1. 9/30 Friction  Text: Chapter 4 section 9  HW 9/30 “Skier” due Thursday 10/3  Suggested Problems: Ch 4: 56, 58, 60, 74, 75, 76, 79, 102  Talk about friction  Lab “Atwood’s Machine” This is about Energy conservation

2. Lab v f 2 = v i 2 + 2a  x I never let you use this and here is why. F net = ma combining we can get: 1 / 2 mv f 2 = 1 / 2 mv i 2 + F net  x This is where the idea of “kinetic energy” comes from! “Potential Energy” of gravity is mgh, or mgy. (must pick y = 0 location) yy  PE = mg  y

3. Lab v f 2 = v i 2 + 2a  x I never let you use this and here is why. F net = ma combining we can get: 1 / 2 mv f 2 = 1 / 2 mv i 2 + F net  x yy  PE = mg  y The System of both blocks has a net loss of PE which is equal to the net gain of KE. Note that both blocks have the same velocity. v v

4. Static Friction The maximum static friction force depends on the type of surface and the contact force between the surfaces. The actual static friction force depends on Newton’s second law! f T,B MAX =  s N T,B note subscripts!  s, called the “coefficient of static friction”, is a number that that is small for slippery surfaces. Note this is not a vector equation as N is 90° to f and only relates their magnitudes.

5. Example A 12 kg block is placed on a ramp angled at 20° with respect to the horizontal. The coefficient of static friction is  s = 0.5. Does the block slip? If not, how big is the frictional force? x y W E,B f R,B N R,B First pick a coordinate system and draw a free body diagram. 20° Normal forces always perpendicular to the contact surface Friction forces always parallel to the contact surface  s = 0.5

6. Example The next step is to divide the weight force into components along the x and y axes. x y W E,B f R,B N R,B WxWx WyWy W E,B = mg = (12)(9.8) = 118 N W x = 118 sin 20° = 40 N 20° W y = 118 cos 20° = 111 N  s = 0.5

7. Example The next step is to divide the weight force into components along the x and y axes. x y W E,B f R,B N R,B WxWx WyWy W E,B = mg = (12)(9.8) = 118 N W x = 118 sin 20° = 40 N 20° W y = 118 cos 20° = 111 N  s = 0.5

8. Example Now we consider the motion in the x and y directions separately. x y W E,B f R,B N R,B WxWx WyWy W x = 40 N W y = 111 N y directionx direction a y = 0a x = ? does it slip? F net,y = N R,B - W y F net,y = ma y N R,B - W y = 0 N R,B = W y = 111 N F net,x = W x - f R,B F net,x = ma x but we don’t know a x W x - f R,B = ma x If the ramp is rough enough the block will not slip, a x = 0, and W x = f. f R,B MAX =  s N R,B = (0.5)(111) = 55.5 N The maximum static friction force exceeds W x so the block remains at rest and f R,B = W x = 40 N.  s = 0.5

9. Kinetic Friction:  Always acts parallel to the surface of contact.  Relative motion of surfaces (sliding)  f a,b =  k N a,b always (Remember, static by second law only)   k is the coefficient of kinetic friction.

10. Kinetic Friction Jack pulls the box at constant speed Draw a FBD of the box. W E,B N F,B a = 0 T S,B f F,B Box Since a = 0, F net = 0 and T S,B = f F,B =  k N F,B v = Jack pulls harder, what happens?

11. Kinetic Friction Jack pulls harder Draw a FBD of the box. W E,B N F,B a = T S,B f F,B Box v = The tension force increases but the kinetic friction force stays the same! Kinetic friction does not depend on velocity or acceleration. f F,B =  k N F,B F net = T S,B - f F,B = ma x y

12. Kinetic Friction Jack pulls the box at constant speed Draw a FBD of the box. W E,B N F,B a = 0 f F,B Box v = How will the string angle change the FBD? 30° Tension force now has a vertical component which means N F,B is smaller. f F,B is proportional to the normal force so it is smaller also.

13. Kinetic Friction Jack pulls the box at constant speed Draw a FBD of the box. W E,B N F,B a = 0 f F,B Box v = The new FBD looks something like this. 30° T S,B F net = 0 so T x = f F,B T y + N F,B = W E,B

14. 36.8 ° A 70 kg skier skis down a slope that is angled 36.8 ° with respect to the horizontal. The coefficient of kinetic friction between the skis and the snow is 0.10. What is the skiers acceleration? Homework Problem If the instantaneous velocity is 16m/s, how far up the slope did he start from rest?