Ch.4 Forces And Newton’s Laws

1. Ch.4 Forces And Newton’s Laws What is a Force? A Push or a Pull Two types of forces: Contact and Non contact Non Contact - Action at a distant force Is the Force a Vector or a Scalar ? Units and Dimensions ? Mass – Measure of Mass is inertia (How does it differ from Weight ?) Higher mass – harder to change the motion, i.e. higher inertia

2. Newton’s First Law Of Motion • An object continues in a state of rest or in a state of motion at a constant speed along a straight line unless compelled to change that state by a net force. • What is the net force? If so, why will a ball rolling on a flat surface stop?

3. Vector sum of all the forces acting on the object is called the net force. ( i.e. both magnitude and direction needs to be considered) What is the net force on the car ?

4. Inertial Reference Frame Newton’s laws are valid in an inertial frame, i.e. the acceleration of the frame should be zero (Constant Velocity). Ideally we do not have such a frame, but earth is a very good approximation.

5. Newton’s Second Law • First law: no net force no change in velocity • Second law: What happens if there is an external force? (net) Acceleration of the object Net Force Acceleration inversely proportional to mass

6. or Only two factors determine the acceleration---Force and Mass Force Units Newton (N) = kg. m/s2 What is 1 Newton ?

7. Newton’s Second Law Of Motion When a net external force acts on an object of mass m, the acceleration a results is directly proportional to the net force and has a magnitude that is inversely proportional to the mass. The direction of the acceleration is the same as the direction of the net force. SI Unit of Force: kg.m/s2 = newton (N)

8. Units for Mass, Acceleration & Force

9. Example 1. Pushing A Car Frictional force = 560 N Push = 275 N+ 395 N = 670 N Net force = 670 N – 560 N, N= 110 N F = ma , 110 N = 1850 kg*a a = 110 N/1850 kg = 0.059 m/s2 (what is the direction ?)

10. 12N θ 5N Forces In 2-D Fx = max Fy = may Two forces are applied onto an object. 1.) 12N towards North, and 2.) 5N towards East, what is the direction of the motion of the object? Resultant force = tan θ = 12/5 =2.4 θ = tan-1(12/5) =67.380

11. ax = ?, ay = ? (d) Example 2. 1300 kg raft P= Force by the man In 65 seconds, where will the boat be if v0x = 0.15 m/s ? (a) A= Force of wind

12. Resultant Force And Acceleration Resultant

13. Displacement Of The Boat In 65 sec For x: v0x = 0.15 m/s, ax = 0.018 m/s2 t = 65 sec, x = ? y = v0yt + (½) ay t2 = 0*65 + (½)*0.011*(65)2 = 23 m x = v0x t + (½)axt2 = 0.15*65 +(1/2)*0.018*(65)2 = 48 m For y: v0y = 0, ay = 0.011 m/s2, t = 65 sec

14. Which of these will cause the acceleration to be doubled ? a) All of the forces acting on the object doubles b) The net force acting on the object doubles c) Both the net force and the mass doubles d) The mass of the object is reduced by a factor of two. a, b, d

15. Newton’s Third Law of Motion • Whenever one body exerts a force on a second body, the second body will exert an oppositely directed force of equal magnitude on the body. • (Action reaction law) Acting on which object ?

16. For the astronaut, mA=92 If P = 36 N, -36 N = 92 aA , i.e. aA = -0.39 m/s2 36 N = 11000 kg as , i.e. as = 0.0033 m/s2 MS = 11000 kg mA = 92 kg Force by Astronaut on Space Ship = P Force on Astronaut by Space ship = P What will be the acceleration of the two objects ?

17. Types Of Forces Fundamental Forces • Gravitational force • Strong Nuclear • Electroweak Electromagnetic force All other forces are Non Fundamental (can be explained by a fundamental force) Maxwell, Unification theory

18. Gravitational Force • Newton’s law of Universal gravitation r center to center G = 6.673 x 10 –11 N m2/kg2 Always attractive , Planetary motion, Satellites

19. W = mg So: Me = 9.799 m/s2 Weight Of An Object • Mass m on the surface of the earth Radius of earth = Re Distance ~ Re (approx)

20. = 1.14*105 N Hubble Telescope • The mass of Hubble telescope is 11600 kg. (Re = 6.38*106 m, Me = 5.98*1024 kg) • (i) What is the weight when resting on earth surface, • (ii) and in orbit 598 km above (r = Re +598 km) R=6.38*106m+598*103m=6.98*106m

21. Check Your Understanding 2 m2>m1, and the net gravitational force acting on the third object is zero. Which of the drawings correctly represents the locations of the object? (d)

22. The Normal Force & Newton’s 3rd Law Table exert a force on the block: FN Net Force on the block: FN – W Normal Force: The force component that a surface exerts on the object (Normal to the contact surface) How can the table exert a force ? (Consider the springs in a mattress)

23. What happens if I push or pull on the block ? What happens to the normal force if the rope has a pull = 15N?

24. Balancing Act

25. Apparent Weight • How do you feel when the elevator suddenly starts going up? FN Feel heavy? FN – W = ma FN = Apparent weight W = True weight

26. Elevator Accelerating Up With “a” Applying 2nd law vertically upward FN – W = ma FN = W+ma Apparent weight is larger than the true weight (W = mg) g = 9.8 m/s2

27. Elevator Accelerating Down With “a” FN – mg = ma ( if downward a = negative) FN = mg +ma What if a = -g ? FN=0

28. Static And Kinetic Friction Two surfaces touching each other Contact surface is not smooth In addition to the normal force, there is a force parallel to the surface FRICTIONAL FORCE

29. fs fsMax Static Friction (fs) As F increases, fs increases from zero up to fsMax • Just before motion fsMax=μs FN (μs= Coefficient of static friction) Units and Dimensions of μs?

30. F fsMax Force Needed To Start The Sled The maximum force needed to just begin to move = need to overcome the max frictional force i.e. F = fsMax = μs FN = μs mg If μs = 0.35, m = 38 kg fsMax= 0.35 * 38 * 9.8 = 130 N What happens once it just starts moving ? Easier to move, i.e. needs force going down i.e. fs changes to fkKinetic Friction

31. Kinetic Friction • fk = μkFN (μk Coefficient of kinetic friction) μk is usually less than μs Example 10: Sled riding : How far does the sled go before stopping ? μk = 0.05

32. fk = μk FN = μk mg Net force on the sled = kinetic frictional force Acceleration What is the meaning of the negative sign? Does the acceleration depend on the mass ? Stopping distance x? v2 = v02 +2ax

33. Tension Mass-less and non-stretch rope Force gets applied to the box undiminished.

34. Traction Of The Foot • T1 and T2 keep the pulley on the foot at rest. i.e. pulley is at equilibrium (Net Force = 0) • Find the

35. Let T1=T2=T, then

36. (Say T1=T2 = T) Since T = 2.2 g (why?) F = 35 N T1 =T2 ( Same rope)

37. Replacing The Engine

39. Setting W=3150N, T2=582N T1=3.30*103N

40. Equilibrium At ConstantVelocity

42. Check Your Understanding 4 Which of the following could lead to equilibrium? a) Three forces act on the object. The forces all point along the same line but may have different directions. b) Two perpendicular forces act on the object. c) A single force acts on the object. d) In none of the situations described in (a), (b) and (c) could the object possibly be in equilibrium. Answer: (a)

43. Non-equilibrium Applications • Non zero net force • Net force – x component, y component • Acceleration in x direction and in y direction

44. Towing A Super Tanker ax=2.0*10-3 D=75.0*103N, m=1.50*108N, R=40.0*103N

46. ax=2.0*10-3 =1.53*105N

47. Hauling A Trailer

48. If the drawbar has no mass T=T` = 28000N

49. The Motion Of A Water Skier What is the net force on skier in each picture?

50. (a) The skier is floating motionless in the water. The skier is floating motionless in the water, so her velocity and acceleration are both zero. Therefore, the net force acting on her is zero, and she is in equilibrium.