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FRICTION

FRICTION. Friction - a force that resists the motion of an object. Acts parallel to the surface and opposite the direction of the motion. Dependent on the type of contact surfaces. Independent of contact area. Equal to applied force when object is at rest or traveling at a constant velocity.

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FRICTION

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  1. FRICTION

  2. Friction- a force that resists the motion of an object • Acts parallel to the surface and opposite the direction of the motion. • Dependent on the type of contact surfaces. • Independent of contact area. • Equal to applied force when object is at rest or traveling at a constant velocity.

  3. TYPES OF FRICTION • Static friction(starting friction)- Friction that is produced by surface projections and bonding. • Kinetic friction(sliding friction)- friction an object has while in motion. • Rolling friction-one object rolls over another • Air resistance- friction produced by the air pushing against something in free fall.

  4. FORMULA Ff = µ FN Ff = frictional force (N) FN = normal force (N) µ-Coefficient of friction *(µ) Has No Units* *See reference table*

  5. ExA. A block of wood rests on a wood desktop. If it has a mass of 10kg, what is the minimum force required to move it? FN = Fg = mg FN = 10kg(9.81m/s2) FN = 98.1N Ff = µ FN = .42(98.1N) = 41.2N • All forces are balanced if at rest or traveling at a constant velocity. • The value for static friction is used because its at rest. FN= 98.1N Ff= 41.2N F= 41.2N Fg= 98.1N

  6. exB-If a 60N force is applied to the block, what will be its acceleration? Fnet = 60N – 41.2N Fnet = 18.8N a = Fnet/m = 18.8N/10kg a = 1.88 m/s2 Ff= 41.2N F = 60N

  7. FORCE VECTOR DIAGRAMS STATIONARY OBJECT: Fg- weight Fg = mg Fn – normal force- force pushing surfaces together Fn = Fg Fn Fg

  8. Applied Parallel force Fn Fp - Parallel force Ff – frictional force Fn – normal force Fg- weight- Fn Ff Fp Fg At constant velocity Fp = Ff

  9. Applied force at an angle Fa – applied force Fp – parallel force Fp = FacosΘ = Ff Fy = FasinΘ Fn = Fg - Fy Fn Fa Fy Fp Ff Fg= mg

  10. A worker pulls a 40kg crate across the floor at a constant velocity. If he applies 214N of force along a rope held at 30 degrees with the ground, find Fg , Fp, Fy , Fn , and μ. Fp = Facosθ = Ff Fp = 214N cos 30 Fp = 185.3N Fg = mg F = 40kg(9.81m/s2) Fg =392.4N Fy = Fa sin θ Fy = 214N sin 30 Fy = 107N

  11. Fn = Fg – Fy Fn = 392.4N – 107N Fn = 285.4N μ = Ff / Fn μ = 185.3N/ 285.4N μ = 0.65

  12. Sliding object on an incline Fg = mg (always perpendicular to the ground) Ff Fn Fn = Fgcosθ (perpendicular to the surface) Ѳ Fp = Ff ( both are parallel to the surface) Fp = Fgsinθ Fp Fn Fg

  13. Ex: A 20kg wooden box is sliding down an incline of 30 degrees at a constant velocity. Find Fg , Fp , Fn and μ. Fg = m g Fg = 20kg(9.81m/s2) Fg = 196N θ Fg Fn Fp Fn = FgcosѲ Fn= 196N cos 30 Fn = 170N Fp = Fg sin Ѳ Fp = 196N sin 30 Fp= 98N µ = Ff / Fn = 98N/170N = .576

  14. COOL MATH TRICK Sin Ѳ = opp/hypcosѲ= adj/hyp µ = Ff = FgsinѲ = opp/hyp = opp/adj Fn FgcosѲadj/hyp µ = tan Ѳ *Only works for objects sliding down an incline*

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