2. Friction is a force that opposes the motion between two surfaces that are in contact  is a force that opposes the motion between two surfaces that are in contact  There are two types  STATIC friction: opposes the start of motion between two surfaces that are not in relative motion.  KINETIC friction: the force between two surfaces that are in relative motion. It always acts in the opposite direction to the motion of the object. Static friction is ALWAYS bigger than kinetic friction

4. Flat Surface Friction Ex: 1 A westward force of 50 N is applied to a 50 kg mass resting on a surface that will exert a frictional force of 16 N. If the force is applied for 5 seconds, what is the resulting velocity? F net = ma 34 N [W] = (50kg)(a) a = 0.68 m/s 2 a = v f – v i t Fnet = Fa + Ff Fnet = 50 N [W] + 16 N [E] Fnet = 34 N [W] 3.4 m/s [W]

5. Flat Surface Friction Ex 2: A 30 kg mass is dragged over a surface with a frictional force of 12 N by pulling on a rope attached to the mass. The rope makes an angle of 30 o to the horizontal and the applied force is 60 N. What is the resulting acceleration? Vertical  No acceleration Horizontal F net = F f + F a(h) F net = -12 N + (cos30 60) F net = 39.96 N (in the positive direction) F net = ma 39.96 = (30)(a) a = 1.3 m/s 2

6. Flat Surface Friction Ex 3: The magnitude of the applied force in Figure 3.71 is 165 N and 30.0 o. If the desk remains stationary, calculate the force of static friction acting on the desk. Desk is stationary so …Fnet = 0 in both x and y HORIZONTAL Fnet = Fa + Ff 0 = (165)(cos 30.0) + Ff Ff = - 143 N [horizontal]

7. Friction  Friction is directly proportional to the weight of an object (F N ) = force due to friction (N) = coefficient of friction = normal force (N) = the Greek letter “mu” “mew” The coefficient of friction is a number (no units) that depends on the two surfaces in contact. The larger the number is, the more “sticky” the situation. So, a very slippery surface (ice) would have a low coefficient of friction.

8. 0.06 – 0.20 0.8 – 1.2 1.0 – 1.6

9. Friction Ex 4: A 12 kg piece of wood is placed on top of another piece of wood. There is 35 N of maximum static friction measured between them. Determine the coefficient of friction between the two pieces of wood. F N = F g = mg F N = (12 kg)(9.84m/s 2 ) F N = 1.2x 10 2 N F f = μ F N 35 N = ( μ )(1.2 x 10 2 N) μ = 0.30 0.30

10. Friction Ex 5: What frictional force must be overcome to start a 50 kg object sliding across a surface with a static coefficient of friction equal to 0.35? F N = F g = mg F N = (50 kg)(9.84m/s 2 ) F N = 490.5 N F f = μ F N F f = (0.35)(490.5 N) F f = 171.675 N 1.7 x 10 2 N

11. Friction Ex 6: A 10 kg box is dragged over a horizontal surface by a force of 40 N. If the box moves with a constant speed of 0.5 m/s, what is the coefficient of kinetic friction for the surface? 0.41

12. Friction Ex 7: A 10 kg box is dragged across a level floor with a force of 60 N. The force is applied at an angle of 30 o above the horizontal. If the coefficient of kinetic friction is 0.20, what is the acceleration of the box? 3.2 m/s 2