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  2. 2. Static and Kinetic Friction
  3. 3. What is Friction?  Force that acts oppose the relative motion of two surfaces  High for dry and rough surfaces  Low for smooth and wet surfaces
  4. 4. What Causes Friction?What Causes Friction?  Friction is the force that opposes the motion between two surfaces that touch.  The surface of any object is rough.  Even an object that feels smooth is covered with tiny hills and valleys.  The contact between the hills of valleys of two surfaces causes them to stick, resulting in friction.
  5. 5.  The amount of friction depends on:  Roughness of the surfaces  Force pushing the surfaces together What Causes Friction?What Causes Friction?
  6. 6.  Kinetic friction occurs when force is applied to an object and the object moves.  Examples: Sliding Friction: pushing an object across a surface Rolling Friction: between wheels and a surface Fluid Friction: opposes the motion of objects traveling through a fluid (air or water) Types of FrictionTypes of Friction
  7. 7.  Static friction occurs when force applied to an object does not cause the object to move. Types of FrictionTypes of Friction
  8. 8. Free Body Diagram Normal Force FN Friction Force ff Applied Force F Gravity Force Fg Fg = mg FN = Fg ff = F
  9. 9. Static Friction frictionstaticoftcoefficien Ff s sNs = ×= µ µ FN fs F Fg The Force of Static Friction keeps a stationary object at rest!
  10. 10. Kinetic Friction frictionkineticoftcoefficien Ff k kNk = ×= µ µ FN fk F Fg Once the Force of Static Friction is overcome, the Force of Kinetic Friction is what slows down a moving object! Motion
  11. 11. Types of Friction I better be safe Ump!! To initiate motion of the box the man must overcome the Force of Static Friction Upon sliding, the baseball player will come to a complete stop due to the Force of Kinetic Friction
  12. 12. Static & Kinetic Friction Coefficients Material Coefficient of Static Friction µS Coefficient of Kinetic Friction µS Rubber on Glass 2.0+ 2.0 Rubber on Concrete 1.0 0.8 Steel on Steel 0.74 0.57 Wood on Wood 0.25 – 0.5 0.2 Metal on Metal 0.15 0.06 Ice on Ice 0.1 0.03 Synovial Joints in Humans 0.01 0.003
  13. 13. PROBLEMS A 75 kg crate is to be pushed up an incline plane 5 m long that makes an angle of 20° with the horizontal. If the coefficient of static friction between the crate and the inclined plane is 0.20, how much force must be given to get it started up the incline? If the coefficient of kinetic friction is 0.15, how much applied force is needed to keep it going at a constant speed up the incline.
  14. 14. Given: m=75 kg; l=5 m; θ=20°; μs=0.20; μk=0.15 Fw=mg Ff Fn=mg(cosθ) Fa θ Fp=mg(sinθ)
  15. 15. )(sin )(cos θ θµµ mgF mgFFFFF p snssfpsfa = ==+= N251.38 N13.138 = = neededisforcemuchthisLEASTATd.threshholtheisthis:note* N51.389=aF )(sin )(cos θ θµµ mgF mgFFFFF p snksfpkfa = ==+= N251.38 N60.103 = = speed.CONSTANTforisThis:note* N98.354=aF
  16. 16. Take into account the last problem. After you have the crate moving up the incline at a constant speed, you want to give it an acceleration of 1.7 m/s2 . What is the new force needed to accomplish this task?
  17. 17. )(sin )(cos θ θµµ mgF mgFFmaFFF p snkkfpkfa = ===−− N251.38 N60.103 = = N48.482=aF maFFF pkfa ++=
  18. 18. A 300 kg sled is pulled at constant speed over a level, horizontal, snow covered surface. The rope that is used to pull the sled makes a 30° angle with the horizontal. If the coefficient of friction is 0.10, find the force required.
  19. 19. Given: m=300 kg; θ=30°; μk=0.10 Fn Fx (cos θ) Ff Fw Fa θ Fy (sin θ) Fax-Ff=max *note: ax is 0 because we are at a constant speed Fay+Fn-Fw=may *note: ay is 0 because the sled never moves up and down

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  • NyageNyori1

    May. 12, 2017

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