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P H Y S I C S Newton’s Laws Overview

P H Y S I C S Newton’s Laws Overview. Review of Newton’s Laws of Motion. 1 st. Objects in motion stay in motion* and objects at rest stay at rest if there is zero net force (balanced) Σ F = m · a (the forces will be unbalanced) Every force has an equal and opposite force

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P H Y S I C S Newton’s Laws Overview

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  1. P H Y S I C S Newton’s Laws Overview
  2. Review of Newton’s Laws of Motion 1st Objects in motion stay in motion* and objects at rest stay at rest if there is zero net force (balanced) ΣF = m·a (the forces will be unbalanced) Every force has an equal and opposite force * straight line/constant speed 2nd 3rd
  3. Inertia Depends on mass More mass  more resistance Less mass  less resistance
  4. Demo: NFL Hits
  5. Equilibrium Equilibrium: Net force is zero (ΣF = 0) ΣFx = 0 ΣFy = 0 FN FEngine FAir Fg
  6. Equilibrium ΣF = 0  Newton’s First Law applies An object in equilibrium can be: in motion (straight line/constant speed) at rest FN Ff Fengine Fair Fg
  7. Terminal Velocity Once the forces of air resistance and gravity become balanced equilibrium is reached No more acceleration
  8. Newton’s Second Law If there is a net force the object will accelerate ΣF = m·a Units: ΣF  net force (N) m  mass (kg) a  acceleration (m/s2)
  9. Newton’s Second Law Equations: F = ma a = F/m m = F / a F = net force m = Mass a= Acceleration
  10. Use one of the equations you just wrote down…
  11. Acceleration Increase acceleration by: Increasing force Decreasing mass
  12. Weight vs Mass Weight Force  Fg Fg= m·g Mass: Amount of matter (does not change) Weight: Pull of gravity (changes)
  13. Weight Force (Fg) g = 9.8 m/s2 g = 1.6 m/s2 g = 26 m/s2 m = 50 kg Fg = 490 N ( 110 lb) m = 50 kg Fg = 80 N ( 18 lb) m = 50 kg Fg = 1300 N ( 292 lb)
  14. In-Class Problem #1 A 2000 kg car has a push force of 5000 N from its engine. If it experiences a friction force of 3000 N determine it’s (a) acceleration, (b) weight and (c) the normal force acting on it. a = 1 m/s2 Fg = 19,600 N FN = 19,600 N
  15. Review of Newton’s Laws of Motion ΣF = 0 ΣF ≠ 0 First Law Second Law a = 0 m/s2 Accelerates at rest in motion* depends on net force depends inversely on mass stays at rest stays in motion* * Straight line/constant speed
  16. Friction Force that resists motion due to imperfections in surfaces MOTION FRICTION
  17. Two Types Static (rest): Keeps object from moving Kinetic (moving): Slows moving object
  18. Friction Force Equation Coefficient of Friction (): Ratio between friction force and normal force:  friction force (N)  coefficient of friction  normal force (N) μs (static) μk (kinetic)
  19. Coefficient of Friction Table
  20. In-Class Problem #2 A 30 kg desk is at rest on the floor. It takes 200 N of force to start it in motion. Determine the static coefficient of friction between the desk and the floor. μs = 0.68
  21. In-Class Problem #3 Once the desk in the previous problem is set in motion the 200 N force continues to be applied. Determine the acceleration of the desk if the coefficient of kinetic friction is 0.52. a = 1.57 m/s2
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