Report

Chapter 4 Forces and Newton’s Laws of Motion Objective • We will be able to explain Newton’s 1st Law of motion 4.1 The Concepts of Force and Mass A force is a push or a pull. Contact forces arise from physical contact . Action-at-a-distance forces do not require contact and include gravity and electrical forces. 4.1 The Concepts of Force and Mass Arrows are used to represent forces. The length of the arrow is proportional to the magnitude of the force. Force is a vector quantity 15 N 5N 4.1 The Concepts of Force and Mass Mass is a measure of the amount of “stuff” contained in an object. (How many atoms are present) 4.2 Newton’s First Law of Motion Newton’s First Law An object continues in a state of rest or in a state of motion at a constant speed along a straight line, unless compelled to change that state by a net force. 4.2 Newton’s First Law of Motion Newton’s First Law The net force is the vector sum of all of the forces acting on an object. 4.2 Newton’s First Law of Motion 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). Individual Forces 4N 10 N Net Force 6N 4.2 Newton’s First Law of Motion Individual Forces Net Force 5N 64 3N 4N 4.2 Newton’s First Law of Motion Inertia is the natural tendency of an object to remain at rest or in motion at a constant speed along a straight line. 4.2 Newton’s First Law of Motion The mass of an object is a quantitative measure of inertia. SI Unit of Mass: kilogram (kg) DEMOS • Cup and Penny demo • Bottle and Paper Demo • Weight and String Demo Question • Which of the following statements can be explained by Newton’s first law? a) When your car suddenly comes to a halt, you lunge forward b) When your car rapidly accelerates, you are pressed backward against the seat c) Both d) Neither Question • An object is moving at a constant velocity. All but one of the following statements could be true. Which one cannot be true? • (a) No forces act on the object. • (b) A single force acts on the object. • (c) Two forces act simultaneously on the object. • (d) Three forces act simultaneously on the object. Question • A cup of coffee is sitting on a table in a recreational vehicle (RV). The cup slides toward the rear of the RV. According to Newton’s first law, which one or more of the following statements could describe the motion of the RV? (A) The RV is at rest, and the driver suddenly accelerates. (B) The RV is moving forward, and the driver suddenly accelerates. (C) The RV is moving backward, and the driver suddenly hits the brakes. • (a) A (b) B (c) C (d) A and B (e) A, B, and C Good Example • The mechanism for locking a seat belt? Question • When we jump into the air why doesn’t the wall come crashing into us? • (Earths Rotation ~ 1000 mph) Mass vs Weight • • • • Mass is NOT Weight Mass is the quantity of matter in an object Weight is the force of gravity on an object They are proportional, Not the same! Weight • Weight = mass x gravity • w = m۰g Question • Does a 2 kg rock have twice as much mass of a 1 kg rock? Twice the inertia? Twice the weight (when weighed in the same location)? • Yes, Yes, Yes Question • Does a liter of molten lead have the same volume as a liter of apple juice? Does sit have the same mass? • Yes, no Question • Why do physicists say mass is more fundamental than weight? • Weight changes based on location, mass doesn’t Question • An elephant and a mouse would both have zero weight in gravity-free space. If they were moving toward you with the same speed would they bump into you with the same force? • No, the elephant is harder to stop, more inertia. Objective • We will define Newton’s 2nd Law and solve problems using the equation F = ma. • We will be able to describe the relationships between forces for 1 or more objects using Newton’s 3rd Law 4.3 Newton’s Second Law of Motion Newton’s Second Law When a net external force acts on an object of mass m, the acceleration that results is directly proportional to the net force and has a magnitude that is inversely proportional to the mass. The direction of the acceleration is the same as the direction of the net force. Demo • Spool of thread Equation • The equation for Force FNet ma FNet a m 4.3 Newton’s Second Law of Motion SI Unit for Force m kg m kg 2 2 N s s This combination of units is called a newton (N). 4.3.1. A net force F is required to give an object with mass m an acceleration a . If a net force 6F is applied to an object with mass 2m, what is the acceleration on this object? a) a b) 2a c) 3a d) 4a e) 6a Question • The same force is applied to different objects. Object 1 has a mass m, object 2 has a mass m/3. Object 2’s acceleration will be ______ times more than object 1. a) 1/3 b) 1/9 c) 3 d) 6 e) 9 4.3 Newton’s Second Law of Motion A free-body-diagram is a diagram that represents the object and the forces that act on it. 4.3 Newton’s Second Law of Motion The net force in this case is: 275 N + 395 N – 560 N = +110 N and is directed along the + x axis of the coordinate system. 4.3 Newton’s Second Law of Motion If the mass of the car is 1850 kg then, by Newton’s second law, the acceleration is F a Net m 110N 2 0.059m s 1850 kg 4.4 The Vector Nature of Newton’s Second Law The direction of force and acceleration vectors can be taken into account by using x and y components. FNet ma is equivalent to F y may F x max 4.5 Newton’s Third Law of Motion Newton’s Third Law of Motion Whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body. Question? • How come I cannot push myself? • If I push on myself with a Force (F), then my body pushes back with a Force (-F). The forces cancel out Question • If I push a book with a force of 20N, and it pushes back with 20 N, then how come the book moves? Why don’t I move? • There are 2 objects. 20 N is enough to overcome friction for the book, but not enough for me. Question? • What pushes a car forward? • The ground does. The car pushes back on the ground. The ground pushes forward on the car. Question? • Scenario: You are on a sail boat and there is no wind. However you have a big industrial fan. What do you have to do to move forward? • Use the fan to blow out of the boat, not on the sail. The fan will cause a force on the air, the air will cause a force on the boat. Question • Explain how a rocket in space works if there is no air to push on? • Rocket pushes on the exhaust, the exhaust pushes on the rocket. Question • A man on a bike and a man in a truck have a head on collision. Which person experiences the greater force? The greater acceleration? • Same force for both. Smaller mass = greater acceleration (bike) Question • Does the Earth pull down on a ball, or does a ball pull up on the Earth? • Both. However, the Earth is much larger and experiences a miniscule acceleration. 4.5 Newton’s Third Law of Motion 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? 4.5 Newton’s Third Law of Motion On thespacecraft F P. On theastronaut F P. P 36 N 2 as 0.0033m s ms 11,000kg P 36 N 2 aA 0.39 m s mA 92 kg Objective • We will be able to draw free body diagrams that include the force due to gravity, friction, tension and the normal. • We will be able to solve problems that involve friction, tension, gravity and the normal forces. 4.6 Types of Forces: An Overview In nature there are two general types of forces, fundamental and nonfundamental. Fundamental Forces 1. Gravitational force 2. Strong Nuclear force 3. Electroweak force 4.6 Types of Forces: An Overview Examples of nonfundamental forces: friction tension in a rope normal or support forces 4.7 The Gravitational Force 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. 4.7 The Gravitational Force For two particles that have masses m1 and m2 and are separated by a distance r, the force has a magnitude given by m1m2 F G 2 r G 6.6731011 N m2 kg2 4.7 The Gravitational Force m1m2 F G 2 r 11 6.67 10 8 1.4 10 N N m kg 2 2 12 kg25 kg 2 1.2 m 4.7 The Gravitational Force 4.7 The Gravitational Force 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. SI Unit of Weight: newton (N) 4.7 The Gravitational Force Relation Between Mass and Weight W G M Em r W mg ME g G 2 r 2 4.7 The Gravitational Force On the earth’s surface: ME g G 2 RE 11 6.67 10 9.80 m s 2 Nm 5.9810 kg kg 6.3810 m 24 2 2 6 2 Objective • We will be able to draw free body diagrams that include the force due to gravity, friction, tension and the normal. • We will be able to solve problems that involve friction, tension, gravity and the normal forces. 4.8 The Normal Force Definition of the Normal Force 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. It is a compression force 4.8 The Normal Force FN 11 N 15 N 0 FN 26 N FN 11 N 15 N 0 FN 4 N 4.8 The Normal Force If the person is standing on a nonaccelerating surface, the normal force is simply the opposite of the gravitation force (weight) F y 0N FN mg FN mg 4.8 The Normal Force Apparent Weight The apparent weight of an object is the reading of the scale. It is equal to the normal force the man exerts on the scale. 4.8 The Normal Force We use this equation if the person or object is on an accelerating platform. F y FN mg ma FN mg ma apparent weight true weight 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface there is a force acting on that object. The component of this force that is parallel to the surface is called the frictional force. Friction is always directed opposite the direction of motion 4.9 Static and Kinetic Frictional Forces When the two surfaces are not sliding across one another the friction is called static friction. 4.9 Static and Kinetic Frictional Forces The magnitude of the static frictional force can have any value from zero up to a maximum value. MAX s s f f f MAX s 0 s 1 s FN is called the coefficient of static friction. 4.9 Static and Kinetic Frictional Forces Note that the magnitude of the frictional force does not depend on the contact area of the surfaces. 4.9 Static and Kinetic Frictional Forces Static friction opposes the impending relative motion between two objects. Kinetic friction opposes the relative sliding motion motions that actually does occur. f k k FN 0 s 1 is called the coefficient of kinetic friction. Question • Question: A person has a choice of either pushing or pulling a sled at a constant velocity, as the drawing illustrates. Friction is present. If the angle is the same in both cases, does it require more or less force to push or pull the sled? • To pull, because the upward component of the pulling force reduces the normal force therefore reducing friction 4.9 Static and Kinetic Frictional Forces 4.9 Static and Kinetic Frictional Forces The sled comes to a halt because the kinetic frictional force opposes its motion and causes the sled to slow down. 4.9 Static and Kinetic Frictional Forces Suppose the coefficient of kinetic friction is 0.05 and the total mass is 40kg. What is the kinetic frictional force? f k k FN k m g 0.0540kg 9.80 m s 20kg 2 4.10 The Tension Force Cables and ropes transmit forces through tension. Tension is a stretching force 4.10 The Tension Force A massless rope will transmit tension undiminished from one end to the other. If the rope passes around a massless, frictionless pulley, the tension will be transmitted to the other end of the rope undiminished. Question • Question: A rope is used in a tug of war between two teams of 5 people each. Both teams are equally strong, so neither team wins. An identical rope is tied to a tree, and the same 10 people just as hard on the loose end as they did in the contest. In both cases, the people pull steadily with no jerking. Which rope sustains the greater tension, a) the rope tied to the tree or b) the rope in the tug of war or c) do the ropes sustain the same tension. When the angle from vertical increases, what happens to the tension force? • Tension always increases as the angle away from vertical increases 4.11 Equilibrium Application of Newton’s Laws of Motion Definition of Equilibrium An object is in equilibrium when it has zero acceleration. F x 0 Fy 0 4.11 Equilibrium Application of Newton’s Laws of Motion Reasoning Strategy • Select an object(s) to which the equations of equilibrium are to be applied. • Draw a free-body diagram for each object chosen above. Include only forces acting on the object, not forces the object exerts on its environment. • Choose a set of x, y axes for each object and resolve all forces in the free-body diagram into components that point along these axes. • Apply the equations and solve for the unknown quantities. 4.11 Equilibrium Application of Newton’s Laws of Motion T1 sin 35 T2 sin 35 0 T1 cos35 T2 cos35 F 0 4.11 Equilibrium Application of Newton’s Laws of Motion 4.11 Equilibrium Application of Newton’s Laws of Motion Force T1 T2 W x component y component T1 sin 10.0 T1 cos10.0 T2 sin 80.0 T2 cos80.0 0 W W 3150 N 4.11 Equilibrium Application of Newton’s Laws of Motion F Fx T1 sin 10.0 T2 sin 80.0 0 y T1 cos 10 .0 T2 cos 80 .0 W 0 The first equation gives sin 80 .0 T T1 2 sin 10 .0 Substitution into the second gives sin 80.0 T cos10.0 T2 cos80.0 W 0 2 sin 10.0 4.11 Equilibrium Application of Newton’s Laws of Motion T2 W sin 80.0 cos 10 . 0 cos 80 . 0 sin 10 . 0 T2 582 N T1 3.3010 N 3 4.12 Nonequilibrium Application of Newton’s Laws of Motion When an object is accelerating, it is not in equilibrium. F x ma x Fy ma y 4.12 Nonequilibrium Application of Newton’s Laws of Motion The acceleration is along the x axis so ay 0 4.12 Nonequilibrium Application of Newton’s Laws of Motion Force T1 T2 D R x component y component T1 cos30.0 T1 sin 30.0 T2 cos30.0 T2 sin 30.0 D 0 R 0 4.12 Nonequilibrium Application of Newton’s Laws of Motion F y T1 sin 30.0 T2 sin 30.0 0 T1 T2 F x m ax T1 cos30.0 T2 cos30.0 D R 4.12 Nonequilibrium Application of Newton’s Laws of Motion T1 T2 T ma x R D 5 T 1 . 53 10 N 2 cos 30.0