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4 Newton’s Laws

4 Newton’s Laws. Force, net-force, mass & inertia Newton’s Laws of Motion Weight, Contact Forces Labeling & Diagramming Hk: 37, 49, 53, 57, 59, 61, 65, 67. Force Concept. 0. Contact Forces Ex: sliding, bouncing Non-Contact Ex: magnetism, gravity /. 2. Inertia.

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4 Newton’s Laws

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  1. 4 Newton’s Laws • Force, net-force, mass & inertia • Newton’s Laws of Motion • Weight, Contact Forces • Labeling & Diagramming • Hk: 37, 49, 53, 57, 59, 61, 65, 67.

  2. Force Concept 0 Contact Forces Ex: sliding, bouncing Non-Contact Ex: magnetism, gravity / 2

  3. Inertia is ‘resistance’ to change in velocity Measurement: Mass SI Unit: Kilogram (Kg) / 0 3

  4. units Force units (SI): newton, N 1N ≈ ¼ lb. 1N = (1kg)(1m/s/s) N/kg = m/s/s 4

  5. 0 Net Force vector sum of all forces acting on an object

  6. 0 Newton’s Laws of Motion 1. An object maintains constant velocity when the Net-Force on it is zero. 2. An object’s acceleration equals the Net-Force on it divided by its mass. 3. Forces always occur in pairs equal in size and opposite in direction. 6

  7. Weight Force

  8. Normal Force Contact Forces • Surfaces in contact are often under compression: each surface pushes against the other. The outward push of each object is called the Normal Force. • If the objects move (even slightly) parallel to their surface the resistance force experienced is called the frictional force.

  9. Tension & Compression • Compressed objects push outward away from their center (aka Normal Force). • Stretched objects pull toward their center. This is called the Tension Force.

  10. Force Label Notation F = general force FN = normal force f = frictional force w = mg = Fg = weight T = tension force / 10

  11. Net Force = change of motion 0 vector sum of all forces acting on an object 11

  12. Problem Solving Template: Two Equations – Two Unknowns 0 12

  13. Force Diagram table force weight force Example: Ball rolls along a smooth level surface 0 velocity

  14. Example of a Force Diagram for a Sled 0 net force equals the mass times its acceleration.

  15. 0 Force Diagrams • Object is drawn as a “point” • Each force is drawn as a “pulling” vector • Each force is labeled • Relevant Angles are shown • x, y axes are written offset from diagram • Only forces which act ON the object are shown

  16. Ex: Ball rolling up & slowing down (Use PHET Vector Addition for net-force) 0 acceleration Fnet upward (decreasing) velocity

  17. 0 Ex. m=3kg, F=86N, 60° below horizontal.

  18. 0 Ex. Continued

  19. Block on Inclined Plane

  20. Ex. Calculate Acceleration of Block on a Frictionless Plane inclined 30°

  21. Ex. Calculate Normal Force on a Block on a Plane inclined 30°

  22. Complete the table below for the sign of the net force. Sketch a motion diagram for each case.

  23. Newton’s 3rd Law of Motion equal-sized oppositely-directed forces Independent of mass Pair-notation x x 23

  24. Newton’s 3rd Law Pair Notation use “x” marks on forces that are 3rd Law pairs. Use “xx” for a different interaction, etc. 24

  25. Force Diagram each object. Which has greater acceleration when released? Spring Force Spring Force x x Acceleration = F/m Acceleration = F/(2m) 25

  26. Motion of Ball Newton’s Second and Third Laws in Operation: Ball hits a large block on a smooth level surface. Force on Ball Force on Block Acceleration of Ball Acceleration of Block

  27. Solving Two Body Problems • Force diagram each object & system (usually with one axis parallel to the acceleration). Use clockwise coordinates for problems with pulleys. • System has a force-pair that cancels out • Solve simplest diagram first, then use this information in another diagram • “ma” is not a force • /

  28. Two Connected Blocks

  29. 4 Summary • Zero net-force; constant velocity • Acceleration = net-force/mass • All forces are pairs • Labeling & diagramming • Solving problems using x, y force template • Solving two body problems • /

  30. 0 Example: Net Force = 0. Block on a surface inclined 30° from horizontal. Applied force F acts 40° below horizontal. Net Force = 0 velocity = constant

  31. Newton’s 2nd Law Examples

  32. A 3kg object sits on a frictionless table. Two horizontal forces act, one is 2N in the y-direction, the other 4N in the x-direction. A top-view diagram will be shown. What is the magnitude of the net-force acting? Fnet 2 2 4

  33. What direction does the 3kg mass accelerate in? Its acceleration is parallel to Fnet by Newton’s 2nd Law. So we need to determine the direction of Fnet. We are in Quadrant I since x and y are both +

  34. What is the magnitude of the acceleration?

  35. Example: A 10kg box is being pushed along a horizontal surface by a force of 15N. A frictional force of 5N acts against the motion. We will want to (a) Calculate the net-force acting and (b) calculate the acceleration of the box. The net-horizontal force determines its x-acceleration The y-acceleration is known to be zero because it remains in horizontal motion, thus The net-force is 10N horizontal (0 vertical) The x-acceleration is:

  36. Coefficients of Friction Ex: Block&Load = 580grams If it takes 2.4N to get it moving and 2.0N to keep it moving

  37. 0 Q1. What are ax and FN if angle is 30?

  38. 0 2) 3kg box at rest on frictionless 30° inclined plane. F acts 40° below horizontal.

  39. 0

  40. 0 Check of Previous Problem:

  41. 0 Q2. 3kg box at rest on frictionless 30° inclined plane. F acts horizontally. Calculate F and Fn.

  42. 5kg 3kg 2kg F=26N F12, surface reaction force 3kg 0 3. Three boxes are pushed by force F along a horizontal frictionless surface. Force diagram object 1 (left box)

  43. 2kg F32, surface reaction force 5kg 0 Diagram object 2: F23, surface reaction force F21, surface reaction force Diagram object 3:

  44. 0 Object1: 3kg Object2: 5kg Object3: 2kg Object1+2+3: 3kg+5kg+2kg

  45. 0 3kg 5kg 2kg Summary: Stimulus=26N Reactions: 18.2N, 5.2N

  46. 5kg 3kg 2kg F=26N 0 Q3. Recalculate problem3 with order switched to 5kg, 3kg, 2kg. 2kg 3kg

  47. 4. Modified Atwood Machine with frictionless plane 0 solve for a and T in terms of m1, m2: Let m1 = 1kg, m2 = 2kg, q = 30°.

  48. 0 Q4. Recalculate problem4 with m1 = 6kg m2 = 1kg. Note that T > (m2)g

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