1. The “Spring Force” If an object is attached to a spring and then pulled or pushed, the spring will exert a force that is proportional to the displacement of the object but opposite in direction. F s = -kx Where k is called the “spring constant” or the “force constant” and describes how stiff the spring is. The units for k are N/m

3. Friction, f A force that always opposes motion Depends on two things: the roughness of the surfaces and how hard they are pressed together. f =  N , mu- the “coefficient of friction” tells how rough the surfaces are. N, the Normal force tells how hard the surfaces are pressed together

4. Example: How large is the frictional force between 2 surfaces if the coefficient of friction is 0.2 and the Normal force is 80 N? f =  N f = 0.2 x 80 f = 16 N

5. There are two kinds of friction: “static friction” (not moving) must be overcome to initiate motion. “kinetic friction” must be overcome while an object is moving Static friction > Kinetic friction

6. Weight = mg Normal force, N F A = 30 Nm = 2 kg  = 0.4a = ? F net = maf =  N N = mg = 2 x 10 = 20N f =  N = 0.4(20) = 8N Horizon: F net = F A - f 30N – 8N = 22N a = F net / m a = 11 m/s 2 FAFA f =  N = 30 N You pull on a box with an applied force of 30 N. The coefficient of friction is 0.4. If the mass of the box is 2 kg, what is its acceleration? 1.Draw the box and all FOUR forces acting on it. 2.Write what you know and don’t know. 3.Write the equations, Fnet = ma and f =  N 4.Calculate the Normal force and the friction force. 5.Calculate the value of the net Force and then the acceleration.

7. Weight = mg Normal force, N F A = 25Nm = 2 kg  = 43  = 0.4 a = ?  F x = maf =  N  F x = -  N + F A cos  = ma (-0.4(36.65) + 25cos 43 ) / 2= a a = 1.81 m/s 2 FAFA f   F y = N – mg - Fsin  = ma = 0 N = mg + Fsin  

8. Weight = mg Normal force, N F A = 15 Nm = 2 kg  = 43  = 0.4a = ?  F x = maf =  N  F x = -  N + F A cos  = ma (-0.4(9.77) + 15cos 43) / 2 = a a = 3.53 m/s 2 FAFA f   F y = N – mg + Fsin  = ma = 0 N = mg – Fsin  

9. Weight = mg Normal force, N If the box is moving at constant speed, there is no acceleration, Therefore the net force must be zero… so the horizontal forces must cancel each other.  N = Fcos  If you push hard enough to just get the box moving, the acceleration is zero in that case also, but the friction is static, not kinetic. FAFA f 

10. Inclines: With friction, case 1: at rest or sliding down   mg Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. N Write Newton’s Second Law for the box.  F = ma f - mgsin  = ma  mgcos  - mgsin  = ma  gcos  - gsin  = a f What is friction? f =  N What is N ? N = mgcos  f =  mgcos 

11. Inclines: With friction, case 2: pushed downward   mg Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. N Write Newton’s Second Law for the box.  F = ma f - mgsin  – F A = ma  mgcos  - mgsin  – F A = ma f What is friction? f =  N f =  mgcos  FAFA

12. Inclines: With friction, case 3: pushed upward   mg Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. N Write Newton’s Second Law for the box.  F = ma F A - mgsin  – f = ma F A  - mgsin  -  mgcos  = ma f What is friction? f =  N f =  mgcos  FAFA

13. Friction along an incline An object placed along an incline will eventually slide down if the incline is elevated high enough. The angle at which it slides depends on how rough the incline surface is. To find the angle where it slides:  Since it doesn’t move: mgsin  =  mgcos  Therefore: Tan  max  =  max, the coefficient of static friction Or The angle  max  = tan -1  max