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. Air Resistance Is a resistance force that acts in the opposite direction of gravitational forces. Force of Gravity Air Resistance Help!! SPLAT!! Force of Gravity Air Resistance v 1 > v 2

4. Static and Kinetic Friction Forces that oppose motion between two surfaces that are touching each other. FxFx Fr F r (non-moving) = Static Friction (F s ) F r (moving) = Kinetic Friction (F k )

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

6. Tile Floor Carpet FkFk FkFk FxFx FxFx

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

8. Coefficient of Kinetic Friction μ k = F k /F n coefficient of kinetic friction μ s = F s, max /F n coefficient of static friction F f = μFn frictional force

9. Sample problem 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? F applied F k = 320 N m = 82 kg g = 9.81 m/s² w = mg w = (82 kg) (9.81 m/s²) w = 804 N w = Fn = 804 N μ k = F k /F n μ k = 320 N/804 N μk = 0.40 FnFn mg

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. F applied = 90 N FkFk mg FnFn 30° g = 9.81 m/s²m = 20.0 kg μk = 0.50 w = mg = 196 N F app,y 30° 90 N F app, x Step 1: Solve for Weight and X & Y components of applied force F app, y = (90N) (sin 30°) = 45.0 N F app, x = (90N) (cos 30°) = 77.9 N

12. Step 2: Find the Kinetic Friction ∑Fy = Fn + w + F app, y = 0 0 = Fn – 196 N + 45.0 N Fn = 196 N – 45 N = 151 N μ k = F k /F n F k = (μ k )(F n ) = (0.50)(151 N) = 75.5 N F app, y = 45.0 NF app, x = 77.9 Nμk = 0.50 w = 196 N F app,y 30° 90 N F app, x

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