1. Dynamics I Motion Along a Line

2. Static Equilibrium • Example: consider the situation below where two ropes hold up a weight: qleft=30oqright = 55o TleftTright W = 100 Nt

3. Dynamic Equilibrium An elevator that is moving upward at 2 m/s. Find the tension T in the cable, if elevator has mass 400 kg

4. Ramps Why is it easier to push something up a ramp than it is to lift it? P

5. Ramps F// = P - mg sin() = mg// F = FN- mg cos() = mg= 0 FN F// Pbalance = mg sin() P  F  W=mg

6. Ramps F// = P – F//- Ff = ma// FN P F// Ff =Fc W=mg

7. Ramps FN= mg cos() P = mg sin() +  mg cos() = mg[sin() +  cos()] FN P Ff =FN W=mg sin() +  cos() < 1 → P < mg

8. Pulleys

9. Pulleys P W = mg P = Tension = W

10. Pulleys P T T P=T, 2T = W; so P=W/2 W

11. Pulleys P

12. Pulleys P W

13. Newton’s law of gravitation G = 6.67 x 10-11 Nm2/kg2

14. Problem 1 • Three masses are each at a vertex of an isosceles right triangle as shown. Write an expression for the force on mass three due to the other two. m1 r r m3 m2

15. Gravity at earth’s surface

16. Gravity: • What is the force of gravity exerted by the earth on a typical physics student? • Typical student mass m = 55kg • g = 9.81 m/s2. • Fg = mg = (55 kg)x(9.81 m/s2 ) • Fg = 540 N = WEIGHT Fg= mg

17. Test your Understanding • An astronaut on Earth kicks a bowling ball and hurts his foot. A year later, the same astronaut kicks a bowling ball on the moon with the same force. His foot hurts... Ouch! • (a) more (b) less (c) the same

18. Test your Understanding • However the weights of the bowling ball and the astronaut are less: • Thus it would be easier for the astronaut to pick up the bowling ball on the Moon than on the Earth. W = mgMoongMoon< gEarth

19. Friction (force on box due to table) Acme Hand Grenades v f f (force on table due to box) The forces shown are an action-reaction pair.

20. Friction... • Friction is caused by the “microscopic” interactions between the two surfaces:

21. Two Kinds of Friction fs • Static friction • Must be overcome in order to budge an object • Present only when there is no relative motion between the bodies, e.g., the box & table top • Kinetic friction • Weaker than static friction • Present only when objects are moving with respect to each other (skidding) FA Objects are still or moving together. Fnet= 0. FA fk Fnet is to the right. a is to the right. v is left or right.

22. Friction Facts • Friction is due to electrostatic attraction between the atoms of the objects in contact. • It can speed you up, slow you down, or make you turn. • It allows you to walk, turn a corner on your bike, warm your hands in the winter, and see a meteor shower. • Friction often creates waste heat. • It makes you push harder / longer to attain a given acceleration. • Like any force, it always has an action-reaction pair.

23. Friction Strength The magnitude of the friction force is proportional to: • how hard the two bodies are pressed together (the normal force, N ). • the materials from which the bodies are made (the coefficient of friction,  ). Attributes that have little or no effect: • sliding speed • contact area

24. Coefficients of Friction • Static coefficient … s. • Kinetic coefficient … k. • Both depend on the materials in contact. • Small for steel on ice or scrambled egg on Teflon frying pan • Large for rubber on concrete or cardboard box on carpeting • The bigger the coefficient of friction, the bigger the frictional force.

25. Friction... • Force of friction acts to oppose motion: • Parallel to surface. • Perpendicular to Normal force. j N F i ma fF mg

26. Model for Sliding Friction The “heavier” something is, the greater the friction will be...makes sense!

27. Dynamics i :F KN = ma j :N = mg so FKmg = ma j N ma F i K mg mg

28. Test your Understanding • A box of mass m1 = 1.5 kg is being pulled by a horizontal string having tension T = 90 N. It slides with friction (mk= 0.51) on top of a second box having mass m2 = 3 kg, which in turn slides on a frictionless floor. • What is the acceleration of the second box ? (a) a = 0 m/s2 (b) a = 2.5 m/s2 (c) a = 3.0 m/s2 T slides with friction (mk=0.51) m1 a = ? m2 slides without friction

29. Solution • First draw FBD of the top box: N1 m1 f = mKN1 = mKm1g T m1g

30. Solution Newtons 3rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2. m1 f1,2 = mKm1g f2,1 m2

31. Solution • Now consider the FBD of box 2: N2 f2,1 = mkm1g m2 m1g m2g

32. mKm1g = m2a a = 2.5 m/s2 Solution Finally, solve F = ma in the horizontal direction: f2,1 = mKm1g m2

33. Inclined Plane with Friction: • Draw free-body diagram: ma KN j N  mg  i

34. i mg sin KN= ma j N = mg cos mg sin Kmgcos= ma Inclined plane... • Consider i and j components of FNET=ma : KN ma j N  a / g = sin Kcos   mg mg cos  i mg sin 

35. Static Friction... • So far we have considered friction acting when something moves. • We also know that it acts in un-moving “static” systems: • In these cases, the force provided by friction will depend on the forces applied on the system. j N F i fF mg

36. Static Friction... • Just like in the sliding case except a = 0. i :FfF = 0 j :N = mg • While the block is static:fFF j N F i fF mg

37. Static Friction... j N F i fF mg

38. Static Friction... S is discovered by increasing F until the block starts to slide: i :FMAXSN = 0 j :N = mg S FMAX / mg j N FMAX i Smg mg

39. Test Your Understanding • A box of mass m =10.21 kg is at rest on a floor. The coefficient of static friction between the floor and the box is ms = 0.4. • A rope is attached to the box and pulled at an angle of q = 30o above horizontal with tension T = 40 N. • Does the box move? (a) yes (b) no (c) too close to call T q m static friction (ms= 0.4)

40. y x Solution Pick axes & draw FBD of box: Apply FNET = ma y: N + T sin q - mg = maY= 0 N = mg - T sin q = 80 N N T x: T cosq - fFR = maX fFR q The box will move if T cosq - fFR > 0 m mg

41. y x T cosq = 34.6 N fMAX = msN = (.4)(80N) = 32 N Solution y: N = 80 N x: T cosq - fFR = maX The box will move if T cosq - fFR > 0 N T fMAX = msN q m So T cosq > fMAXand the box does move mg

42. Static Friction: • We can also consider S on an inclined plane. • In this case, the force provided by friction will depend on the angle  of the plane. 

43. j i Static Friction... • The force provided by friction, fF, depends on . ma = 0 (block is not moving) fF mg sin ff  (Newton’s 2nd Law along x-axis) N  mg 

44. j i Static Friction... • We can find sby increasing the ramp angle until the block slides: mg sin ff In this case: ffSN  SmgcosM SN mg sinMSmgcosM N mg M Stan M 

45. Friction Facts • Since fF=N , the force of friction does not depend on the area of the surfaces in contact. • By definition, it must be true that S > K for any system (think about it...).

46. Friction vsApplied force fF = SN fF = KN fF fF = FA FA

47. Problem: Box on Truck • A box with mass m sits in the back of a truck. The coefficient of static friction between the box and the truck is S. • What is the maximum acceleration a that the truck can have without the box slipping? S m a

48. Problem: Box on Truck • Draw Free Body Diagram for box: • Consider case where fF is max...(i.e. if the acceleration were any larger, the box would slip). N j i fF = SN mg

49. Problem: Box on Truck Use FNET = ma for both i and j components • I : SN = maMAX • J: N = mg aMAX= S g N j aMAX i fF = SN mg

50. Forces and Motion • An inclined plane is accelerating with constant acceleration a. A box resting on the plane is held in place by static friction. What is the direction of the static frictional force? S a Ff Ff Ff (a) (b) (c)