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Chapter 9 Statics

Chapter 9 Statics. Two 50 lb. children sit on a see-saw. Will they balance in all three cases?. Review of Forces: Ex 1. A dentist places braces on a person’s teeth that exert 2.00 N in each direction as shown. Calculate the net force on the teeth.

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Chapter 9 Statics

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  1. Chapter 9 Statics

  2. Two 50 lb. children sit on a see-saw. Will they balance in all three cases?

  3. Review of Forces: Ex 1 A dentist places braces on a person’s teeth that exert 2.00 N in each direction as shown. Calculate the net force on the teeth.

  4. Look at each force separately and resolve it into x and y components. Fx = (2.00 N)(cos 20o) = 1.88 N Fy = (2.00 N)(sin 20o) = 0.684 N For the other force: Fx = -1.88 N Fy = 0.684 N 20 o 70 o 2.0 N

  5. The x-components cancel and the y-components add SFx = 1.88 N – 1.88 N = 0 SFy = 0.684 N + 0.684 N = 1.37 N 2.0 N 2.0 N Fnet = 1.37 N

  6. Review of Forces: Ex 2 Calculate the force exerted by the traction device shown below.

  7. Calculate the x and y-components: Top force: Fx = (200 N)(cos 37o) = 160 N Fy = (200 N)(sin 37o) = 120 N Bottom force: Fx = (200 N)(cos 37o) = 160 N Fy = (200 N)(sin 37o) = 120 N (downward) SFy = 120 N –120 N = 0 SFx = 160 N + 160 N = 320 N Why doesn’t the patient slide out of the bed?

  8. Review of Torque Mr. Saba can’t quite budge this rock. How can he increase his torque so he can move it? • = FR

  9. Conditions for Static Equilibrium St = 0 SF = 0

  10. SF = 0: Ex 1 A 90.0 kg man cannot do a pull-up. His best efforts produce a scale reading of 23 kg. What force is he exerting? SFx = 0 SFy = FB + Fs – mg Since he is not moving SF = 0

  11. 0 = FB + FS – mg FB = mg – FS FB = (90 kg)(9.8 m/s2) – (23 kg)(9.8 m/s2) FB = 660 N

  12. SF = 0: Ex 2 Calculate the F1 and F2 assuming the chandelier is not moving. SFx = 0 SFy = 0 SFx = 0 = F2 – (F1)(cos 60o) SFy = 0 = (F1)(sin 60o) - mg

  13. 0 = F2 – (F1)(cos 60o) two eqns, two unknowns 0 = (F1)(sin 60o) – mg 0 = (F1)(sin 60o) – (200 kg)(9.8 m/s2) 0 = (F1)(0.866) – 1960 N 1960 N = (F1)(0.866) F1 = 2260 N 0 = F2 – (F1)(cos 60o) F2 = (F1)(cos 60o) F2 = (2263)(cos 60o) = 1130 N

  14. A 200 g mass is hung by strings as shown. Calculate the tension in string A and string B. (A = 1.13 N, B = 1.51 N)

  15. St = 0: Example 1 Two children sit on a see-saw as shown. The board of the see-saw has a mass of 2.00 kg centered at the pivot. Where should the 25.0 kg child sit so that they are in perfect balance? ? 2.50 m 30.0 kg 25.0 kg

  16. Let’s pick the pivot as the origin St = 0 0=(245 N)(x) – (294 N)(2.5 m) – (FN)(0) + (mg)(0) 0 = (245 N)(x) – (294 N)(2.5 m) (245 N)(x) = 735 m-N x = 3.0 m FN 2.5 m x mg 294 N 245 N

  17. A crane lifts a 1000.0 kg car as shown in the picture below, 8.00 m from the pivot. Calculate the force that the motor of the crane must provide 2.00 m from the pivot. (39,200 N)

  18. A teacher (56.0 kg) sits at the end of a 4 m long see saw. Where should a 203 kg owlbear sit to perfectly balance the seesaw? (55 cm)

  19. SF = 0 and St = 0: Example 1 A heavy printing press is placed on a large beam as shown. The beam masses 1500 kg and the press 15,000 kg. Calculate the forces on each end of the beam.

  20. SFy = 0 and St = 0 (we’ll ignore Fx) SFy = 0 = F1 + F2 - (1500 kg)(g) – (15,000 kg)(g) 0 = F1 + F2 – 1.617 X 105 N F1 + F2 = 1.617 X 105 N We’ll choose F1 as the pivot St = 0 0 = (F1)(0 m) – (1500kg)(g)(10m) - (15,000 kg)(g)(15 m) + (F2)(20 m) 0 = -2.352 X 106 m-N + (F2)(20 m) (F2)(20 m) = 2.352 X 106 m-N

  21. F1 + F2 = 1.617 X 105 N (F2)(20 m) = 2.352 X 106 m-N (F2) = 2.352 X 106 m-N/(20m) = 1.176 X 105 N F1 + 1.176 X 105 N = 1.617 X 105 N F1 = 4.41 X 104 N

  22. SF = 0 and St = 0: Example 3 The beam below has a mass of 1200 kg. Calculate F1 and F2 for the cantilever as shown. Assume the center of gravity is at 25 m. F1 F2 20.0 m 30.0 m (1200 kg)(g)

  23. SFy = 0 and St = 0 SFy = 0 = F1 + F2 - (1200 kg)(g) 0 = F1 + F2 – 11,760 N F1 + F2 = 11,760 N We’ll choose F1 as the pivot St = 0 0 = (F1)(0 m) + (F2)(20m) - (11,760 N)(25 m) 0 = (F2)(20m) – 294,000 N (F2)(20m) = 294,000 N

  24. F1 + F2 = 11,760 N (F2)(20m) = 294,000 N F2 = 294,000 N/20 m = 14,700 N F2 = 14,700 N F1 = 11,760 N - F2 F1 = 11,760 N - 14,700 N F1 = -2940 N (we picked the wrong direction for Force 1)

  25. The following system is completely in balance. Calculate the mass of Mass C and the value of the force at the support point (S). (217 g, 3.34 N)

  26. SF = 0 and St = 0: Example 4 A sign of mass M = 280 kg is suspended from a 25.0 kg beam that is 2.20 m long. The angle between the sign and the wire is 30o. Calculate FH and FT, the forces at the hinge and the tension in the wire.

  27. SFx = 0 0 = FHcosq - FTcos30o 0 = FHcosq – 0.866FT SFy = 0 0 = FHsinq + FTsin30o – (25 kg)(g) – (280 kg)(g) 0 = FHsinq + 0.5FT – 2989 N

  28. St = 0 (choose hinge as origin) 0 = (25kg)(g)(1.1m) + (280kg)(g)(2.2m) – (FT)(sin30o)(2.2m) T hree equations, three unknowns 0 = FHcosq – 0.866FT 0 = FHsinq + 0.5FT – 2989 N 0 = 6306 – 1.1FT FT = 5730 N, FH = 4960 N, q = 1.43o

  29. SF = 0 and St = 0: Example 5 Calculate the force on the hinge and the tension in the cord for the following sign. The beam is 2.00 m long, though the sign is hung at 1.80 m.

  30. SFx = 0 0 = FHcosq – FTcos20o • = FHcosq – 0.940FT SFy = 0 0 = FHsinq + FTsin20o – (20 kg)(g) – (150 kg)(g) 0 = FHsinq + 0.342FT– 1666N

  31. St = 0 (choose hinge as origin) 0 = (20kg)(g)(1m) + (150kg)(g)(1.8m) – (FT)(sin30o)(2m) T hree equations, three unknowns • = FHcosq – 0.940FT 0 = FHsinq + 0.342FT– 1666N 0 = 2842 – 0.684FT FT = 4155 N, FH = 3914 N, q = 3.59o

  32. SF = 0 and St = 0: Example 6 5.0 m 15.0 m A crane is designed to hold a maximum of 10,000 kg (M). The top crossbeam has a mass of 450 kg. Calculate the forces exerted by the post (Fp) and the angled beam (FA). 35o FA Fp M

  33. FAx q Fpy Fp FAy FA Fpx

  34. SFx =0 0 = FAcos35o - FPsinq 0 = 0.819FA - FPsinq SFy =0 0 = FA sin35o – Fp cosq – mg – Mg 0 = FA sin35o – Fp cosq – (450 kg)(g) –(10,000kg)(g) 0 = 0.574FA – Fp cosq – 1.02 X 105 N

  35. St = 0 (choose top of post as origin) 0 = (450 kg)(g)(10m) – (10,000 kg)(g)(20m) + (FA)(sin35o)(15m) 0 = 2.00 X 106 – 8.6 FA Three Equations, three unknowns 0 = 0.819FA - FPsinq 0 = 0.574FA – Fp cosq – 1.02 X 105 N 0 = 2.00 X 106 – 8.6 FA FA = 2.34 X 105 N, Fp = 1.95 X 105 N, q = 80.7o

  36. The Ladder: Example 1 A 5-m long ladder leans against a wall at a point 4 m above the ground. The ladder has a mass of 12.0 kg and is uniform. Assume the wall is frictionless, but the ground is not. Calculate the force from the wall (Fw) and the force from the ground (FG).

  37. 5 m Working with the Triangle 52 = 42 + x2 x2 = 52 – 42 x = 3 sin ? = 4/5 ? = 53o 4 m ? x 37o 5 m 4 m 53o 3 m

  38. SFx = 0 0 = FGx – Fw 0 = Fgcosq - Fw SFy = 0 0 = FGy – mg 0 = Fgsinq - 118

  39. Fw 53o 37o Back to the ladder Tilting the Ladder FG 37o 53o mg Fw FG 53o 37o mg

  40. Fw FG Fwsin53o Calculating the Torque forces Choose the point at ground as the pivot St = 0 0 = (5m)(Fw)(sin53o) – (2.5m)(mg)(sin 37o) 0 = 3.99Fw - 177 m-N 53o 37o mgsin37o mg

  41. Three equations, three unknowns 0 = Fgcosq - Fw 0 = Fgsinq - 118 0 = 3.99Fw - 177 q = 69.3o Fg = 126 N Fw = 44.3 N

  42. The Ladder: Example 2 Mr. Fredericks (56.0 kg) leans a 3.00 m, 20.0 kg ladder against his house at an angle of 65.0o with the ground. He can safely climb 2.50 m up the ladder before it slips. Calculate the coefficient of friction between the ground and the ladder.

  43. SFx = 0 0 = Fgcosq - Fw SFy = 0 0 = Fgsinq – (20.0 kg)(g) – (56.0 kg)(g) 0 = Fgsinq – 744.8

  44. Choose the point at ground as the pivot St = 0 0 = (20)(g)(1.5)(sin25o) + (56)(g)(2.5m)(sin 25o) - (FW)(3)(sin 65o) 0 = 704 – 2.72Fw

  45. Three equations, three unknowns 0 = Fgcosq - Fw 0 = Fgsinq – 744.8 0 = 704 – 2.72Fw FW = 259 N, Fg = 788 N, q = 70.8o

  46. Dealing with Friction Fgx = Fgcos 70.8o Fgx = 259 N = Ffr Fgy = Fg sin 70.8o Fgy = 744 N = FN Ffr = mFN 259 = m744 m = 0.35

  47. The Ladder: Example 3 A man who weighs 800 N climbs to the top of a 6 meter ladder that is leaning against a smooth wall at an angle of 60°. The non-uniform ladder weighs 400 N and its center of gravity is 2 meters from the foot of the ladder. What must be the minimum coefficient of static friction between the ground and the foot of the ladder if it is not to slip?

  48. SFx = 0 0 = Fw - Fgcosq SFy = 0 0 = Fgsinq – 800 – 400 0 = Fgsinq – 1200

  49. Fw 60o Working with the Triangle Tilt the Ladder 30o 800 N 30o 400 N FG 60o Fw FG 60o 30o 30o 800 N 400 N

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