Dynamics (Newton s Laws) - The Causes of Motion

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AP Physics Dynamics (Newton s Laws) - The Causes of Motion Introduction: This unit introduces the most basic of all scientific concepts, the interaction between forces and matter. You should understand from the outset that any time two or more objects interact, the interaction causes a force on each object. There are no exceptions. In your earlier science courses, a force was defined as a push or a pull. This is correct but now you need to understand that a force on a body is caused by an interaction between bodies. We isolate and then study the forces acting on one body without regard to the origin of the forces. This gets confusing for the beginning physics student. You need to spend a lot of time thinking about, sorting through, and organizing in your mind the information learned in this unit. Sir Isaac Newton had to go through the same process in developing his laws of motion. You have an advantage you have over three hundred years of the experiences of all of the physics to follow. Force is a vector quantity; it always has direction and magnitude. A force applied to a body can do two things; it can alter the dimension or shape of the body or it can alter the state of motion of the body. Natural forces known to scientists are gravitational, electromagnetic, and nuclear forces. Dynamics is the study of forces which cause motion. Performance Objectives: Upon completion of the readings and activities of this unit and when asked to respond either orally or on a written test, you will: State the first law of motion and display a clear understanding of its universality and implications. Give examples of what happens to an object when no external net force acts on it. Use the words force, mass, weight, and inertia in their correct scientific meanings. State the second law of motion, and display a clear understanding of its universality and implications. Be able to apply the law to determine the results of forces. Solve problems involving this law. Understand the rationale behind the definition of the Newton. Recognize the relationship between the second law and the unit of force. Distinguish between weight and mass. Explain the nature of weight as a force. Use the second law to determine the mass. Demonstrate an understanding of the meaning of new force. Use the concept of net force to solve problems. State the observations regarding sliding friction. Compute the coefficient of friction. Solve problems involving frictional forces. State the third law, and display a clear understanding of its universality and implications. Distinguish between forces applied to a body and forces being applied by the body. Be able to isolate bodies to solve problems. Incorporate the problems solving techniques learned in this unit with ideas learned in the kinematics and the vector units. Textbook Reference: Physics For Scientists and Engineers: Chapter 4 In the beginning there was Aristotle. And objects at rest tended to remain at rest. And objects in motion tended to come to rest. And soon everything was at rest and God saw that boring. Then God created Newton. - Dr. William Baker, President Bell Laboratories (1978) All forces result from interactions. There are no exceptions. When two bodies interact, each body exerts on the other a force. The two forces are called and action-reaction pair. The forces of an action-reaction pair are (1) equal in magnitude, (2) opposite in direction, (3) act on two different bodies and (4) do not cancel each other.

Introductory Questions: 1. When you push against the wall, does it push against you? How can you tell? 2. Your weight is the result of a gravitational force of the earth on your body. What is the corresponding reaction force? 3. As you stand on the floor, does the floor exert an upward force against your feet? How much force does it exert? 4. Suppose a brick is suspended from a rigid support by a suitable length of cord. a.) What downward force acts on the brick? b.) If this force is the action force, what is the corresponding reaction force? 5. a.) What upward force acts on the suspended brick in Question 4? b.) If this force is the action force, what is the corresponding reaction force? For most of this unit, we will isolate one body and look at the forces acting on that one body. The forces we will discuss in this unit are called contact forces. The interacting bodies will be in contact with each other. The only exception will be the force of gravity which is the result of a body interacting with the earth s gravitational field. We will study field forces later in this course. Newton s First Law of Motion: An object at rest remains at rest, and an object in straight line motion continues in motion with constant velocity (same speed and same direction) unless acted upon by a net external force. This is also known as the Law of Inertia. 8. In terms of inertia, what is the disadvantage of a lightweight camera when snapping the shutter? Why do most photographers prefer a massive tripod? 9. In tearing a paper towel or plastic bag from a roll, why is a sharp jerk more effective than a slow pull? 10. Why will the coin drop into the glass when a force accelerates the card? See diagram 11. Why does the downward motion and sudden stop of the hammer tighten the hammerhead? 12. Why is it that a slow continuous increase in the downward force breaks the string above the massive ball, but a sudden increase breaks the lower string? 13. An astronaut in space has a weightless anvil. Is it more difficult, less difficult or just as difficult to shake the anvil back and forth in space as it is on earth? 10. 11. 12. 13. Every body preserves in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed there on. - Isaac Newton, Principia Mathematica Conceptual Questions - First Law: 6. A ball is rolled across the top of a table and slowly comes to a stop. Considering Newton s first law of motion, explain why the ball stops. How could the ball have remained in motion? 7. Why do you fall backward on a bus when it accelerates from rest? Who do you fall forward when the driver decelerates to rest? 14. Many automobile passengers suffer neck injuries when struck by cars from behind. How does Newton s law of inertia apply here? How do headrests help guard against this type of injury? 15. Most car ads now include mileage ratings, on for highway driving and one for city driving. Why is the city driving rating always less than the highway driving rating?

16. If a ball moving with a velocity of 20 cm/s has no net force act on it, its velocity after 5.0 s will be? Newton s Second Law of Motion: The acceleration of an object is directly proportional to the net external force acting on the object and inversely proportional to the mass of the object on which the net external force acts. The acceleration is always in the direction of the net force. This is also known as the Law of Acceleration. Second Law Exercises: 17. A net force gives a 2 kg mass an acceleration of 5.0 m/s 2. What is the magnitude of the force? 10 N 18. A net force of 30.0 N gives a stone an acceleration of 4.0 m/s 2. What is the mass of the stone? 7.5 kg 19. A net force of 25 N is applied to a 2.0 kg mass. What is the acceleration of the mass? 12.5 m/s 2 20. What net force gives a 1.0 kg mass an acceleration of 9.8 m/s 2? 21. Determine the weight of a 4.8 kg mass. 47 N 22. A small yacht weighs 4900 N. What is its mass in kilograms? 500 kg 23. A car has a mass of 1200 kg. What is the weight of the car? What net external force must be applied to the car to accelerate the car along a level highway at the rate of 4.0 m/s 2 (neglecting friction)? What acceleration would this force produce if a 750 N frictional force were present? 11,760 N 4800 N 3.375 m/s 2 More Second Law Problems, But This Time The Net Force Is Not Given: The equation net applied opposing is a vector equation. Keep in mind that this is a vector equation and you must keep track of the directions. The applied force is in the direction of the motion and the opposing force is any force acting on the body in a direction opposite to the motion. However, if we rewrite this equation for only one dimension, the direction is indicated with a plus or a minus sign. Usually we will look at the sum of the components in the x and y directions. Recall from your study of vectors that all the vectors in the x- direction can be added algebraically by choosing one direction to be positive and the opposite direction negative. The equation is then written as F net = F applied F opposing. Commit this equation to memory. Say it to yourself each time you solve a problem. When solving problems, choose a direction in which you think the object will move and consider that the positive direction. Consider all the forces in that direction applied forces. The opposing forces are any forces acting on the body in a direction opposite to the motion. When moving bodies are in contact and sliding in opposite directions relative to each other, there is an opposing force due to friction. When there is relative motion between a fluid and a body that moves through the fluid, the opposing force is called the drag force. If a body falls through air (which is a fluid), the drag force is called air resistance or air friction. If a body falls far enough through air, the air resistance eventually equals the weight of the body, so the net force and the acceleration on the body are zero. The body then falls at a constant velocity called the terminal velocity. 24. Determine the acceleration that a force of 25.0 N gives to a 4.0 kg mass. The friction force to overcome is 5.0 N. 5.0 m/s 2 25. A car located on a level highway has a mass of 400.0 kg. The friction force opposing the motion of the car is 750 N. What acceleration will an applied force of 2350 N produce on the car? 4.0 m/s 2 26. A rubber ball weighs 4.9 N. a.) What is its mass? b.) At what rate is the ball accelerated straight up if a 68.0 N force is applied to it in that direction? 0.5 kg 128 m/s 2 27. A rocket weighs 9800.0 N. a.) What is its mass? b.) What applied force gives it a vertical acceleration of 4.00 m/s 2? 1000.0 kg 13,800 N 28. What applied force accelerates a 20.0 kg stone straight up at 10.0 m/s 2? 396 N 29. A force of 90.0 N is exerted straight up on a stone that weighs 7.35 N. Calculate a.) the mass of the stone. b.) the net force acting on the stone. c.) the acceleration of the stone. 0.75 kg 82.65 N 110 m/s 2 30. A falling bowling ball has a mass of 2.0 kg. The upward force of air resistance is 11.6 N. What is the acceleration of the bowling ball? -4.0 m/s 2

31. A physics student weighing 600.0 N plans to escape a burning building by sliding down an improvised rope made of bed sheets tied together. The maximum upward force that the sheets can exert without tearing is 300.0 N. a.) Can the student slide down at constant speed? b.) What is the least acceleration with which the student can slide down the rope? -4.9 m/s 2 The Normal Force: When the two interacting bodies are in contact with each other, the forces that each exerts on the other is called the normal force. Normal is the mathematical term meaning perpendicular. In the free-body diagram, draw the normal force perpendicular to the surface in contact with the body being considered. A common misconception is that the normal force is always equal to the weight of the body. Sketch in your notes the possible configurations for a normal force that is net equal to the weight. 32. A person weighing 490.0 N stands on a scale in an elevator. a.) What does the scale read when the elevator is at rest? b.) The elevator starts to ascend and accelerates the person upward at 2.0 m/s 2. What is the reading on the scale now? c.) When the elevator reaches a desirable speed, it no longer accelerates. What is the reading on the scale as the elevator rises uniformly? d.) The elevator begins to slow down as it reaches the proper floor. Do the scale readings increase or decrease? e.) The elevator starts to descend. Does the scale reading increase or decrease? f.) What does the scale read if the elevator descends at a constant speed? g.) If the cable snapped and the elevator fell freely, what would the scale read? itself has a mass of 500.0 kg. Tensile strength tests show that the cable supporting the elevator can tolerate a maximum force of 29,600 N. What is the greatest upward acceleration that the elevator s motor can produce without breaking the cable? 5.0 m/s 2 36. The mass of an elevator plus occupants is 950.0 kg. The tension in the cable is 10,950 N. a.) At what rate is the elevator accelerated upward? b.) What is the normal force experienced by a 75 kg passenger when the elevator is experiencing the maximum upward acceleration? 1.7 m/s 2 862.5 N Conceptual Questions - Second Law: 37. If you find a body that is not moving even though we know it to be acted on by a force. What inference can we draw? 38. How does the weight of a falling body compare to the air resistance just before it reaches terminal velocity? After? 39. Suppose you place a ball in the middle of a wagon and then accelerate the wagon forward. Describe the motion of the ball relative to the ground and to the wagon. 40. Why is it that a basketball dropped from the top of the Peachtree Plaza Hotel will hit the ground at the same speed as if it were dropped from the twentieth floor? (After about 49.5 m, its terminal speed of about 20 m/s is reached). 41. What is the acceleration of a rock at the top of its trajectory when thrown straight upward? Is your answer consistent with Newton s Second Law? 33. An elevator of mass 1000.0 kg is supported by a cable that can sustain a force of 12,000.0 N. What is the greatest upward acceleration that can be given the elevator without breaking the cable? 2.2 m/s 2 34. The mass of an elevator plus its occupants is 750.0 kg. The tension in the cable is 8950 N. At what rate does the elevator accelerate upward? 2.1 m/s 2 35. Safety engineers estimate that an elevator can hold 20 persons of 75 kg average mass. The elevator

Forces at an Angle to the Motion Remember Resolution of a Vector! 42. A 20.0 kg sled is pulled along level ground. The sled s rope makes an angle of 60.0 degrees with the snow-covered ground and pulls on the sled with a force of 180 N. Find the acceleration of the sled if the friction force to be overcome is 15 N. 3.8 m/s 2 43. A 10.0 kg block is pulled up a frictionless incline that makes an angle of 53 with the horizontal. a) If the person pulling on the string can exert 48.3 N of force, what is the acceleration of the block? b) What is the magnitude and direction of the force exerted on the block by the inclined plane? 3.0 m/s 2 59 N 44. A 110 kg crate is pushed at constant speed up the frictionless 34 ramp shown. a) What horizontal force is required? b) What is the force exerted by the ramp on the crate? 730 N 1300 N Remember Those Kinematic Formulas? 45. The instruments attached to a weather balloon have a mass of 5.0 kg. a.) The balloon is released on a calm day and exerts an upward force of 89 N on the instruments. At what rate does the balloon with the instruments accelerate straight up? b.) After 10.0 seconds of acceleration, the weather balloon instruments are released automatically. What is the magnitude and direction of their velocity at the instant of their release? c.) What net force acts on the instruments after their release? d.) What time elapses before the instruments begin to fall straight down? +8.0 m/s 2 +80 m/s -49 N 8.2 s 46. An artillery shell has a mass of 8.0 kg. The shell is fired from the muzzle of a gun with a speed of 700.0 m/s. The gun barrel is 3.5 m long. What is the average force on the shell while it is in the gun barrel? 5.6 x 10 5 N 48. A car weighing 9800.0 N travels at 30.0 m/s. a.) What braking force brings it to rest in 100.0 m? b.) in 10.0m? 4500 N 45000 N 49. A rocket that weighs 7840 N on earth is fired. The force of propulsion is 10,440 N. Determine: a.) the mass of the rocket. b.) the upward acceleration of the rocket. c.) the velocity of the rocket at the end of 10 seconds. 800 kg 3.25 m/s 2 32.5 m/s 50. A 60.0 kg sled is coasting (F net = 0 N) with a constant velocity of 10.0 m/s over smooth ice. It enters a rough stretch of ice 6.0 m long in which the force of friction is 120 N. With what speed does the sled emerge from the rough stretch of ice. 8.7 m/s 51. A 75 kg paratrooper jumping out of the back of an airplane quickly acquires a velocity of 60.0 m/s as he falls toward the ground and then opens his parachute. After falling an additional 30.0 m, his velocity has been reduced to 8.0 m/s. a.) What is the acceleration of the paratrooper while his fall is being checked? b.) What is the applied force exerted by the parachute? 58.9 m/s 2 5153 N 52. A 100 kg rocket sled was moving at a constant speed v. A 5000 N force acting for 20.0 seconds opposite to the direction of its motion, slowed the sled down to one-third v. a.) What was the value of v? b.) How far did the sled move while it was slowing down? 1500 m/s 20 km 53. The driver of a 600.0 kg sports car, heading directly for a railroad crossing, 300.0 m away, applies the brakes in a panic stop. The car is moving at 40.0 m/s and the brakes can supply a force of 1500 N. a.) How fast will the car be moving when it reaches the crossing? b.) Will the driver escape collision with a freight which, at the instant the brakes are applied, is still blocking the road and still requires 11.0 s to clear the crossing? 10 m/s The driver escapes collision because it takes 12.0 seconds to reach the tracks. 47. A racing car has a mass of 700.0 kg. It starts from rest and travels 120 m in 2.0 seconds. What is the force applied to it? (Ignore friction) 4.2 x 10 4 N

Two or More Body Problems: This is analogous to the situation where two bodies were moving when you did the kinematics problems. If two bodies are moving, then you will need two equations to describe their motions. If three bodies are moving, then three equations are needed. Write an equation of motion for each moving body. Solve the equations simultaneously for the unknown variables. Although two bodies are moving, the two bodies are attached and therefore move with the same speed and the same acceleration. The two bodies are treated as one system. Usually it is necessary to determine the acceleration of the system first: F net a system = m total To find the tension in the cord connecting the bodies, isolate one body and solve for the tension using Newton s Second Law. 54. Two blocks are in contact on a frictionless horizontal table. A horizontal force is applied to one block, as shown in the figure below. a) If m 1 = 2.3 kg, m 2 = 1.2 kg and F = 3.2 N, find the force of contact between the two blocks. 1.1 N b) Show that: if the same force F is applied to m 2 rather than to m 1, that the force of contact between the blocks is 2.1 N, which is not the same value derived in (a). Explain. 55. The figure below shows three crates with masses m 1 = 45.2 kg, m 2 = 22.8 kg and m 3 = 34.3 kg on a horizontal frictionless surface. a) What horizontal force F is needed to push the crates to the right, as one unit, with an acceleration of 1.32 m/s 2? 135 N b) Find the force exerted by m 2 on m 3. 45.3 N c) By m 1 on m 2. 75.4 N 56. Three blocks are connected, as shown in the figure below, on a horizontal frictionless table and pulled to the right with a force T 3 = 6.5 N. If m 1 = 1.2 kg, m 2 = 2.4 kg and m 3 = 3.1 kg, calculate (a) the acceleration of the system and (b) the tensions T 1 and T 2. 0.97 m/s 2 T 1 = 1.2 N T 2 = 3.5 N 57. The two blocks in the figure to the right are connected by a heavy, uniform rope of mass 4 kg. An upward force of 200 N is applied as shown. a) What is the acceleration of the system? 2.7 m/s 2 b) What is the tension at the top of the heavy rope? 112.5 N c) What is the tension at the midpoint of the rope? 87.5 N 58. If the figure to the right, the frictional force between the 30 kg block and the table is negligible. a) If the peg is removed, what is the acceleration of the system? b) How long would it take the block to hit the pulley? c) What is the tension in the cord while the block is moving? d) What is the tension in the cord after the block ceases to move? 1.4 m/s 2 0.65 sec 42 N 49 N 59. Objects of mass 5.0 kg and 2.0 kg are connected by a light cord that passes over a horizontal frictionless rod. a.) What is the acceleration of the system? b.) What is the tension in the cord on the 5.0 kg side? c.) What is the tension in the cord on the 2.0 kg side? 4.2 m/s 2 b = c = 28 N 60. A cord connecting objects of mass 10.0 kg and 5.0 kg passes over a light frictionless pulley. a.) What is the acceleration of the system? b.) What is the tension in the cord? 3.27 m/s 2 65.3 N

61. Bob and Joe, two construction workers on the roof of a building, are about to raise a bucket of nails from the ground by means of a rope passing over a pulley 16 m above the ground. Bob has a mass of 100.0 kg and Joe has a mass of 80.0 kg. The bucket s mass is 40.0 kg and the mass of the nails is 80.0 kg. They slip off the roof and the following unfortunate sequence of events takes place: Bob and Joe, hanging on the same rope, strike the ground just as the bucket of nails hits the pulley. Un-nerved by his fall, Bob lets go of the rope, and the falling bucket of nails pulls Joe up to the roof where he cracks his head against the pulley but manages to hang on. However, the bottom falls out of the bucket when it struck the ground, and the empty bucket rises as Joe returns to the ground. Finally, Joe has had enough and lets go of the rope and remains on the ground only to be hit in the head by the empty bucket. Ignoring the possible mid-air collisions which merely add insult to injury, how long did it take for this little drama to unfold? (To make calculations easier, use g = -10 m/s s ) 12.9 s 62. A 110 kg man lowers himself to the ground from a height of 12 m by holding on to a rope passed over a frictionless pulley and attached to a 74 kg sandbag. a.) With what speed does the man hit the ground? b.) Is there anything he could do to reduce the speed with which he hits the ground? 6.8 m/s Yes, he could climb the rope while falling 63. A 10 kg monkey is climbing a massless rope attached to a 15 kg log over a (now get this frictionless) tree limb. a.) With what minimum acceleration must the monkey climb up the rope so that it can raise the 15 kg log off the ground? b.) If, after the log has been raised off of the ground, the monkey stops climbing and hangs on to the rope. What will now be the monkey s acceleration? c.) What is the tension in the rope now that the monkey has stopped climbing and is hanging on? 4.9 m/s 2 2.0 m/s 2 upward 120 N 64. A man of mass 80.0 kg stands on a platform of mass 40.0 kg. He pulls on a rope that is fastened to the platform and runs over a pulley on the ceiling. With what force does he have to pull in order to give himself and the platform and upward acceleration of 1.0 m/s 2? 648 N 65. A block of mass 6.0 kg resting on a horizontal frictionless surface is connected to a hanging 4.0 kg block by a cord passing over a frictionless pulley. Find the tension in the cord and the acceleration of the blocks. 3.92 m/s 2 23.52 N 66. A 6.0 kg block resting on a horizontal surface which is not frictionless is connected to a hanging block of 4.0 kg mass by a cord passing over a light, frictionless pulley. When the system is released, the blocks have an acceleration of 2.0 m/s 2. a.) Find the net force on each mass. b.) Find the tension in the cord. 12 N 8 N 31.2 N 67. A block of mass m 1 = 3.70 kg on a frictionless inclined plane of angle θ = 28.0 is connected by a cord over a small frictionless, massless pulley to a second block of mass m 2 = 1.86 kg hanging vertically as in the diagram below. a) What is the acceleration of each block? b) Find the tension in the cord. 0.217 m/s 2 17.8 N Newton s Third Law of Motion: Whenever one body exerts a force on another, the second body exerts on the first a force of equal magnitude in the opposite direction. This is also known as the law of action and reaction or the law of interaction. According to this law, there is no such thing as a single force. A body can produce a force only if there is some other body to exert its force upon. FORCES ALWAYS OCCUR IN PAIRS! an action force and a reaction force. The action and reaction forces are not on the same body. Conceptual Questions -Third Law 1. When you push against a wall does it push against you? Try this sometime while wearing roller skates. 2. Do you find it easier to walk on a carpeted floor than one having a polished smooth surface? Why? 3. Use Newton's third law to explain why when standing on a weighing scale you cannot decrease your weight by pulling upward on your boot straps. 4. Your weight is the result of a gravitational force of the earth on your body. What is the corresponding reaction force?

5. As you stand on a floor, does the floor exert an upward force against your feet? How much force does it exert? Why are you not moved upward by this force? 6. If you walk on a log that is floating in the water, the log moves backward. Why? 7. Why can you exert greater force on the pedals of a bicycle if you pull up on the handlebars? 8. Two people of equal mass attempt a tug-of-war with a 12-meter rope while standing on frictionless ice. When they pull on the rope, they slide toward each other. How do their accelerations compare, and how far does each person slide before meeting? 9. An astronaut on a "space-walk" finds that the rope connecting him to the space capsule has broken. Using a special space pistol, the astronaut manages to get back to the capsule. Explain. 10. Newton's third law states that action and reaction forces are always equal and opposite. If this is true, why don't they always cancel one another and leave no unbalanced force acting on any body? 11. Suppose a brick is suspended from a rigid support by a suitable length of cord. a) What downward force acts on the brick? b) If this force is the action force, what is the corresponding reaction force? 12. a) What upward force acts on the suspended brick in Question 11? b) If this force is the action force, what is the corresponding reaction force? 13. Two people pull upon a light wagon with equal force in opposite directions, the wagon will not move. Is this an example of action and reaction pairs? Explain.