850kg)a 560 N 275 N a 0.059m / s 2 The direction of force and acceleration vectors can be taken into account by using x and y components. F ma is equivalent to F y may F x max 11 9/12/2016 Working a Newton’s 2nd Law Problem 17 N 12 N F F 12 2 17 2 20.8 N 17 N F ma 12 N 20.8 N (27kg)a a 0.77m / s 2 Assignment Read pg. 75-80 Do pg. 97-98 Questions #3,4 Problems #1,2,3,6 12 9/12/2016 Chapter 4 Test Monday 9/19 Gravity as an accelerating force A very commonly used accelerating force is gravity. Here is gravity in action. The acceleration is g. 13 9/12/2016 Gravity as an accelerating force In the absence of air resistance, gravity acts upon all objects by causing the same acceleration…g. Gravity as an accelerating force The pulley lets us use gravity as our accelerating force… but a lot slower than free fall. Acceleration here is a lot lower than g. 14 9/12/2016 The problem of weight Are weight and mass the same thing? •No. Weight can be defined as the force due to gravitation attraction. •W = mg Flat surfaces – 1 D The normal force is one component of the force that a surface exerts on an object with which it is in contact – namely, the component that is perpendicular to the surface. FN = mg for objects resting on horizontal surfaces. FN mg 15 9/12/2016 John places a block on the top of a flat table. The block has a mass of 1.5kg. What is the normal force the table applies to the block? FN mg FN (1.5kg)(10m / s 2 ) FN 15 N John places a block on the top of a flat table. The block has a mass of 1.5kg. He then proceeds to push down on the block with a force of 11N. What is the normal force the table applies to the block? FN 15N 11N 26 N 16 9/12/2016 John places a block on the top of a flat table. The block has a mass of 1.5kg. He then proceeds to pull up on the block with a force of 11N. What is the normal force the table applies to the block? FN 15N 11N 4 N How much tension must a rope withstand in order to accelerate a 1000kg object up with an acceleration of 0.65 m/s2? Tension W ma T=W+ma T 10000 N (1000kg)(.65m / s 2 ) T 10000 N 650 N T 10650 N W=mg=10000 N 17 9/12/2016 Will pushes a 10 kg block on a frictionless floor at a 30o angle below the horizontal with a force of 150 N. a) F cos ma a) What is the acceleration? 150 cos 30o (10kg)a b) What is the normal force? 129.9 N (10kg)a a 13m / s 2 Fcos b) FN F sin W Fsin FN 150 sin 30o 100 N F FN 75 100 FN 175 N W=mg=100 N T sin 10o T sin 10o W 2T sin 10o 500 N T sin 10o 250 N T Tsin10 250 N 1440 N sin 10o Tsin10 W=mg=500 N 18 9/12/2016 normal force is Ramps – 2 D The perpendicular to angled ramps as well. It’s always equal to the component of weight perpendicular to the surface. N = mgcos N F=mgsin mg What is the normal force for a 5.0 kg block on a ramp with a 15o angle with the horizontal? FN mg cos FN (5.0kg)(10m / s 2 ) cos 15o FN 48 N 19 9/12/2016 How long will it take a 5.0 kg block to slide down a frictionless 20 m long ramp that is at a 15o angle with the horizontal? F mg sin ma mg sin a g sin a (10m / s 2 ) sin 15 2.6m / s 2 1 x vot at 2 2 1 20m 0 (2.6m / s 2 )t 2 2 20 1.3t 2 15.38 t 2 t 3.9 s V>0 V=0 A>0 A=0 Elevator Ride – going up! Heavy feeling Normal feeling V>0 A=0 V>0 A<0 Normal feeling Light feeling N N N N mg mg mg mg Ground floor Just starting up Between floors Arriving at top floor 20 9/12/2016 V<0 V=0 A<0 A=0 Elevator Ride – going down! Normal feeling V<0 A=0 V<0 A>0 Heavy feeling Normal feeling Light feeling N N N N mg mg mg mg Top floor Beginning descent Between floors Arriving at Ground floor A person weighing 800N rides the elevator up to the 4th floor from the 1st floor. The elevator has an acceleration of 1.0m/s2, how heavy does the person feel when the elevator begins to go up? W m( g a ) W (80kg)(10 1) W 880 N 21 9/12/2016 A person stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.75 of the person’s regular weight. Calculate the acceleration of the elevator, and find the direction of the acceleration? .75W m( g a ) .75mg mg ma .75 g g a .75(10) 10 a 7.5 10 a 2.5 a a 2.5m / s 2 Assignment Read pg. 81-94 Do pg. 98-103 Problems #10,12,15,16,24,25,72 22 9/12/2016 Chapter 4 Test Monday 9/19 Magic Pulleys N mg T -x T m1 mg m2 x 23 9/12/2016 Mass 1 (10 kg) rests on a frictionless table connected by a string to Mass 2 (5 kg). Find (a) the acceleration of each block and, (b) the tension in the connecting string. a)T m2 g (5kg)(10m / s 2 ) 50 N b) F ma 50 N (5kg 10kg)a 50 N (15kg)a a 3.3m / s 2 m1 m2 Chapter 4 Test Monday 9/19 24 9/12/2016 Friction •The force that opposes a sliding motion. •Enables us to walk, drive a car, etc. •Due to microscopic irregularities in even the smoothest of surfaces. There are two types of friction Static friction •exists before sliding occurs •Kinetic friction •exists after sliding occurs •In general fk <= fs 25 9/12/2016 Friction and the Normal Force •The frictional force which exists between two surfaces is directly proportional to the normal force. •That’s why friction on a sloping surface is less than friction on a flat surface. Applied forces affect normal force. friction applied force normal weight N = applied force 26 9/12/2016 Static Friction •fs sN • fs : static frictional force (N) • s: coefficient of static friction • N: normal force (N) •Static friction increases as the force trying to push an object increases… up to a point! A force diagram illustrating Static Friction Normal Force Frictional Force Applied Force Gravity 27 9/12/2016 A force diagram illustrating Static Friction Normal Force Bigger Applied Force Bigger Frictional Force Gravity A force diagram illustrating Static Friction The forces on the book are now UNBALANCED! Normal Force Frictional Force Gravity Even Bigger Applied Force Static friction cannot get any larger, and can no longer completely oppose the applied force. 28 9/12/2016 Kinetic Friction •fk = kN • fk : kinetic frictional force (N) • k: coefficient of kinetic friction • N: normal force (N) •Kinetic friction (sliding friction) is generally less than static friction (motionless friction) for most surfaces. A force of 48.0N is required to start a 5.0-kg box moving across a horizontal concrete floor. a) What is the coefficient of static friction between the box and the floor? b) If the 48.0 N force continues, the box accelerates at 0.70 m/s 2. What is the coefficient of kinetic friction? a) f s 48 N s FN FN mg (5kg)(10m / s 2 ) 50 N 48 N s (50 N ) s .96 b)netF ma (5kg)(0.70m / s 2 ) 3.5 N netF netF F f k 3.5 48 f k f k 44.5 N k FN k 44.5 / 50 0.89 29 9/12/2016 What is the maximum acceleration a car can undergo if the coefficient of static friction between the tires and the road is 0.80? F fs F s FN ma s (mg ) a s g a (0.80)(10m / s 2 ) a 8m / s 2 Chapter 4 Test Monday 9/19 30 9/12/2016 Magic Pulleys N mg T -x T m1 mg m2 x Mass 1 (10 kg) rests on a frictionless table connected by a string to Mass 2 (5 kg). Find (a) the acceleration of each block and, (b) the tension in the connecting string. a)T m2 g (5kg)(10m / s 2 ) 50 N b) F ma 50 N (5kg 10kg)a 50 N (15kg)a a 3.3m / s 2 m1 m2 31 9/12/2016 Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg) as shown. What must the minimum coefficient of static friction be to keep Mass 1 from slipping? FN m1 g (10kg)(10m / s 2 ) 100 N appliedF m2 g 50 N s FN appliedF s (100 N ) 50 N s .5 m1 m2 Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). If s = 0.3 and k = 0.2, what is a) the acceleration and b) the tension in the string? F m g (10kg)(10m / s 2 ) 100 N N 1 k FN .2(100) 20 N appliedF 50 N NetF 50 N 20 N 30 N F ma 30 N (15kg)a a 2m / s 2 b)T m2 g Fk 50 N 20 N 70 N m1 m2 32 9/12/2016 Newton’s Third Law of Motion Any time a force is exerted on an object, that force is caused by another object. Newton’s third law: Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first. Newton’s Third Law of Motion A key to the correct application of the third law is that the forces are exerted on different objects. Make sure you don’t use them as if they were acting on the same object. 33 9/12/2016 Newton’s Third Law of Motion Rocket propulsion can also be explained using Newton’s third law: hot gases from combustion spew out of the tail of the rocket at high speeds. The reaction force is what propels the rocket. Note that the rocket does not need anything to “push” against. Suppose that the magnitude of the force is 36 N. If the mass of the spacecraft is 11,000 kg and the mass of the astronaut is 92 kg, what are the accelerations? spacecraft P ma 36 N (11000kg)a a 0.0036m / s 2 astronaut P ma 36 N (92kg)a a 0.39m / s 2 34 9/12/2016 T sin 10o T sin 10o W 2T sin 10o 500 N T sin 10o 250 N T Tsin10 250 N 1440 N sin 10o Tsin10 W=mg=500 N An automobile engine has a weight W, whose magnitude is W=3150N. This engine is being positioned above an engine compartment, as Figure 4.29a illustrates. To position the engine, a worker is using a rope. Find the tension T1 in the supporting cable and the tension T2 in the positioning rope. x components :right left T2 sin 80o T1 sin 10o T2 .18T1 y components : up down T1 cos 10o W T2 cos 80o .985T1 3150 .174T 2 .985T1 3150 .174(.18T 1) .985T1 3150 .031T 1 .954T1 3150 T1 3303 N T2 (.18)(3303) 595 N 35 9/12/2016 Assignment Do pg. 100-104 Problems #27,30,37,41,45,50,78 Chapter 4 Test Monday 9/19 36 9/12/2016 Newton’s Third Law of Motion In nature there are two general types of forces, fundamental and nonfundamental. Fundamental Forces 1. Gravitational force 2. Strong Nuclear force 3. Electroweak force 37 9/12/2016 Newton’s Law of Universal Gravitation Every particle in the universe exerts an attractive force on every other particle. A particle is a piece of matter, small enough in size to be regarded as a mathematical point. The force that each exerts on the other is directed along the line joining the particles. For two particles that have masses m1 and m2 and are separated by a distance r, the force has a magnitude given by mm F G 12 2 r G 6.673 1011 N m2 kg 2 38 9/12/2016 Definition of Weight The weight of an object on or above the earth is the gravitational force that the earth exerts on the object. The weight always acts downwards, toward the center of the earth. On or above another astronomical body, the weight is the gravitational force exerted on the object by that body. 39