1 / 182

100 90 80 70 vel 60

100 90 80 70 vel 60 (m/s) 50 40 30 20 10. 1 2 3 4 5 6 7 8 9 10 time (sec). Ch7 – More Forces .

redford
Download Presentation

100 90 80 70 vel 60

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. 100 90 80 70 vel 60 (m/s) 50 40 30 20 10 1 2 3 4 5 6 7 8 9 10 time (sec)

  2. Ch7 – More Forces • Equilibrium - net force = 0 • In this chapter there can be 3 or more forces present in a problem. • Equilibrium results in an object that is stationary (static) • ormoving at constant speed (dynamic). • Ex1) What is the tension in the rope? 1 kg

  3. Ch7 – More Forces • Equilibrium - net force = 0 • In this chapter there can be 3 or more forces present in a problem. • Equilibrium results in an object that is stationary (static) • ormoving at constant speed (dynamic). • Ex1) What is the tension in the rope? Fnet = Fg – FT 0 = 10N – FT FT = 10N FT 1 kg Fg

  4. Ex2) What is the tension in each rope? 1 kg

  5. Ex2) What is the tension in each rope? Fg FT FT Fnet = Fg – 2FT 0 = 10N – 2FT Ft = 5N 1 kg

  6. Ex3) What is the tension in each rope if they are pulled to an angle of 45°? 45° 1 Kg

  7. Ex3) What is the tension in each rope if they are pulled to an angle of 45°? 45° Fg Fnet = Fg – 2FTy 0 = 10N – 2(FT ∙ sinθ) 0 = 10N – 2(FT ∙ sin45°) FT = 7N FTyFTy FTFT FT FTy 1 Kg 45° FTx FTy= FT ∙ sinθ FTx= FT ∙ cosθ

  8. Ex 4) A 168 N sign is held by 2 ropes that make 22˚ angles as shown. What is the tension in each? 22˚ 22˚ OPEN

  9. Ex 4) A 168 N sign is held by 2 ropes that make 22˚ angles as shown. What is the tension in each? FT FTy FTy FT Fnet = Fg − 2FTy 0 = 168N – 2(FT ∙ sinθ) 0 = 168N – 2(FT ∙ sin22˚) FT = 224N 22˚ 22˚ OPEN Fg

  10. Ex 5) A 168N sign is held by 2 ropes as shown. What is the tension in rope 2 if FT1 = 86N? FT2 FT1 = 86N 20˚ 60˚ Open

  11. Ex 5) A 168N sign is held by 2 ropes as shown. What is the tension in rope 2 if FT1 = 86N? FTy2 FTy1 Fg Ch7 HW#1 1 – 5 (Sep paper) FT2 FT1 = 86N Fnet = Fg− FT1y− FT2y 0 = 168N − FT1∙sinθ1 − FT2∙sinθ2 0 = 168N − 86N∙sin20˚ − FT2∙sin60˚ FT2 = 160N 20˚ 60˚ Open

  12. Lab7.1 – Vector Addition - due in tomorrow - Ch7 HW#1 due at beginning of period

  13. Ch7 HW#1 1 – 5 What is the tension in each rope? Fg FT FT 20 kg

  14. Ch7 HW#1 1 – 5 1. What is the tension in each rope? Fnet = Fg – 2FT 0 = 200N – 2FT Ft = 100N Fg FT FT 20 kg

  15. 2. What is the tension in each rope? 45° Fg FTyFTy FTFT 20kg

  16. 2. What is the tension in each rope? Fnet = Fg − 2FTy 0 = 200N – 2(FT ∙ sinθ) 0 = 200N – 2(FT ∙ sin45˚) 45° FT = 141N Fg FTyFTy FTFT 20kg

  17. 3. What is the tension in each rope? Fg FTyFTy FTFT 50kg

  18. 3. What is the tension in each rope? Fnet = Fg − 2FTy 0 = 500N – 2(FT ∙ sinθ) 0 = 500N – 2(FT ∙ sin15˚) FT = 965N Fg FTyFTy FTFT 50kg

  19. 4. What is the tension in each rope? 45° 60° Fg FTy1 FTy2 FT1 =259N FT2 50kg

  20. 4. What is the tension in each rope? Fnet = Fg − FTy1 – FTy2 0 = 500N – (FT1 ∙ sinθ) – (FT2 ∙ sinθ) 0 = 500N – (259N ∙ sin45°) – (FT2 ∙ sin60°) 45° 60° FT2 = 366N Fg FTy1 FTy2 FT1 =259N FT2 50kg

  21. 4. What is the tension in each rope? 40° 50° Fg FTy1 FTy2 FT1 =400N FT2 75kg

  22. 4. What is the tension in each rope? Fnet = Fg − FTy1 – FTy2 0 = 750N – (FT1 ∙ sinθ) – (FT2 ∙ sinθ) 0 = 500N – (400N ∙ sin40°) – (FT2 ∙ sin50°) 40° 50° FT2 = 643N Fg FTy1 FTy2 FT1 =400N FT2 75kg

  23. Ch7.2 – Inclined Planes

  24. Ch7.2 – Inclined Planes FN FN = Fg Fg θ

  25. Ch7.2 – Inclined Planes FN FN = Fg Fg FN Fg|| = Fg.sinθ θ Fg||Fg ┴ = Fg.cosθ Fg┴ FgFg ┴ = FN Fg|| θ

  26. Ex1) A 100N block is placed on a 40˚ inclined plane. Find Fg ┴ and Fg|| . 40°

  27. Ex1) A 100N block is placed on a 40˚ inclined plane. Find Fg ┴ and Fg|| . FN Fg|| Fg┴ Fg Fg ┴ = Fg.cosθFg|| = Fg.sinθ = 100N.cos40° = 100N.sin40° = 77N = 64N 40°

  28. Ex2) A 50.0kg mass is placed on a 25° inclined plane. What is the magnitude of the force tending to make it slide down the incline? What is the magnitude of the normal force acting on it? 25°

  29. Ex2) A 50.0kg mass is placed on a 25° inclined plane. What is the magnitude of the force tending to make it slide down the incline? What is the magnitude of the normal force acting on it? a. Fg|| = Fg.sinθ = 500N.sin25° FN = 211N b. FN = Fg ┴ = Fg.cosθ Fg|| = 500N.cos25° = 453N Fg┴ Fg 25°

  30. Ex3) A 20.0kg mass is placed on a frictionless inclined planed angled at 42°. What is its acceleration down the incline? 42°

  31. Ex3) A 20.0kg mass is placed on a frictionless inclined planed angled at 42°. What is its acceleration down the incline? FN Fg|| Fg┴ Fg Fnet = Fg|| m.a = Fg.sinθ m.a = m.g.sinθ (mass cancels out of the equation, didn’t Galileo already tell us mass doesn’t a = g.sinθ affect accl?) a = (9.8m/s2).sin42° a = 6.6 m/s2 42°

  32. Ex4) A 35.0kg mass is placed on an inclined plane. The angle is increased until the mass is just about to slip. The maximum angle reached before it slips is 45°. What is the coefficient of friction between the mass and plane? 45°

  33. Ex4) A 35.0kg mass is placed on an inclined plane. The angle is increased until the mass is just about to slip. The maximum angle reached before it slips is 45°. What is the coefficient of friction between the mass and plane? FN Ff,s Fg|| Fg┴ Fg Fnet = Fg|| – Ff,s m.a = Fg.sinθ – μs.FN 0 = Fg.sinθ – μs.Fg┴ 0 = Fg.sinθ – μs. Fg.cosθ 0 = sinθ – μs.cosθ Ch7 HW#2 6 – 9 45°

  34. Ch7 HW#2 6 – 9 A 91N block is placed on a 35˚ inclined plane. Find Fg ┴ and Fg|| . Fg Fg ┴ = Fg.cosθFg|| = Fg.sinθ 35°

  35. Ch7 HW#2 6 – 9 A 91N block is placed on a 35˚ inclined plane. Find Fg ┴ and Fg|| . Fg Fg ┴ = Fg.cosθFg|| = Fg.sinθ = 91N.cos35° = 91N.sin35° = 52N = 75N 35°

  36. 7) A 25kg mass is placed on a 50° inclined plane. What is the magnitude of the force tending to make it slide down the incline? What is the magnitude of the normal force acting on it? a. Fg|| = Fg.sinθ b. FN = Fg┴ 50°

  37. 7) A 25kg mass is placed on a 50° inclined plane. What is the magnitude of the force tending to make it slide down the incline? What is the magnitude of the normal force acting on it? a. Fg|| = Fg.sinθ = 250N.sin50° FN = 192N b. FN = Fg┴ = Fg.cosθ Fg|| = 250N.cos50° Fg┴ = 161N Fg 50°

  38. 8. A 12kg mass is placed on a frictionless inclined planed angled at 32°. What is its acceleration down the incline? Fg 32°

  39. 8. A 12kg mass is placed on a frictionless inclined planed angled at 32°. What is its acceleration down the incline? FN Fg|| Fg┴ Fg Fnet = Fg|| m.a = Fg.sinθ m.a = m.g.sinθ a = g.sinθ a = (9.8m/s2).sin32° a = 5.3 m/s2 32°

  40. 9. A 50kg mass is placed on an inclined plane. The angle is increased until the mass is just about to slip. The maximum angle reached before it slips is 35°. What is the coefficient of friction between the mass and plane? Fg 35°

  41. 9. A 50kg mass is placed on an inclined plane. The angle is increased until the mass is just about to slip. The maximum angle reached before it slips is 35°. What is the coefficient of friction between the mass and plane? FNFf,s Fg|| Fg┴ Fg Fnet = Fg|| – Ff,s m.a = Fg.sinθ – μs.FN 0 = Fg.sinθ – μs.Fg┴ 0 = Fg.sinθ – μs. Fg.cosθ 0 = sinθ – μs.cosθ 35°

  42. Ch7.2 cont – More Incline Planes Ex5) A trunk weighing 562N is on a 30° incline plane. a. If there’s no friction, what is its accl down the incline? Fg b. How fast will it be going after 4 sec? 30°

  43. Ch7.2 cont – More Incline Planes Ex5) A trunk weighing 562N is on a 30° incline plane. a. If there’s no friction, what is its accl down the incline? FN Fnet = Fg|| m.a = Fg.sinθ m.a = m.g.sinθ a = g.sinθ Fg|| a = (9.8m/s2).sin30° Fg┴ a = 5 m/s2 Fg b. How fast will it be going after 4 sec? vi = 0m/s vf = ? t = 4s a = _____ 30°

  44. Ch7.2 cont – More Incline Planes Ex5) A trunk weighing 562N is on a 30° incline plane. a. If there’s no friction, what is its accl down the incline? FN Fnet = Fg|| m.a = Fg.sinθ m.a = m.g.sinθ a = g.sinθ Fg|| a = (9.8m/s2).sin30° Fg┴ a = 5 m/s2 Fg b. How fast will it be going after 4 sec? vi = 0m/s vf = ? t = 4s a = 5 m/s2 vf = vi + a.t = 0 + (5m/s2)(4s) = 20m/s 30°

  45. c. If the trunk were placed on an incline where the coefficient of kinetic friction was 0.15, what would be the acceleration of the trunk? FN Ff,k Fg|| Fg┴ Fg 30°

  46. c. If the trunk were placed on an incline where the coefficient of kinetic friction was 0.15, what would be the acceleration of the trunk? FN Fnet = Fg|| – Ff,k Ff,km.a = Fg.sinθ – µk∙FN m.a = m.g.sinθ – µk∙Fg┴ m.a = m.g.sinθ – µk∙ m.g.cosθ Fg|| a = (9.8m/s2).sin30° Fg┴ – (.15)∙(9.8m/s2).cos30° Fg a = 3m/s2 30°

  47. Ex6) A 50kg wheel barrow is pushed up a 15° incline at constant speed. If the total friction against the wheel barrow is 100N, what force is required to push it? 15° Ch7 HW #3 10 – 12

  48. Ex6) A 50kg wheel barrow is pushed up a 15° incline at constant speed. If the total friction against the wheel barrow is 100N, what force is required to push it? F FN Fg|| Ff,k Fg┴ Fg Fnet = F – Fg|| – Ff 0 = F – Fg.sinθ – 100N 0 = F – 500N.sin15° – 100N F = ____N 15° Ch7 HW #3 10 – 12

  49. Lab7.2 – Inclined Planes - due tomorrow - Ch7 HW#3 10 – 13 due at beginning of period

  50. Ch7 HW#3 10 – 12 A 6kg block is placed on a 10˚ inclined plane. Find Fg ┴ and Fg|| . Fg Fg ┴ = Fg.cosθFg|| = Fg.sinθ 10°

More Related