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Newton’s  Laws of Motion Topic 4 (cont.)
Lecture Outline ,[object Object],[object Object],[object Object],[object Object]
Weight and Normal Force ,[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object]
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?
[object Object],[object Object],[object Object],[object Object]
Free-Body Diagrams ,[object Object],[object Object]
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?
[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],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
[object Object],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
[object Object],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
[object Object],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?
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!
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!
[object Object],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
[object Object],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
[object Object],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
[object Object],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
[object Object],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
[object Object],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
[object Object],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
[object Object],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
[object Object],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
[object Object],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
[object Object],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
[object Object],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
[object Object],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
[object Object],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
[object Object],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
[object Object],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.
[object Object],ConcepTest 4.16a   Tension I 1)  0 N 2)  50 N 3)  100 N 4)  150 N 5)  200 N
[object Object],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
[object Object],1)  0 N 2)  50 N 3)  100 N 4)  150 N 5)  200 N ConcepTest 4.16b   Tension II
[object Object],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
[object Object],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
[object Object],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

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07 newton's law of motion

  • 1. Newton’s Laws of Motion Topic 4 (cont.)
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  • 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?
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  • 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?
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  • 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. 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!
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Notas del editor

  1. 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.
  2. 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 .
  3. 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.
  4. 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.
  5. 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).
  6. 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.
  7. 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.
  8. 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.
  9. 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 .
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