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Transcript
Newton’s Laws of Motion Chapter 4 Newton’s First Law of Motion Force A force is a push or pull. An object at rest needs a force to get it moving; a moving object needs a force to change its velocity. Force Force is a vector, having both magnitude and direction. The magnitude of a force can be measured using a spring scale. Newton’s First Law of Motion This is Newton’s first law, which is often called the law of inertia: Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it. Newton’s First Law of Motion It may seem as though it takes a force to keep an object moving. Example: Push your book across a table—when you stop pushing, it stops moving. Why? Your book stops because the force of friction stops it. It doesn’t take a force to keep an object moving in a straight line—it takes a force to change its motion. ConcepTest 4.1b Newton’s First Law II A hockey puck slides on ice at constant velocity. What is the net force acting on the puck? 1) more than its weight 2) equal to its weight 3) less than its weight but more than zero 4) depends on the speed of the puck 5) zero ConcepTest 4.1b Newton’s First Law II A hockey puck slides on ice at constant velocity. What is the net force acting on the puck? 1) more than its weight 2) equal to its weight 3) less than its weight but more than zero 4) depends on the speed of the puck 5) zero The puck is moving at a constant velocity, and therefore it is not accelerating. Thus, there must be no net force acting on the puck. ConcepTest 4.1c Newton’s First Law III You put your book on the bus seat next to you. When the bus stops suddenly, the book slides forward off the seat. Why? 1) a net force acted on it 2) no net force acted on it 3) it remained at rest 4) it did not move, but only seemed to 5) gravity briefly stopped acting on it ConcepTest 4.1c Newton’s First Law III You put your book on the bus seat next to you. When the bus stops suddenly, the book slides forward off the seat. Why? 1) a net force acted on it 2) no net force acted on it 3) it remained at rest 4) it did not move, but only seemed to 5) gravity briefly stopped acting on it The book was initially moving forward (because it was on a moving bus). When the bus stopped, the book continued moving forward, which was its initial state of motion, and therefore it slid forward off the seat. Follow-up: What is the force that usually keeps the book on the seat? ConcepTest 4.4a Off to the Races I From rest, we step on the gas of our Ferrari, providing a force F for 4 secs, speeding it up to a final speed v. If the applied force were only ½F, how long would it have to be applied to reach the same final speed? 1) 16 s 2) 8 s 3) 4 s 4) 2 s 5) 1 s F v ConcepTest 4.4a Off to the Races I From rest, we step on the gas of our Ferrari, providing a force F for 4 secs, speeding it up to a final speed v. If the applied force were only ½F, how long would it have to be applied to reach the same final speed? 1) 16 s 2) 8 s 3) 4 s 4) 2 s 5) 1 s In the first case, the acceleration acts over time T = 4 s to give velocity v = aT. In the second case, the force is half, therefore the acceleration is also half, so to achieve the same final speed, the time must be doubled. F v Mass Mass is the measure of inertia of an object. In the SI system, mass is measured in kilograms. Mass is not weight! Mass is a property of an object. Weight is the force exerted on that object by gravity. Example: If you go to the Moon, whose gravitational acceleration is about 1/6 g, you will weigh much less. Your mass, however, will be the same. Newton’s Second Law of Motion Newton’s Second Law of Motion Newton’s second law is the relation between acceleration and force. Acceleration is proportional to force and inversely proportional to mass. It takes a force to change either the direction or the speed of an object. More force means more acceleration; the same force exerted on a more massive object will yield less acceleration. Newton’s Second Law of Motion Since force is a vector, so F ma is true along each coordinate axis. The unit of force in the SI system is the newton (N). Note that the pound is a unit of force, not of mass, and can therefore be equated to Newtons but not to kilograms. ConcepTest 4.3 Truck on Frozen Lake A very large truck sits on a frozen lake. Assume there is no friction between the tires and the ice. A fly suddenly smashes against the front window. What will happen to the truck? 1) it is too heavy, so it just sits there 2) it moves backward at constant speed 3) it accelerates backward 4) it moves forward at constant speed 5) it accelerates forward ConcepTest 4.3 Truck on Frozen Lake A very large truck sits on a frozen lake. Assume there is no friction between the tires and the ice. A fly suddenly smashes against the front window. What will happen to the truck? 1) it is too heavy, so it just sits there 2) it moves backward at constant speed 3) it accelerates backward 4) it moves forward at constant speed 5) it accelerates forward When the fly hit the truck, it exerted a force on the truck (only for a fraction of a second). So, in this time period, the truck accelerated (backward) up to some speed. After the fly was squashed, it no longer exerted a force, and the truck simply continued moving at constant speed. ConcepTest 4.8 On the Moon An astronaut on Earth kicks a bowling ball and hurts his foot. A year later, the same astronaut kicks a bowling 1) more 2) less 3) the same ball on the Moon with the same force. His foot hurts... Ouch! ConcepTest 4.8 On the Moon An astronaut on Earth kicks a bowling ball and hurts his foot. A year later, the same astronaut kicks a bowling 1) more 2) less 3) the same ball on the Moon with the same force. His foot hurts... The masses of both the bowling ball and the astronaut remain the same, so his foot feels the same resistance and hurts the same as before. Follow-up: What is different about the bowling ball on the Moon? Ouch! ConcepTest 4.5 Force and Mass A force F acts on mass M for a time interval T, giving it a final speed v. If the same force acts for the same time on a different mass 2M, what would be the final speed of the bigger mass? 1) 4v 2) 2v 3) v 4) 1 2 v 5) 1 4 v ConcepTest 4.5 Force and Mass A force F acts on mass M for a time interval T, giving it a final speed v. If the same force acts for the same time on a different mass 2M, what would be the final speed of the bigger mass? 1) 4v 2) 2v 3) v 4) 1 2 v 5) 1 4 v In the first case, the acceleration acts over time T to give velocity v = aT. In the second case, the mass is doubled, so the acceleration is cut in half; therefore, in the same time T, the final speed will only be half as much. Follow-up: What would you have to do to get 2M to reach speed v ? ConcepTest 4.6 Force and Two Masses A force F acts on mass m1 giving acceleration a1. The same force acts on a different mass m2 giving acceleration a2 = 2a1. If m1 and m2 are glued together and the same force F acts on this combination, what is the resulting acceleration? F F F m1 a1 m2 m2 m1 a2 = 2a1 a 3 3 4 3 2 a1 3) 1 2 a1 4) 4 3 a1 5) 2 3 a1 1) 2) a1 ConcepTest 4.6 Force and Two Masses A force F acts on mass m1 giving acceleration a1. The same force acts on a different mass m2 giving acceleration a2 = 2a1. If m1 and m2 are glued together and the same force F acts on this combination, what is the resulting acceleration? F m1 3 4 3 2 a1 3) 1 2 a1 4) 4 3 a1 5) 2 3 a1 1) 2) a1 a1 F = m1 a1 a2 = 2a1 m2 F F = m2 a2 = (1/2 m1 )(2a1 ) Mass m2 must be ( 1 m1) because its 2 acceleration was 2a1 with the same force. Adding the two masses together gives ( 32 )m1, leading to an F m2 m1 a3 acceleration of ( 32 )a1 for the same applied force. F = (3/2)m1 a3 => a3 = (2/3) a1 Newton’s Third Law of Motion 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 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. Newton’s Third Law of Motion Helpful notation: the first subscript is the object that the force is being exerted on; the second is the source. Newton’s Third Law of Motion Conceptual Example 4-5: Third law clarification. Michelangelo’s assistant has been assigned the task of moving a block of marble using a sled. He says to his boss, “When I exert a forward force on the sled, the sled exerts an equal and opposite force backward. So how can I ever start it moving? No matter how hard I pull, the backward reaction force always equals my forward force, so the net force must be zero. I’ll never be able to move this load.” Is he correct? Weight Weight is the force exerted on an object by gravity. Close to the surface of the Earth, where the gravitational force is nearly constant, the weight of an object of mass m is: where The Normal Force An object at rest must have no net force on it. If it is sitting on a table, the force of gravity is still there; what other force is there? The force exerted perpendicular to a surface is called the normal force. It is exactly as large as needed to balance the force from the object. (If the required force gets too big, something breaks!) Using Free-Body Diagrams with Newton’s Laws 1. Draw a sketch. 2. For one object, draw a free-body diagram, showing all the forces acting on the object. Make the magnitudes and directions as accurate as you can. Label each force. If there are multiple objects, draw a separate diagram for each one. 3. Resolve vectors into components. 4. Apply Newton’s second law to each component. 5. Solve. Problem Solving—A General Approach 1. Read the problem carefully; then read it again. 2. Draw a sketch, and then a free-body diagram. 3. Choose a convenient coordinate system. 4. List the known and unknown quantities; find relationships between the knowns and the unknowns. 5. Estimate the answer. 6. Solve the problem without putting in any numbers (algebraically); once you are satisfied, put the numbers in. 7. Keep track of dimensions. 8. Make sure your answer is reasonable. ConcepTest 4.9a Going Up I A block of mass m rests on the floor 1) N > mg of an elevator that is moving upward 2) N = mg at constant speed. What is the 3) N < mg (but not zero) relationship between the force due 4) N = 0 to gravity and the normal force on 5) depends on the size of the elevator the block? v m ConcepTest 4.9a Going Up I A block of mass m rests on the floor 1) N > mg of an elevator that is moving upward 2) N = mg at constant speed. What is the 3) N < mg (but not zero) relationship between the force due 4) N = 0 to gravity and the normal force on 5) depends on the size of the elevator the block? The block is moving at constant speed, so it must have no net force on it. The forces v on it are N (up) and mg (down), so N = mg, just like the block at rest on a table. m ConcepTest 4.9b Going Up II A block of mass m rests on the floor of an elevator that is accelerating upward. What is the relationship between the force due to gravity and the normal force on the block? 1) N > mg 2) N = mg 3) N < mg (but not zero) 4) N = 0 5) depends on the size of the elevator a m ConcepTest 4.9b Going Up II A block of mass m rests on the floor of an elevator that is accelerating upward. What is the relationship between the force due to gravity and the normal force on the block? 1) N > mg 2) N = mg 3) N < mg (but not zero) 4) N = 0 5) depends on the size of the elevator The block is accelerating upward, so it must have a net upward force. The forces on it are N (up) and mg (down), so N must be greater than mg in order to give the net upward force! Follow-up: What is the normal force if the elevator is in free fall downward? N m a>0 mg S F = N – mg = ma > 0 \ N > mg ConcepTest 4.10 Normal Force Below you see two cases: a physics student pulling or pushing a sled with a force F that is applied at an angle q. In which case is the normal force greater? 1) case 1 2) case 2 3) it’s the same for both 4) depends on the magnitude of the force F 5) depends on the ice surface Case 1 Case 2 ConcepTest 4.10 Normal Force Below you see two cases: a physics student pulling or pushing a sled with a force F that is applied at an angle q. In which case is the normal force greater? 1) case 1 2) case 2 3) it’s the same for both 4) depends on the magnitude of the force F 5) depends on the ice surface Case 1 In case 1, the force F is pushing down (in addition to mg), so the normal force needs to be larger. In case 2, the force F is pulling up, against gravity, so the normal force is lessened. Case 2 ConcepTest 4.12 Climbing the Rope When you climb up a rope, 1) this slows your initial velocity, which is already upward the first thing you do is pull 2) you don’t go up, you’re too heavy down on the rope. How do 3) you’re not really pulling down—it just seems that way you manage to go up the rope by doing that?? 4) the rope actually pulls you up 5) you are pulling the ceiling down ConcepTest 4.12 Climbing the Rope When you climb up a rope, 1) this slows your initial velocity, which is already upward the first thing you do is pull 2) you don’t go up, you’re too heavy down on the rope. How do 3) you’re not really pulling down—it just seems that way you manage to go up the rope by doing that?? 4) the rope actually pulls you up 5) you are pulling the ceiling down When you pull down on the rope, the rope pulls up on you!! It is actually this upward force by the rope that makes you move up! This is the “reaction” force (by the rope on you) to the force that you exerted on the rope. And voilá, this is Newton’s Third Law. ConcepTest 4.16a Tension I You tie a rope to a tree and you pull on the rope with a force of 100 N. What is the tension in the rope? 1) 0 N 2) 50 N 3) 100 N 4) 150 N 5) 200 N ConcepTest 4.16a Tension I You tie a rope to a tree and you pull on the rope with a force of 100 N. What is the tension in the rope? 1) 0 N 2) 50 N 3) 100 N 4) 150 N 5) 200 N The tension in the rope is the force that the rope “feels” across any section of it (or that you would feel if you replaced a piece of the rope). Because you are pulling with a force of 100 N, that is the tension in the rope. ConcepTest 4.16b Tension II Two tug-of-war opponents each 1) 0 N pull with a force of 100 N on 2) 50 N opposite ends of a rope. What is 3) 100 N the tension in the rope? 4) 150 N 5) 200 N ConcepTest 4.16b Tension II Two tug-of-war opponents each 1) 0 N pull with a force of 100 N on 2) 50 N opposite ends of a rope. What is 3) 100 N the tension in the rope? 4) 150 N 5) 200 N This is literally the identical situation to the previous question. The tension is not 200 N !! Whether the other end of the rope is pulled by a person, or pulled by a tree, the tension in the rope is still 100 N !! ConcepTest 4.16c Tension III You and a friend can 1) you and your friend each pull on opposite ends of the rope each pull with a force 2) tie the rope to a tree, and you both pull from the same end of 20 N. If you want to rip a rope in half, 3) it doesn’t matter—both of the above are equivalent what is the best way? 4) get a large dog to bite the rope ConcepTest 4.16c Tension III You and a friend can each pull with a force of 20 N. If you want to rip a rope in half, what is the best way? 1) you and your friend each pull on opposite ends of the rope 2) tie the rope to a tree, and you both pull from the same end 3) it doesn’t matter—both of the above are equivalent 4) get a large dog to bite the rope Take advantage of the fact that the tree can pull with almost any force (until it falls down, that is!). You and your friend should team up on one end, and let the tree make the effort on the other end. ConcepTest 4.14a Collision Course I 1) the car A small car collides with a large truck. Which experiences the greater impact force? 2) the truck 3) both the same 4) it depends on the velocity of each 5) it depends on the mass of each ConcepTest 4.14a Collision Course I 1) the car A small car collides with a large truck. Which experiences the greater impact force? 2) the truck 3) both the same 4) it depends on the velocity of each 5) it depends on the mass of each According to Newton’s Third Law, both vehicles experience the same magnitude of force. ConcepTest 4.14b Collision Course II In the collision between the car and the truck, which has the greater acceleration? 1) the car 2) the truck 3) both the same 4) it depends on the velocity of each 5) it depends on the mass of each ConcepTest 4.14b Collision Course II In the collision between the car and the truck, which has the greater acceleration? We have seen that both vehicles experience the same magnitude of force. But the acceleration is given by F/m so the car has the larger acceleration, because it has the smaller mass. 1) the car 2) the truck 3) both the same 4) it depends on the velocity of each 5) it depends on the mass of each ConcepTest 4.15a Contact Force I If you push with force F on either the heavy box (m1) or the light box (m2), in which of the two cases below is the contact force between the two boxes larger? 1) case A 2) case B 3) same in both cases A m2 F m1 B m2 m1 F ConcepTest 4.15a Contact Force I If you push with force F on either the heavy box (m1) or the light box (m2), in which of the two cases below is the contact force between the two boxes larger? 1) case A 2) case B 3) same in both cases The acceleration of both masses together A is the same in either case. But the contact force is the only force that accelerates m1 in case A (or m2 in case B). Because m1 is m2 F m1 the larger mass, it requires the larger B contact force to achieve the same acceleration. m2 m1 F ConcepTest 4.15b Contact Force II Two blocks of masses 2m and m are in contact on a horizontal frictionless surface. If a force F is applied to mass 2m, what is the force on mass m ? 1) 2F 2) F 3) 4) 5) 1 2F 1 F 3 1 F 4 F 2m m ConcepTest 4.15b Contact Force II Two blocks of masses 2m and m are in contact on a horizontal frictionless surface. If a force F is applied to mass 2m, what is the force on mass m ? 1) 2F 2) F 3) 4) 5) 1 2F 1 F 3 1 F 4 The force F leads to a specific acceleration of the entire system. In order for mass m to accelerate at the same rate, the force on it must be smaller! How small?? Let’s see... F 2m m ConcepTest 4.11 On an Incline Consider two identical blocks, one resting on a flat surface and the other resting on an incline. For which case is the normal force greater? 1) case A 2) case B 3) both the same (N = mg) 4) both the same (0 < N < mg) 5) both the same (N = 0) ConcepTest 4.11 On an Incline Consider two identical blocks, one resting on a flat surface and the other resting on an incline. For which case is the normal force greater? 1) case A 2) case B 3) both the same (N = mg) 4) both the same (0 < N < mg) 5) both the same (N = 0) In case A, we know that N = W. y In case B, due to the angle of the incline, N < W. In fact, we N f can see that N = W cos(q). q q W Wy x Motion with Variable Acceleration Motion with Constant Acceleration If an object is moving with constant acceleration a, then: v(t ) adt v(t ) v0 at x(t ) v(t )dt x(t ) x0 v0t at 1 2 2 Variable Acceleration If acceleration changes as a function of time, the previous definitions still apply, but the resulting equations of motion may look…well…nasty! Example: Suppose an object has an acceleration a(t ) Ae .t Find the objects position function. A t t v(t ) a(t )dt A e dt v(t ) e b A t x(t ) v(t )dt ( e b)dt A t x(t ) 2 e bt c Velocity-Dependent Forces: Drag When an object moves through a fluid, it experiences a drag force that depends on the velocity of the object. For small velocities, the force is approximately proportional to the velocity; for higher speeds, the force is approximately proportional to the square of the velocity. Terminal Velocity If the drag force on a falling object is dependent on its velocity, the object gradually slows until the drag force and the gravitational force are equal. Then it falls with constant velocity, called the terminal velocity. Problem 4.17 – Can cars “stop on a dime”? Calculate the acceleration of a 1400-kg car if it can stop from 35 km/h on a dime (diameter = 1.7 cm). How many g’s is this? What is the force felt by the 68-kg driver? First and foremost, we need to convert all given quantities to SI units. 35 km/h ≈ 9.7222222 m/s; 1.7 cm = 0.017 m Problem 4.17 (cont.) We were not told that the acceleration varies, so we will assume it is constant. Since we are assuming constant acceleration, we can actually use the projectile motion equation to solve the problem. Since we are not given time, but rather a distance, we can use the equation: v 2 v02 2a( x x0 ) Solving for acceleration, we get: v 2 v02 a 2( x x0 ) Problem 4.17 (cont.) Now we just substitute our given values, and find that: 3 2 a 2.8 10 m/s Dividing that result by 9.80 m/s2, we see that: a 2.8 10 g ' s 2 So, what is the force felt by the driver? Using Newton’s second law, we learn that the driver experienced 1.9105 N. Question 4.24 – Bear Wants my Lunch Question 4.24 – Bear Wants my Lunch Here we are not concerned with the acceleration of the backpack, but rather how much force does the person need to apply to keep the backpack at a certain height. y T1 T2 q q x w We are not told where the back pack is along the rope, or what angle each side of the rope makes with the horizontal, so we are going to assume the back pack is midway along the rope, and that the angle of the rope is the same on both sides. Question 4.24 – Bear Wants my Lunch We always assume frictionless conditions (unless otherwise stated), so the branch through which the men is pulling the rope lets the rope slide freely. Furthermore, we always assume that the rope does not stretch, so the magnitude of the tension on the rope does not change as it bends around the branch. Therefore: | T1 || F | Also, since it is the same rope all through out: | T2 || T1 || F | Question 4.24 – Bear Wants my Lunch Before we apply Newton’s laws to this problem, it is a good idea to write down all vectors as the sum of their components. T1 F cos q ˆi F sin q ˆj T2 F cos q ˆi F sin q ˆj w mg ˆj Now we can apply Newton’s 2nd law to the vertical components to get our answer: F y ma 2F sin q mg 0 mg F 2 sin q Question 4.24 – Bear Wants my Lunch mg F 2 sin q Notice that the angle q appears in the denominator. As the person lifts the backpack higher, the angle q becomes smaller, which means that sin(q ) becomes smaller, which in turn makes F larger. Therefore, the force needed to raise the backpack increases as the backpack gets higher, and in fact, it is impossible to pull it hard enough so that it doesn’t sag at all.