1 / 53

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:

eden
Download Presentation

Newton’s Second Law

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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)

More Related