According to stationary observer

2. accela R is reaction force = reading on scales R mg Spring scales Measured weight in an accelerating Reference Frame According to stationary observer F = ma Taking “up” as +ve R - mg = ma R = m(g + a) If a = 0  R = mg normal weight If a is +ve R = m(g + a) weight increase If a is -ve R = m(g - a) weight decrease

3. R is reaction force = reading on scales R mg According to traveller F = ma R - mg = ma BUT in his ref. frame a = 0! so R = mg!! How come he still sees R changing when lift accelerates? Didn’t we say the laws of physics do not depend on the frame of reference? Only if it is an inertial frame of reference! The accelerating lift is NOT!

4. Total Force Force #1 Force #2 Force #1 Force #2 Forces are Vectors so Directions are Important Total Force = 0 Forces Add Forces Cancel!

5. there's a fraction too much.... Friction

6. Why doesn’t Mick Doohan fall over? Friction provides the central force mg In the rest reference frame

8. What is Friction • Surfaces between two materials are not even • Microscopically the force is atomic •  Smooth surfaces have high friction • Causes wear between surfaces •  Bits break off • Lubrication separates the surfaces

9. The Source of Friction between two surfaces

10. Static and Sliding (Dynamic) Friction • Static frictional force: when nothing is sliding • Sliding frictional force: when surfaces are sliding • Static frictional forces always greater than sliding ones • Lubrication provides microscopic rollers between surfaces

11. Static Friction If no force F No friction forcef N Surface with friction F f mg As F increases friction f increases in the opposite direction. Therefore Total Force on Block = zero does not move As F continues to increase, at a critical point most of the (“velcro”) bonds break and f decreases rapidly. Fis now greater thanf and slipping begins

12. N Surface with friction F f f depends on surface properties. Combine these properties into a coefficient of frictionm fmN m is usually < 1 Static f < or = ms N Kinetic f = mk N

13. fmax Slipping begins (fmax = sN ) f f < fmax (= kN ) Coefficient of Kinetic friction < Coefficient of Static friction F Static friction Kinetic friction

14. N f F q mg cosq q mg sinq mg thus mS = At qcrit F = f mg sin qcrit = f = mS N = mS mg cosqcrit mS = tan qcrit Independent of m, or g. Property of surfaces only

15. Coefficient of Friction

16. N N F1 F2 fcrit2 fcrit1 mg mg Making the most of Friction A F1 > F2 fcrit = S N B F1 = F2 fcrit = S mg C F1 < F2 Friction force does not depend on area!

17. So why do Petrol Heads use fat tyres? tribophysics To reduce wear? Tyres get hot and sticky which effectively increases . The wider the tyre the greater the effect? The truth! Friction is not as simple as people think

18. acceleration Driving Torque Force of road on Tyre Force of Tyre on road What force drives the car?

19. v vo N d f mg Stopping Distance depends on friction Braking force Friction road/tyres v2 =vo2 + 2a(x-xo) 0 = vo2 + 2ad Max value of ais when f is max. fmax = sN = smg F = ma  -fmax= mamax -smg = mamax amax = - sg

20. Thus since dmin depends on v2!! Take care!! If v0 = 90 kph (24 m s-1) and m = 0.6 ==> d = 50 m!!

21. Taking a curve on Flat surface N v Fcent r mg Fcent is provided by friction. If no slippingthe limit is when Fcent = fs(limit) = sN = smg So that So for a given s (tyre quality) and given r there is a maximum vel. for safety. If s halves, safe v drops to 70%….take care! Does not depend on m

22. Lateral Acceleration of 4.5 g The lateral acceleration experienced by a Formula-1 driver on a GP circuit can be as high as 4.5 g This is equivalent to that experienced by a jet-fighter pilot in fast-turn manoeuvres.

23. Albert Park GP circuit N mv2/R mg R = 70 m V=55 m s-1 Central force provided by friction. mv2/R = N = mg  = v2/Rg   = 4.3 • for racing tyres is ~ 1 (not 4!). How can the car stay on the road?

24. Soft rubber Grooved tread Are these just for show, or advertising?

25. 200 km/h

26. p= mv =momentum Another version of Newton #2 Momentum p transferred over a time t gives a force:- F is a measure of how much momentum is transferred in time Dt

27. @ 200 kph v = 55 m s-1 = v m Distance travelled in 1 sec @ velocity v = v x A m3 Volume of air hitting each spoiler (area A) in 1 sec Area A m2 A ~ 0.5 m2 = x v x A kg mass of air (density )hitting each spoiler in 1 sec  ~ 1 kg m-3 = x v2x A kg m s-1 Momentum of air hitting each spoiler in 1 sec If deflected by 900, mom change in 1 sec F ~ 3 x 104 N ~ 3 Tonne! Newton says this is the resulting force mv2/R = N = mg mv2/R = N =  (m + 3000) g

28. Big Consequences! (USGS)

29. That's all folks

30. VISCOUS DRAG FORCE What is it? like fluid friction a force opposing motion as fluid flows past object Assumptions low viscosity (like air) turbulent flow

31. v C is the Drag coefficient. It incorporates specifics like shape, surface texture etc. What does the drag force depend on? D  velocity (v2) D  effective area (A) D  fluid density (r) D  rA v2 D= ½ C rA v2

32. Vm Area A Fluid of density In 1 sec a length of V metres hits the object Volume hitting object in 1 sec. =AV Mass hitting object in 1 sec. =  AV momentum (p) transferred to object in 1 sec. = ( AV)V Force on object = const AV2

34. V=0 mg D V mg D V mg SF = mg - D SF = mg -1/2CrAv2 D increases as v2 until SF=0 i.e. mg= 1/2CrAv2

35. D dv = m mg - D dt mg dv + r - = 2 m 1/2C Av mg 0 dt F = mg –D ma = mg -D

38. When entertainment defies reality

39. Calculate: Drag force on presidents wife Compare with weight force Could they slide down the wire? D= ½ CrAv2 Assume C = 1 v = 700 km h-1

40.  Calculate: The angle of the cable relative to horizontal. Compare this with the angle in the film (~30o) D= ½ CrAv2 Assume C = 1 v = 700 km h-1

41. In working out this problem you will prove the expression for the viscous drag force