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5 Additional Applications of Newton’s Laws

5 Additional Applications of Newton’s Laws. Friction Drag Forces Motion Along a Curved Path The Center of Mass Hk: 31, 43, 53, 57, 67, 81, 91, 101. Friction. Surface Force opposing relative motion Component of Contact Force (other component is the Normal Force)

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5 Additional Applications of Newton’s Laws

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  1. 5 Additional Applications of Newton’s Laws • Friction • Drag Forces • Motion Along a Curved Path • The Center of Mass • Hk: 31, 43, 53, 57, 67, 81, 91, 101.

  2. Friction • Surface Force opposing relative motion • Component of Contact Force (other component is the Normal Force) • Characterized by coefficients (mu) • Static (varies 0 to max) • Kinetic (~ constant) • Rolling (~ one tenth of kinetic)

  3. 0 Surface Dependence of Friction

  4. Normal Force Dependence of Friction • Contact area ~ to Normal Force • Frictional Force ~ Normal Force • Summary: • Two factors affect frictional force; • Surface composition • Normal Force • /

  5. Equations for Friction 0 5

  6. Example Friction Ex. 10kg block. FN = weight = mg = 98N. Static coef. = 0.50; Kinetic coef. = 0.30. 0 6

  7. 0 Block at rest. Draw a Force Diagram for the block.

  8. 5kg 3kg 2kg F=26N fk 0 Three boxes are pushed by force F with v > 0 along a horizontal surface with mk = 0.291.

  9. Maximum angle block remains at rest: Derive the Angle of Repose relation:

  10. Atwood with Friction. m1=1kg m2=2kg. Kinetic friction = 0.5.

  11. Drag Forces

  12. Motion Along a Curved Path • Force required turn and to change speed • Coordinates usually used are F/B (tangential) and L/R (radial) • Sum forces tangential = mass x tangential acceleration • Sum forces L/R (radial centripetal) = mass x centripetal acceleration • /

  13. 0 What is the fastest speed the car can go without sliding? Assume the car has m = 1200kg andms = 0.92.

  14. A block loops the loop. Which force diagram is correct for when it passed point D?

  15. Center of Mass Definition

  16. Center of Mass Acceleration

  17. Center of Mass when Net External Force is Zero • Zero Net Force implies center of mass acceleration is also zero. So if CM originally at rest, it remains at rest. If CM moving, its velocity remains same. • Example: two people standing on ice push off one another • /

  18. Ex. Center of Mass. A 100kg person walks 6 feet forward in a 50kg canoe. How far did he move relative to the shore?

  19. Summary • Friction depends on Surface Composition and Normal Force • Drag Force vary with speed • Force required to move along curved path even at constant speed • Center of Mass stays same when only internal forces operate

  20. Can you stop in time? Buggy rolls.You slide.

  21. Given m = 75kg, M = 20kg, D = 3.5m, vo =1.1m/s. What frictional coefficient is needed? Insert values, determine ax.

  22. Diagramming Refresher:

  23. Accelerating with F2WD. Stopping with 4W Disc-Brakes

  24. fs A 3kg box at rest on level surface with ms = 0.55. What is the largest F acting 60° below horizontal for which the box remains at rest? (60° is close to maximum angle)

  25. Relative vs. Absolute Velocity vpc vpg P

  26. Which force diagram applies to the object at Point B?

  27. Which force diagram applies to the object at Point C?

  28. Assume mass = 1.2kg and radius = 45cm.If speed at Point D is 3.6m/s, what is the size of the normal force acting at Point D? -cen +cen

  29. +cen -cen Q. Assume mass = 1.2kg and radius = 45cm.If speed at Point B is 5.1m/s, what is the size of the normal force acting at Point B?

  30. Net Given: T = 50N, q = 30°, r = 1mFind: mg and v.

  31. Net The speed is now 6.5m/s and r = 1.0m. Angle, tension, mass? For example, if m = 1.0kg, then T = 42.3N.

  32. Net Q. The speed of a mass on a string of length L is 6.5m/s. The radius r = 2.0m. Find angle, tension, mass, and L. For example, if m = 1.0kg, then T = 23.9N.

  33. Practice Q: What is F such that 0.5kg block stays at rest if all surfaces are frictionless?

  34. Banked Turn:

  35. Banked Turn:

  36. Modified Atwood Machine with friction. Objects are in CW motion. Let m1 = 2kg, m2 = 3kg, q = 30°, sliding friction coeff. 0.44

  37. Q. Recalculate last problem with m1 = 6kg m2 = 1kg. (All else remaining the same)

  38. Practice Q: Find the variable relationships.

  39. Figures

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