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Chapter 7

Chapter 7. Forces and Motion In Two Dimensions. Equilibrium. An object is in equilibrium when the Net Force on the object is zero. F Net = 0 Acceleration = 0 Constant Velocity Velocity = 0. Equilibrium in 2 Dimensions. F x = 0 F y = 0. Static Equilibrium.

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Chapter 7

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  1. Chapter 7 Forces and Motion In Two Dimensions

  2. Equilibrium An object is in equilibrium when the Net Force on the object is zero. • FNet = 0 • Acceleration = 0 • Constant Velocity • Velocity = 0

  3. Equilibrium in 2 Dimensions • F x = 0 • F y = 0

  4. Static Equilibrium Objects are not moving. • FNet = 0 • Acceleration = 0 • Velocity = 0

  5. 20N 20N 50N 50N Static Equilibrium FNet = 0 Fx = 0 Fy = 0

  6. 60° 60° 2kg

  7. F1 F2 2kg Fg = mg Fg = 20N Fg = 2kgx9.8m/s2 Fg = 20N

  8. x direction F1 F2 Fx = -F1x + F2x 0 = -F1x + F2x 60° 60° F1x = F2x F1cos60°= F2cos 60° F1 = F2 = F Fg = 20N

  9. y direction F1 F2 Fy = F1y + F2y - Fg 0 = F1y + F2y - Fg 60° 60° Fg = F1y + F2y Fg = F1sin60°+ F2sin 60° Fg = Fsin60°+ Fsin 60° Fg = 20N Fg = 2Fsin60°

  10. __Fg__ 2sin60° _20N__ 2sin60° = F = F Fg = 2Fsin60° 11.5N = F

  11. Homework Finish Work Sheet

  12. Ft Fg FP 50kg

  13. Ft Fg FP 50kg

  14. Ftx Fty Ft FP 30° Fg = mg Fg = 50kg·9.8m/s2 Fg Fg = 500N

  15. X - Components Y- Components Fg sin30° Ft = Fy = Fpy - Fg Fx = Fpx - Ftx 0= Fp - Ftx 0= Fpy - Fg Fpx = Ftx Fty = Fg Ftsin30° = Fg Fp = Ftcos30°

  16. X - Components Y- Components Fg sin30° 500N sin30° Ft = Ft = Fp = Ftcos30° Fp = 1000cos30° Fp = 866N Ft = 1000N

  17. Homework Finish Work Sheet

  18. FN Fg Incline Plane FN=Fg

  19. FN Ff Fg Incline Plane θ

  20. FN Ff θ Fg

  21. Fy =FN – Fgcosθ Fx =Ff – Fgsinθ FN = Fgcosθ FN Ff θ Fg

  22. FN Ff Fg θ

  23. FN Ff θ Fg θ

  24. Fy =FN – Fgcosθ Fx =Ff – Fgsinθ Fgx Fgy FN = Fgcosθ FN θ θ Ff Fg

  25. Problem In a block/inclined plane system, the inclined plane makes an angle of 60° with the ground. The coefficient of friction is 0.5. If the block has a mass of 1.02kg, what is the net force on the block? What is the blocks acceleration?

  26. FN Ff Fg Incline Plane 1.02kg Ff=μFN θ=60° =10N Fg=mg Fg=1.02kg·9.8m/s2

  27. FN Ff θ Fg θ

  28. Fy =FN – Fgy Fx =Ff – Fgx Fx =Ff – Fgsinθ Fy =FN – Fgcosθ FN Fgx Ff Fgy 60° Fg

  29. Fy = FN – Fgcosθ 0 = FN – Fgcosθ FN = Fgcosθ FN = 10cos60° FN = 5N

  30. Fx =Ff – Fgsinθ Fx =μFN – Fgsinθ Ff = μFN Fx =0.5·5N– 10N·sin60° Fx =2.5– 8.7N Fx = –6.2N Net Force of 6.2N down the incline!!

  31. FNet m = a 6.2N 1.02kg = a FNet = ma 6.1m/s2= a 6.1m/s2 down the incline!!

  32. Projectile Motion

  33. Projectile Motion

  34. Horizontal Component Vertical Component Projectile Motion

  35. Projectile Motion Projectile motion is the combination of two independent motions, the motion in the x direction and the motion in the y direction. These two motions are usually independent of each other.

  36. x component Projectile Motion y component

  37. Problem Solving Strategy 1. Break up the problem into two interconnected one-dimensional problems. y component x component

  38. Problem Solving Strategy 2. Vertical motion (y component) is exactly that of an object being dropped or thrown straight up or down. (g - gravity!!!!!)

  39. Problem Solving Strategy 3. Horizontal motion (x component) is the same as solving constant velocity problems.

  40. Problem Solving Strategy 4. Vertical (y) and horizontal (y) components are connected by the variable time (t). Solving for time in one dimension, x or y, automatically gives you the time for the other dimension.

  41. *Basic Equations* v = v0 + at d = d0 +1/2(v+v0)t d = d0 +v0t +½at2 v2 = v02 +2a(d-d0)

  42. Problem: A ball is kicked horizontally, with a velocity of 25m/s, off a 122.5m high cliff. How far from the cliff did the ball land?

  43. v = 25 m/s y component y = 122.5 m x component Sketch the Problem

  44. 2(-122.5m) (-9.8m/s2) y component (up is +) d = d0 +v0t +½at2 d = ½at2 -122.5m = ½(-9.8m/s2)t2 = t2

  45. 2(-122.5m) (-9.8m/s2) y component (up is +) √ = t 5.0s = t Use this to solve for distance in the x direction!!!

  46. x component d = d0 + v0t +½at2 d = v0t d = (25m/s)(5s) d = 125m

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