1. Physics 2011 Lecture 4: Newton’s Laws S.Norr

2. Sir Isaac Newton Born: 1642 Died: 1727 Philosophiae Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy) (1687)

3. In Principia Mathematica: Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. FFa Law 2: For any object, F NET =  F = ma FF Law 3: Forces occur in pairs: F A-B = - F B-A (For every action there is an equal and opposite reaction.)

4. Netwon’s First Law: Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. An Object in motion tends to stay in motion unless acted upon by an external force The application of an external force results in an Acceleration of the object

5. What makes a good Inertial Reference Frame? Acceleration must be negligible. Is Duluth, MN a good IRF? Consider the UCM equations:

6. Calculating the Centripetal Acceleration at Duluth: Duluth is on the surface of Earth, rotating with a period of 1 Day, with a radius of less than ½ of Earth’s Diameter: T = 1 Day * 24 Hrs * 3600 Sec/Hr = 86400 Sec R = approx. 6 x 10 6 meters Duluth is a fairly good IRF

7. BUT….Earth orbits the sun… T = 1 year = 365 * 24* 3600 = 31.6 Msec R = 150 x 10 9 meters a = 0.006 m/s 2 - So, again a very small acceleration DULUTH is an IRF

8. Newton’s Second Law The net resultant Force (the graphical sum of all forces) acting on a Body is equivalent to the product of its Mass and the Acceleration it is experiencing. F net = m*a In other words, any imbalance of forces on an object causes the object to accelerate. That acceleration is directly proportional to the net force and inversely proportional to its mass.

9. Defining Forces: Forces can be a push or a pull The SI unit of Force is the Newton (1 kg*m/s 2 ) Forces act on a Body with a magnitude at some direction: A VECTOR! (hold for applause)

10. Combining Forces: Forces add and subtract just like vectors, as one might expect: F NET (x,y,z) = F 1 (x,y,z) + F 2 (x,y,z) = [ F 1 x + F 2 x] Î + etc

11. Adding Force Vectors Graphically

12. Superposition of Forces Superposition states that total reaction from multiple disturbances is the algebraic sum of the individual reactions from each disturbance. Thus: –Any force vector can be replaced by its component vectors, all acting at the same point –Any number of forces acting at the same point can be replaced by a single Resultant Force equal to the vector sum of the individual forces.

13. Component Vectors

14. Components and 2 nd Law FaComponents of F = ma : F X = ma X F Y = ma Y F Z = ma Z Suppose we know m and F X, we can solve for a X and apply the things we learned about kinematics over the last few weeks:

15. Example: Pushing a Box on Ice. iA skater is pushing a heavy box (mass m = 100 kg) across a sheet of ice (horizontal & frictionless). He applies a force of 50 N in the i direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10m ? d F v m a i

16. Calculations: Start with F = ma. Or a = F / m. –Recall that v 2 - v 0 2 = 2a(x - x 0 ) from Chap.2 –So v 2 = 2Fd / m ; where: F = 50 N, d = 10 m, m = 100 kg Thus, V = +/- 3.2 m/s 2 and we’ll discard the –tive solution d F v m a i

17. Review: Newton's Laws Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. Fa FF Law 2: For any object, F NET = ma Where F NET =  F action-reactionF F F Law 3: Forces occur in action-reaction pairs, F A,B = - F B,A. Where F A,B is the force acting on object A due to its interaction with object B and vice-versa.

18. The Free Body Diagram FaNewton’s 2nd Law says that for an object F = ma. for an object.Key phrase here is for an object. FaSo before we can apply F = ma to any given object we isolate the forces acting on this object:

19. The Free Body Diagram... Consider the following case –What are the forces acting on the plank ? P = plank F = floor W = wall E = earth F F W,P F F P,W F F P,F F F P,E F F F,P F F E,P

20. The Free Body Diagram... Consider the following case –What are the forces acting on the plank ? Isolate the plank from the rest of the world. F F W,P F F P,W F F P,F F F P,E F F F,P F F E,P

21. The Free Body Diagram... The forces acting on the plank should reveal themselves... F F P,W F F P,F F F P,E

22. Aside... In this example the plank is not moving... –It is certainly not accelerating! Fa F –So F NET = ma becomes F NET = 0 –This is the basic idea behind statics, which we will discuss in a few weeks. FFF F P,W + F P,F + F P,E = 0 F F P,W F F P,F F F P,E

23. Example Example dynamics problem: x A box of mass m = 2 kg slides on a horizontal frictionless floor. A force F x = 10 N pushes on it in the x direction. What is the acceleration of the box? Fi F = F x i a a = ? m y x

24. Example... Draw a picture showing all of the forces F F F B,F F F F,B F F B,E F F E,B y x

25. Example... Draw a picture showing all of the forces. Isolate the forces acting on the block. F F F B,F F F F,B F g F B,E = mg F F E,B y x

26. Example... Draw a picture showing all of the forces. Isolate the forces acting on the block. Draw a free body diagram. F F F B,F gmggmg y x

27. Example... Draw a picture showing all of the forces. Isolate the forces acting on the block. Draw a free body diagram. Solve Newton’s equations for each component. – F X = ma X – F B,F - mg = ma Y F F F B,F gmggmg y x

28. Example... F X = ma X – So a X = F X / m = (10 N)/(2 kg) = 5 m/s 2. F B,F - mg = ma Y – But a Y = 0 – So F B,F = mg. Normal ForceThe vertical component of the force of the floor on the object (F B,F ) is often called the Normal Force (N). Since a Y = 0, N = mg in this case. FXFX N mg y x

29. Example Recap FXFX N = mg mg a X = F X / m y x

30. Normal Force A block of mass m rests on the floor of an elevator that is accelerating upward. What is the relationship between the force due to gravity and the normal force on the block? m (a) N > mg (b) N = mg (c) N < mg a

31. Solution m N mg All forces are acting in the y direction, so use: F total = ma N - mg = ma N = ma + mg therefore N > mg a

32. Tools: Ropes & Strings Can be used to pull from a distance. TensionTension (T) at a certain position in a rope is the magnitude of the force acting across a cross- section of the rope at that position. –The force you would feel if you cut the rope and grabbed the ends. –An action-reaction pair. cut T T T

33. Tools: Ropes & Strings... Consider a horizontal segment of rope having mass m: –Draw a free-body diagram (ignore gravity). xUsing Newton’s 2nd law (in x direction): F NET = T 2 - T 1 = ma So if m = 0 (i.e. the rope is light) then T 1 =  T 2 T1T1 T2T2 m ax

34. Tools: Ropes & Strings... An ideal (massless) rope has constant tension along the rope. If a rope has mass, the tension can vary along the rope – For example, a heavy rope hanging from the ceiling... We will deal mostly with ideal massless ropes. T = T g T = 0 TT

35. Tools: Ropes & Strings... The direction of the force provided by a rope is along the direction of the rope: mg T m Since a y = 0 (box not moving), T = mg

36. Force and acceleration A fish is being yanked upward out of the water using a fishing line that breaks when the tension reaches 180 N. The string snaps when the acceleration of the fish is observed to be is 12.2 m/s 2. What is the mass of the fish? m = ? a = 12.2 m/s 2 snap ! (a) 14.8 kg (b) 18.4 kg (c) 8.2 kg

37. Solution: Draw a Free Body Diagram!! T mg m = ? a = 12.2 m/s 2 l Use Newton’s 2nd law in the upward direction: F TOT = ma T - mg = ma T = ma + mg = m(g+a)