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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 .

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100 90 80 70 vel 60

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  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°

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