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Warm-up

15°. Fn. mg. Fn. Fx. Fx. mg. Warm-up. Rose is sliding down an ice covered hill inclined at an angle of 15 ° with the horizontal. If Rose and the sled have a combined mass of 54.0 kg, what is the force pulling them down the hill? (neglect friction). m = 54 kg. g = 9.81 m/s ². θ = 15°.

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Warm-up

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  1. 15° Fn mg Fn Fx Fx mg Warm-up Rose is sliding down an ice covered hill inclined at an angle of 15° with the horizontal. If Rose and the sled have a combined mass of 54.0 kg, what is the force pulling them down the hill? (neglect friction) m = 54 kg g = 9.81 m/s² θ= 15° Sin θ = Fx/mg Fx = mg sin θ Fx = (54 kg)(9.81 m/s²) sin 15° θ = 15° Fx = 137 N

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

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

  4. Fx mg Fr (Air Resistance) Air Resistance • Can also be a horizontal force

  5. Fr Fa Static and Kinetic Friction • Are forces that oppose motion between two surfaces that are touching each other. Fr (non-moving) =Static Friction (Fs) Fr(moving) =Kinetic Friction (Fk) Fk = ∑F

  6. The force of friction is proportional to the size and mass of an object that you are pushing across a surface. Fk Fapplied Fk Fapplied The amount of friction that occurs between an object and the surface can depend on the type of surface that the object is in contact with.

  7. Fk Fk Fapplied Fapplied Tile Floor Carpet

  8. The quantity that expresses the dependence of frictional forces on the particular surface the object is in contact with is called thecoefficient of friction (µ). It is the ratio between the normal force and the force of friction between two surfaces.

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

  10. Practice 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

  11. Practice Problem 2 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? (Pg. 146)

  12. 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

  13. 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

  14. 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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