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Frictional Forces

Frictional Forces. Chapter 4, Section 4 Pg. 141-149. Friction. There are two types of frictional forces in the physical world that effect us. - Static/Kinetic Friction - Air Resistance. Force of Gravity. Force of Gravity. Air Resistance.

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Frictional Forces

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  1. Frictional Forces Chapter 4, Section 4 Pg. 141-149

  2. Friction • There are two types of frictional forces in the physical world that effect us. - Static/Kinetic Friction - Air Resistance

  3. Force of Gravity Force of Gravity Air Resistance Air Resistance SPLAT!! Air Resistance • Is a resistance force that acts in the opposite direction of gravitational forces. Help!! v1 > v2

  4. Fr Fx Static and Kinetic Friction • Forces that oppose motion between two surfaces that are touching each other. Fr (non-moving) =Static Friction (Fs) Fr(moving) =Kinetic Friction (Fk)

  5. Frictional force depends on the size and mass of the object moving across a surface. Fk Fx Fk Fx The amount of friction occurring between an object and the surface depends on the surface type.

  6. Fk Fk Fx Fx Tile Floor Carpet

  7. The value that expresses the dependence of frictional forces on the surface type the object is in contact with is called thecoefficient of friction (µ). It is the ratio between the force of friction and the normal force.

  8. Coefficient of Kinetic Friction μk = Fk/Fncoefficient of kinetic friction μs = Fs,max/Fncoefficient of static friction Ff = μFn frictional force

  9. Sample problem Fn Fk = 320 N Fapplied mg While redecorating her apartment, Suzy slowly pushes an 82 kg cabinet across a wooden dining room floor, which resists the motion with a force of friction of 320 N. What is the coefficient of kinetic friction between the cabinet and the floor? m = 82 kg g = 9.81 m/s² μk = Fk/Fn w = mg μk = 320 N/804 N w = (82 kg) (9.81 m/s²) w = Fn = 804 N μk = 0.40 w = 804 N

  10. Sample Problem 2 (Pg. 146) A student moves a box of books by attaching a rope to the box by pulling with a force of 90 N at an angle 30° to the horizontal. The box of books has a mass of 20.0 kg and the coefficient of kinetic friction between the bottom of the box and the sidewalk is 0.50. What is the acceleration of the box?

  11. Step 1: Solve for Weight and X & Y components of applied force Fn Fk 90 N Fapp,y Fapplied = 90 N mg 30° Fapp, x m = 20.0 kg g = 9.81 m/s² μk = 0.50 30° w = mg = 196 N Fapp, y = (90N) (sin 30°) = 45.0 N Fapp, x = (90N) (cos 30°) = 77.9 N

  12. Step 2: Find the Kinetic Friction 90 N Fapp,y 30° Fapp, x Fapp, y = 45.0 N μk = 0.50 Fapp, x = 77.9 N w = 196 N ∑Fy = Fn + w + Fapp, y = 0 0 = Fn – 196 N + 45.0 N Fn = 196 N – 45 N = 151 N μk = Fk/Fn Fk = (μk)(Fn) = (0.50)(151 N) = 75.5 N

  13. Step 3: Solve for acceleration 90 N Fapp,y 30° Fapp, x m = 20.0 kg Fk = 75.5 N ∑F = ma Fapp, x = 77.9 N a = ∑F/m a = (Fapp, x – Fk) / m a = (77.9 N – 75.5 N) / 20.0 kg a = 0.12 m/s² to the right

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