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# 07 newton's law of motion

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• Figure 4-15. Caption: Example 4–6. (a) A 10-kg gift box is at rest on a table. (b) A person pushes down on the box with a force of 40.0 N. (c) A person pulls upward on the box with a force of 40.0 N. The forces are all assumed to act along a line; they are shown slightly displaced in order to be distinguishable. Only forces acting on the box are shown. Answer: The box is always at rest, so the forces must always add to zero. In (a), the normal force equals the weight; in (b), the normal force equals the weight plus the 40.0 N downward force; in (c), the normal force equals the weight minus the 40.0 N upward force.
• Figure 4-16. Caption: Example 4–7. The box accelerates upward because F P &gt; mg. Answer: Now there is a net upward force on the box of 2.0 N, and the box accelerates upward at a rate of 0.20 m/s 2 .
• Figure 4-17. Caption: The acceleration vector is shown in gold to distinguish it from the red force vectors. Her weight is always mg, but the normal force is 0.8 mg, which the scale reads as 52 kg. The scale reads her true weight, or a mass of 65 kg.
• Figure 4-19. Caption: Example 4–9:Two force vectors act on a boat. In this example, the net force on the boat has a magnitude of 53.3 N and acts at an 11 ° angle to the x axis.
• Figure 4-20. Caption: Example 4–10. Which is the correct free-body diagram for a hockey puck sliding across frictionless ice? If the velocity is constant, there is no net force, and (b) is correct. If the puck is slowing down, the force is opposite to the direction of motion, which is illustrated in (c).
• Figure 4-21. Caption: (a) Pulling the box, Example 4–11; (b) is the free-body diagram for the box, and (c) is the free-body diagram considering all the forces to act at a point (translational motion only, which is what we have here). Answer: (a) The free-body diagram is shown in part (b); the forces are gravity, the normal force, and the force exerted by the person. The forces in the vertical direction cancel – that is, the weight equals the normal force plus the vertical component of the external force. The horizontal component of the force, 34.6 N, accelerates the box at 3.46 m/s 2 . (b) The vertical component of the external force is 20.0 N; the weight is 98.0 N, so the normal force is 78.0 N.
• Figure 4-22. Caption: Example 4–12. (a) Two boxes, A and B, are connected by a cord. A person pulls horizontally on box A with force F P = 40.0 N. (b) Free-body diagram for box A. (c) Free-body diagram for box B. Answer: Free-body diagrams for both boxes are shown. The net force on box A is the external force minus the tension; the net force on box B is the tension. Both boxes have the same acceleration. The acceleration is the external force divided by the total mass, or 1.82 m/s 2 . The tension in the cord is the mass of box B multiplied by the acceleration, or 21.8 N.
• Figure 4-23. Caption: Example 4–13. (a) Atwood’s machine in the form of an elevator–counterweight system. (b) and (c) Free-body diagrams for the two objects. Answer: Each mass has two forces on it, gravity pulling downward and the tension in the cable pulling upward. The tension in the cable is the same for both, and both masses have the same acceleration. Writing Newton’s second law for each mass gives us two equations; there are two unknowns, the acceleration and the tension. Solving the equations for the unknowns gives a = 0.68 m/s 2 and F T = 10,500 N.
• Figure 4-26. Caption: Example 4–16. (a) Box sliding on inclined plane. (b) Free-body diagram of box. Answer: On an incline (or any surface), the normal force is perpendicular to the surface and any frictional forces are parallel to the surface. (a) The normal force is equal to the component of the weight perpendicular to the incline, or mg cos θ . (b) The force causing the acceleration is the component of the weight parallel to the incline; therefore the acceleration is g sin θ . (c) The normal force is 85 N and the acceleration is 4.9 m/s 2 .
• ### 07 newton's law of motion

1. 1. Newton’s Laws of Motion Topic 4 (cont.)
2. 2. Lecture Outline <ul><li>Weight and Normal Force </li></ul><ul><li>Free-Body Diagram </li></ul><ul><ul><li>Tension in flexible cord </li></ul></ul><ul><ul><li>Incline Plane </li></ul></ul>
3. 3. Weight and Normal Force <ul><li>An object’s weight is the force exerted on it by gravity. </li></ul><ul><li>Any object on Earth is under the influence of Gravitational Force @ Force of gravity - weight </li></ul><ul><li>From Newton’s Second Law: </li></ul><ul><li>Here, g is the acceleration of gravity, g = 9.80 ms -2 </li></ul><ul><li>Weight therefore has the same units as force – Newton. </li></ul>
4. 4. <ul><li>As long as we are on earth, we are under the influence of gravitational force </li></ul><ul><li>An object at rest 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? </li></ul><ul><li>Normal Force - The force exerted perpendicular to a surface </li></ul><ul><li>Normal force = Gravitational Force </li></ul>
5. 5. <ul><li>Example 4-6: Weight, normal force, and a box. </li></ul><ul><li>A friend has given you a special gift, a box of mass 10.0 kg with a mystery surprise inside. The box is resting on the smooth (frictionless) horizontal surface of a table. </li></ul><ul><li>Determine the weight of the box and the normal force exerted on it by the table. </li></ul><ul><li>(b) Now your friend pushes down on the box with a force of 40.0 N. Again determine the normal force exerted on the box by the table. </li></ul><ul><li>(c) If your friend pulls upward on the box with a force of 40.0 N, what now is the normal force exerted on the box by the table? </li></ul>
6. 6. Example 4-7: Accelerating the box. What happens when a person pulls upward on the box in the previous example with a force greater than the box’s weight, say 100.0 N?
7. 7. <ul><li>Example 4-8: Apparent weight loss. </li></ul><ul><li>A 65-kg woman descends in an elevator that briefly accelerates at 0.20 g downward. She stands on a scale that reads in kg. </li></ul><ul><li>During this acceleration, what is her weight and what does the scale read? </li></ul><ul><li>(b) What does the scale read when the elevator descends at a constant speed of 2.0 m/s? </li></ul>
8. 8. Free-Body Diagrams <ul><li>Forces acted on an object, all the forces can be added and represented as a single force, net force – resultant force . </li></ul><ul><li>Force is a vector, can be added as a vector </li></ul>
9. 9. A hockey puck is sliding at constant velocity across a flat horizontal ice surface that is assumed to be frictionless. Which of these sketches is the correct free-body diagram for this puck? What would your answer be if the puck slowed down?
10. 10. <ul><li>Example 4-11: Pulling the mystery box. </li></ul><ul><li>Suppose a friend asks to examine the 10.0-kg box you were given previously, hoping to guess what is inside; and you respond, “Sure, pull the box over to you.” She then pulls the box by the attached cord along the smooth surface of the table. The magnitude of the force exerted by the person is F P = 40.0 N, and it is exerted at a 30.0° angle as shown. Calculate </li></ul><ul><li>the acceleration of the box </li></ul><ul><li>(b) the magnitude of the upward force F N exerted by the table on the box. </li></ul>
11. 11. <ul><li>Tension in a flexible cord </li></ul><ul><li>When a flexible cord pull an object, the cord is said to be under tension </li></ul><ul><li>The force exerted on the object is the tension, F T </li></ul>
12. 12. <ul><li>Example 4-12: Two boxes connected by a cord. </li></ul><ul><li>Two boxes, A and B, are connected by a lightweight cord and are resting on a smooth table. The boxes have masses of 12.0 kg and 10.0 kg. A horizontal force of 40.0 N is applied to the 10.0-kg box. Find </li></ul><ul><li>the acceleration of each box </li></ul><ul><li>the tension in the cord connecting the boxes. </li></ul>
13. 13. <ul><li>Example 4-13: Elevator and counterweight (Atwood’s machine). </li></ul><ul><li>A system of two objects suspended over a pulley by a flexible cable is sometimes referred to as an Atwood’s machine. Here, let the mass of the counterweight be 1000 kg. Assume the mass of the empty elevator is 850 kg, and its mass when carrying four passengers is 1150 kg. For the latter case calculate </li></ul><ul><li>the acceleration of the elevator and </li></ul><ul><li>the tension in the cable. </li></ul>
14. 14. <ul><li>Inclines </li></ul><ul><li>Gravity accelerate object vertically (y axis) </li></ul><ul><li>Object put on an inclined plane also accelerate, but at a certain angle </li></ul><ul><li>Take x axis to point along the incline (acceleration) </li></ul><ul><li>Take y axis perpendicular to the plane (Normal Force) </li></ul>
15. 15. <ul><li>Example 4-16: Box slides down an incline. </li></ul><ul><li>A box of mass m is placed on a smooth incline that makes an angle θ with the horizontal. </li></ul><ul><li>Determine the normal force on the box. </li></ul><ul><li>Determine the box’s acceleration. </li></ul><ul><li>Evaluate for a mass m = 10 kg and an incline of θ = 30°. </li></ul>
16. 16. <ul><li>What can you say about the force of gravity F g acting on a stone and a feather? </li></ul>ConcepTest 4.7a Gravity and Weight I 1) F g is greater on the feather 2) F g is greater on the stone 3) F g is zero on both due to vacuum 4) F g is equal on both always 5) F g is zero on both always
17. 17. <ul><li>What can you say about the force of gravity F g acting on a stone and a feather? </li></ul>The force of gravity (weight) depends on the mass of the object!! The stone has more mass, therefore more weight. ConcepTest 4.7a Gravity and Weight I 1) F g is greater on the feather 2) F g is greater on the stone 3) F g is zero on both due to vacuum 4) F g is equal on both always 5) F g is zero on both always
18. 18. <ul><li>What can you say about the acceleration of gravity acting on the stone and the feather? </li></ul>1) it is greater on the feather 2) it is greater on the stone 3) it is zero on both due to vacuum 4) it is equal on both always 5) it is zero on both always ConcepTest 4.7b Gravity and Weight II
19. 19. <ul><li>What can you say about the acceleration of gravity acting on the stone and the feather? </li></ul>The acceleration is given by F/m so here the mass divides out. Since we know that the force of gravity (weight) is mg , then we end up with acceleration g for both objects. 1) it is greater on the feather 2) it is greater on the stone 3) it is zero on both due to vacuum 4) it is equal on both always 5) it is zero on both always ConcepTest 4.7b Gravity and Weight II Follow-up: Which one hits the bottom first?
20. 20. 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 ball on the Moon with the same force. His foot hurts... 1) more 2) less 3) the same Ouch!
21. 21. 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. 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 ball on the Moon with the same force. His foot hurts... 1) more 2) less 3) the same Follow-up: What is different about the bowling ball on the Moon? Ouch!
22. 22. <ul><li>A block of mass m rests on the floor of an elevator that is moving upward at constant speed . What is the relationship between the force due to gravity and the normal force on the block? </li></ul>ConcepTest 4.9a Going Up I 1) N > mg 2) N = mg 3) N < mg (but not zero) 4) N = 0 5) depends on the size of the elevator m v
23. 23. <ul><li>A block of mass m rests on the floor of an elevator that is moving upward at constant speed . What is the relationship between the force due to gravity and the normal force on the block? </li></ul>The block is moving at constant speed, so it must have no net force on it. The forces on it are N (up) and mg (down), so N = mg , just like the block at rest on a table. ConcepTest 4.9a Going Up I 1) N > mg 2) N = mg 3) N < mg (but not zero) 4) N = 0 5) depends on the size of the elevator m v
24. 24. <ul><li>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? </li></ul>1) N > mg 2) N = mg 3) N < mg (but not zero) 4) N = 0 5) depends on the size of the elevator ConcepTest 4.9b Going Up II m a
25. 25. <ul><li>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? </li></ul>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 ! 1) N > mg 2) N = mg 3) N < mg (but not zero) 4) N = 0 5) depends on the size of the elevator  F = N – mg = ma > 0  N > mg ConcepTest 4.9b Going Up II Follow-up: What is the normal force if the elevator is in free fall downward? m a > 0 mg N
26. 26. <ul><li>Below you see two cases: a physics student pulling or pushing a sled with a force F which is applied at an angle  . In which case is the normal force greater? </li></ul>ConcepTest 4.10 Normal Force 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
27. 27. <ul><li>Below you see two cases: a physics student pulling or pushing a sled with a force F which is applied at an angle  . In which case is the normal force greater? </li></ul>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 . ConcepTest 4.10 Normal Force 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
28. 28. <ul><li>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? </li></ul>ConcepTest 4.11 On an Incline 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) A B
29. 29. <ul><li>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? </li></ul>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) N W W y x y f   ConcepTest 4.11 On an Incline In Case A , we know that N = W . In Case B , due to the angle of the incline, N < W . In fact, we can see that N = W cos(  ). A B
30. 30. <ul><li>When you climb up a rope, the first thing you do is pull down on the rope. How do you manage to go up the rope by doing that?? </li></ul>ConcepTest 4.12 Climbing the Rope 1) this slows your initial velocity, which is already upward 2) you don’t go up, you’re too heavy 3) you’re not really pulling down – it just seems that way 4) the rope actually pulls you up 5) you are pulling the ceiling down
31. 31. <ul><li>When you climb up a rope, the first thing you do is pull down on the rope. How do you manage to go up the rope by doing that?? </li></ul>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 4.12 Climbing the Rope 1) this slows your initial velocity, which is already upward 2) you don’t go up, you’re too heavy 3) you’re not really pulling down – it just seems that way 4) the rope actually pulls you up 5) you are pulling the ceiling down
32. 32. <ul><li>In outer space, a bowling ball and a ping-pong ball attract each other due to gravitational forces. How do the magnitudes of these attractive forces compare? </li></ul>1) the bowling ball exerts a greater force on the ping-pong ball 2) the ping-pong ball exerts a greater force on the bowling ball 3) t he forces are equal 4) t he forces are zero because they cancel out 5) there are actually no forces at all ConcepTest 4.13a Bowling vs. Ping-Pong I F 12 F 21
33. 33. <ul><li>In outer space, a bowling ball and a ping-pong ball attract each other due to gravitational forces. How do the magnitudes of these attractive forces compare? </li></ul>The forces are equal and opposite by Newton’s Third Law! ConcepTest 4.13a Bowling vs. Ping-Pong I 1) the bowling ball exerts a greater force on the ping-pong ball 2) the ping-pong ball exerts a greater force on the bowling ball 3) t he forces are equal 4) t he forces are zero because they cancel out 5) there are actually no forces at all F 12 F 21
34. 34. <ul><li>In outer space, gravitational forces exerted by a bowling ball and a ping-pong ball on each other are equal and opposite . How do their accelerations compare? </li></ul>1) they do not accelerate because they are weightless 2) accels. are equal, but not opposite 3) accelerations are opposite, but bigger for the bowling ball 4) accelerations are opposite, but bigger for the ping-pong ball 5) accelerations are equal and opposite ConcepTest 4.13b Bowling vs. Ping-Pong II F 12 F 21
35. 35. <ul><li>In outer space, gravitational forces exerted by a bowling ball and a ping-pong ball on each other are equal and opposite . How do their accelerations compare? </li></ul>The forces are equal and opposite -- this is Newton’s Third Law!! But the acceleration is F/m and so the smaller mass has the bigger acceleration . 1) they do not accelerate because they are weightless 2) accels. are equal, but not opposite 3) accelerations are opposite, but bigger for the bowling ball 4) accelerations are opposite, but bigger for the ping-pong ball 5) accelerations are equal and opposite ConcepTest 4.13b Bowling vs. Ping-Pong II Follow-up: Where will the balls meet if they are released from this position? F 12 F 21
36. 36. <ul><li>A small car collides with a large truck. Which experiences the greater impact force? </li></ul>ConcepTest 4.14a Collision Course I 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
37. 37. <ul><li>A small car collides with a large truck. Which experiences the greater impact force? </li></ul>ConcepTest 4.14a Collision Course I 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 According to Newton’s Third Law, both vehicles experience the same magnitude of force.
38. 38. <ul><li>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? </li></ul>ConcepTest 4.16a Tension I 1) 0 N 2) 50 N 3) 100 N 4) 150 N 5) 200 N
39. 39. <ul><li>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? </li></ul>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). Since you are pulling with a force of 100 N , that is the tension in the rope. ConcepTest 4.16a Tension I 1) 0 N 2) 50 N 3) 100 N 4) 150 N 5) 200 N
40. 40. <ul><li>Two tug-of-war opponents each pull with a force of 100 N on opposite ends of a rope. What is the tension in the rope? </li></ul>1) 0 N 2) 50 N 3) 100 N 4) 150 N 5) 200 N ConcepTest 4.16b Tension II
41. 41. <ul><li>Two tug-of-war opponents each pull with a force of 100 N on opposite ends of a rope. What is the tension in the rope? </li></ul>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 !! 1) 0 N 2) 50 N 3) 100 N 4) 150 N 5) 200 N ConcepTest 4.16b Tension II
42. 42. <ul><li>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? </li></ul>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 ConcepTest 4.16c Tension III
43. 43. <ul><li>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? </li></ul>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. 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 ConcepTest 4.16c Tension III