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
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) yy PE = mg y
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 yy 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
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.
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
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
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
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
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.
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?
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
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.
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
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 What is the skiers acceleration? Homework Problem If the instantaneous velocity is 16m/s, how far up the slope did he start from rest?