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Drawing Pictures

Drawing Pictures. 2.1.2A Free Body Diagrams and Net Force. Free Body Diagrams. Equilibrium. VECTOR diagrams Show ALL FORCES acting on an object Must be properly labeled. F N. F g. Determining Net Force. equilibrium  net force = 0  acceleration = 0

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Drawing Pictures

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  1. Drawing Pictures 2.1.2A Free Body Diagrams and Net Force

  2. Free Body Diagrams Equilibrium • VECTOR diagrams • Show ALL FORCES acting on an object • Must be properly labeled FN Fg

  3. Determining Net Force • equilibrium  net force = 0  acceleration = 0 • MOTIONLESS or CONSTANT VELOCITY • net force ≠ 0 • SPEEDING UPor SLOWING DOWN

  4. Determining Net Force – Top View • Consider each dimension separately then combine using Pythagorean Theorem. 2 kg object X Y +7.0 N F1 = +7.0 N +7.0 N 0 N Net force = +7.0 N a = 3.5 m/s2 Not in notes

  5. Determining Net Force – Top View • Consider each dimension separately then combine using Pythagorean Theorem. 2 kg object X Y +7.0 N -3.0 N F1 = +7.0 N F2 = -3.0 N +4.0 N 0 N Net force = +4.0 N a = 2.0 m/s2 Not in notes

  6. Determining Net Force – Top View • Consider each dimension separately then combine using Pythagorean Theorem. 2 kg object X Y F3 = +3.0 N +7.0 N -3.0 N +3.0 N F1 = +7.0 N F2 = -3.0 N +4.0 N +3.0 N Net force = 5.0 N a = 2.5 m/s2 Not in notes

  7. Determining Net Force – Top View • Consider each dimension separately then combine using Pythagorean Theorem. 2 kg object X Y F3 = +3.0 N +7.0 N -3.0 N +3.0 N -5.0 N F1 = +7.0 N F2 = -3.0 N +4.0 N -2.0 N Net force = 4.5 N a = 2.25 m/s2 F4 = -5.0 N Notes

  8. Example #1 • A 30 newton force and a 20 newton force act concurrently on an object. • What are the minimum/maximum net force can these two forces produce? At what angle between the forces does each happen? • What net force is produced when the angle between the two forces is 90°? min = 10 N occurs at 180° max = 50 N occurs at 0° 36 N

  9. Determining Net Force – Side View • Consider each dimension separately. • On a flat surface total vertical force = 0. 2 kg object X Y +8.0 N -Fg +FN F = +8.0 N +8.0 N 0 N frictionless Net force = +8.0 N a = 4.0 m/s2 Not in notes

  10. Determining Net Force – Side View • Consider each dimension separately. • On a flat surface total vertical force = 0. 2 kg object Hor. Vert. +8.0 N -2.0 N -Fg +FN F = +8.0 N Ff = -2.0 N +6.0 N 0 N Net force = +6.0 N a = 3.0 m/s2 Notes

  11. Example #2 • A 50 kilogram object is pushed along a flat, frictionless surface with a constant force of 100 newtons. • Sketch this object. Include all forces with labels and quantities. • What rate of acceleration will the crate experience?

  12. Determining Net Force – Vertical • Object moving straight up or down  total horizontal force = 0. 20 N object in freefall Hor. Vert. -20 N 0 N -20 N Net force = -20 N a = -9.81 m/s2 Fg = -20 N Not in notes

  13. Example #3 • What is the net force acting on a 3000 newton rocket if its engine produces an upward thrust of 4500 newtons? Fthrust 1500 N upward Fg

  14. End of 2.1.2A - PRACTICE

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