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Newton’s Second Law

Newton’s Second Law. Chapter 6. The Second Law. Force = mass X acceleration S F = ma S F = 0 or S F = ma -Still object -Accelerating object -Obj. at constant velocity. Sum of all the forces acting on a body Vector quantity. The Second Law. Situation One:

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Newton’s Second Law

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  1. Newton’s Second Law Chapter 6

  2. The Second Law Force = mass X acceleration SF = ma SF = 0 or SF = ma -Still object -Accelerating object -Obj. at constant velocity Sum of all the forces acting on a body Vector quantity

  3. The Second Law Situation One: Non-moving Object • Still has forces Force of the material of the rock Force of gravity http://alfa.ist.utl.pt/~vguerra/Other/Rodin/thinker.jpg

  4. The Second Law Situation Two: Moving Object: Constant Velocity SF = 0 Fpedalling = Fair + Ffriction Fpedalling Fair Ffriction http://2003.tour-de-france.cz/images/foto/05-07-2003/armstrong4.jpg

  5. The Second Law Situation Two: Moving Object: Accelerating SF = ma ma = Fpedalling – Fair - Ffriction Fpedalling Fair Ffriction http://2003.tour-de-france.cz/images/foto/05-07-2003/armstrong4.jpg

  6. The Second Law Unit of Force = the Newton SF=ma SF = (kg)(m/s2) 1 N = 1 kg-m/s2 (MKS) 1 Newton can accelerate a 1 kg object from rest to 1 m/s in 1 s.

  7. Equilibrium • No motion • Constant velocity BOTH INDICATE NO ACCELERATION SF=0

  8. Three ropes are tied together for a wacky tug-of-war. One person pulls west with 100 N of force, another south with 200 N of force. Calculate the magnitude and direction of the third force. ? 100 N 200 N

  9. A car with weight 15,000 N is being towed up a 20o slope (smooth) at a constant velocity. The tow rope is rated at 6000 N. Will it break?

  10. Accelerating Systems • SF=ma • Must add up all forces on the object

  11. What Force is needed to accelerate a 5 kg bowling ball from 0 to 20 m/s over a time period of 2 seconds?

  12. Calculate the net force required to stop a 1500 kg car from a speed of 100 km/h within a distance of 55 m. 100 km/h = 28 m/s v2 = vo2 + 2a(x-xo) a = (v2 - vo2)/2(x-xo) a = 02 – (28 m/s)2/2(55m) = -7.1 m/s2

  13. A 1500 kg car is pulled by a tow truck. The tension in the rope is 2500 N and the 200 N frictional force opposes the motion. The car starts from rest. a. Calculate the net force on the car b. Calculate the car’s speed after 5.0 s

  14. A 500.0 gram model rocket (weight = 4.90 N) is launched straight up from rest by an engine that burns for 5 seconds at 20.0 N. • Calculate the net force on the rocket • Calculate the acceleration of the rocket • Calculate the height and velocity of the rocket after 5 s • Calculate the maximum height of the rocket even after the engine has burned out.

  15. Calculate the sum of the two forces acting on the boat shown below. (53.3 N, +11.0o)

  16. Mass vs. Weight Mass • The amount of matter in an object/INTRINSIC PROPERTY • Independent of gravity • Measured in kilograms Weight • Force that results from gravity pulling on an object • Weight = mg (g = 9.8 m/s2)

  17. Mass vs. Weight • Weight = mg is really a re-write of F=ma. • Weight is a force • g is the acceleration (a) of gravity • Metric unit of weight is a Newton • English unit is a pound

  18. A 60.0 kg person weighs 1554 N on Jupiter. What is the acceleration of gravity on Jupiter?

  19. Elevator at Constant Velocity a= 0 SF = FN – mg ma = FN – mg 0 = FN – mg FN = mg Suppose Chewbacca has a mass of 102 kg: FN = mg = (102kg)(9.8m/s2) FN = 1000 N FN mg a is zero

  20. Elevator Accelerating Upward a = 4.9 m/s2 SF = FN – mg ma = FN – mg FN = ma + mg FN = m(a + g) FN=(102kg)(4.9m/s2+9.8 m/s2) FN = 1500 N FN mg a is upward

  21. Elevator Accelerating Downward a = -4.9 m/s2 SF = FN – mg ma = FN – mg FN = ma + mg FN = m(a + g) FN=(102kg)(-4.9m/s2+9.8 m/s2) FN = 500 N FN mg a is down

  22. At what acceleration will he feel weightless? FN = 0 SF = FN – mg ma = FN – mg ma = 0 – mg ma = -mg a = -9.8 m/s2 Apparent weightlessness occurs if a > g FN mg

  23. A 10.o kg present is sitting on a table. Calculate the weight and the normal force. FN Fg = W

  24. Suppose someone leans on the box, adding an additional 40.0 N of force. Calculate the normal force.

  25. Now your friend lifts up with a string (but does not lift the box off the table). Calculate the normal force.

  26. What happen when the person pulls upward with a force of 100 N? SF = FN+ Fp – mg SF = 0 +100.0N – 98N = 2.0N ma = 2N a = 2N/10.0 kg = 0.2 m/s2 Fp = 100.0 N Fg = mg = 98.0 N

  27. Free Body Diagrams: Ex. 3 A person pulls on the box (10.0 kg) at an angle as shown below. Calculate the acceleration of the box and the normal force. (78.0 N) Fp = 40.0 N 30o FN mg

  28. Friction • Always opposes the direction of motion. • Proportional to the Normal Force (more massive objects have more friction) FN Ffr Ffr Fa FN Fa mg mg

  29. Friction Static -opposes motion before it moves (ms) • Generally greater than kinetic friction • Fmax = Force needed to get an objct moving Fmax = msFN Kinetic - opposes motion while it moves (mk) • Generally less than static friction Ffr = mkFN

  30. Friction and Rolling Wheels Rolling uses static friction • A new part of the wheel/tire is coming in contact with the road every instant B A

  31. Braking uses kinetic friction Point A gets drug across the surface A

  32. A 50.0 kg wooden box is pushed across a wooden floor (mk=0.20) at a steady speed of 2.0 m/s. • How much force does she exert? (98 N) • If she stops pushing, calculate the acceleration. (-1.96 m/s2) • Calculate how far the box slides until it stops. (1.00 m)

  33. A 100 kg box is on the back of a truck (ms = 0.40). The box is 50 cm X 50 cm X 50 cm. a. Calculate the maximum acceleration of the truck before the box starts to slip.

  34. Your little sister wants a ride on her sled. Should you push or pull her?

  35. Inclines What trigonometric function does this resemble?

  36. Inclines FN mg q

  37. Inclines FN q mgcosq mg q mgsinq

  38. Inclines FN Ffr mgsinq mgcosq q

  39. A 50.0 kg file cabinet is in the back of a dump truck (ms = 0.800). • Calculate the magnitude of the static friction on the cabinet when the bed of the truck is tilted at 20.0o (170 N) • Calculate the angle at which the cabinet will start to slide. (39o)

  40. Given the following drawing: • Calculate the acceleration of the skier. (snow has a mk of 0.10) (4.0 m/s2) • Calculate her speed after 4.0 s? (16 m/s)

  41. First we need to resolve the force of gravity into x and y components:

  42. FGy = mgcos30o FGx = mgsin30o The pull down the hill is: FGx = mgsin30o The pull up the hill is: Ffr= mkFN Ffr= (0.10)(mgcos30o)

  43. SF = pull down – pull up SF = mgsin30o– (0.10)(mgcos30o) ma = mgsin30o– (0.10)(mgcos30o) ma = mgsin30o– (0.10)(mgcos30o) a = gsin30o– (0.10)(gcos30o) a = 4.0 m/s2 (note that this is independent of the skier’s mass)

  44. To find the speed after 4 seconds: v = vo + at v = 0 + (4.0 m/s2)(4.0 s) = 16 m/s

  45. Suppose the snow is slushy and the skier moves at a constant speed. Calculate mk SF = pull down – pull up ma = mgsin30o– (mk)(mgcos30o) ma = mgsin30o– (mk)(mgcos30o) a = gsin30o– (mk)(gcos30o)

  46. Since the speed is constant, acceleration =0 0 = gsin30o– (mk)(gcos30o) (mk)(gcos30o) = gsin30o mk= gsin30o= sin30o = tan30o =0.577 gcos30o cos30o

  47. Drag D ≈ ¼Av2 D = drag force A = Area V = velocity Fails for • Very small particles (dust) • Very fast (airplanes) • Water and dense fluids

  48. Finding Acceleration with Drag Derive the formula for the acceleration of a freefalling object including the drag force.

  49. Terminal Speed • Find the formula for terminal speed (a=0) for a freefalling body • Calculate the terminal velocity of a person who is 1.8 m tall, 0.40 m wide, and 75 kg. (64 m/s)

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