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Forces and Newtons Laws of Motion The most important chapter of physics EVAR.

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Presentation on theme: "Forces and Newtons Laws of Motion The most important chapter of physics EVAR."— Presentation transcript:

1 Forces and Newtons Laws of Motion The most important chapter of physics EVAR

2 What is a Force? A force is a push or a pull Contact forces: – Boat pulling a water skier – Pushing a car with a dead battery Noncontact forces: – Gravitational attraction of moon to earth – Magnetic attraction/repulsion – Electrostatic attraction/repulsion The SI unit of force is the Newton (N)

3 Newtons First Law An object at rest will stay at rest. An object in motion will stay in motion in a straight line unless acted upon by an external net force. Take a moment to contemplate this truth of the universe. Now, lets look at the individual pieces of that statement….

4 Deconstructing Newtons First Law …An object at rest will stay at rest… – Things dont just move spontaneously. If an object is just sitting there, it will continue to sit there unless something else pushes or pulls it. …An object in motion will tend to stay in motion in a straight line… – The reason you are used to seeing things slow down in the real world is because they are usually acted upon by friction (or air resistance). In the absence of friction, however, objects would continue in a straight line forever until they hit something.

5 Deconstructing Newtons First Law …by some external net force… – external means it comes from outside. Forces from within do not count as external forces – A net force is the resultant force if you have several different forces acting on an object. – If there is only one force, then the net force is that force. – If there are several, you add them together as vectors to find the one final resultant force. – In order to show that, we need to know that….

6 What else can we say about forces? Forces are vectors – That means they have magnitude and direction A group of forces acting on the same object will add together graphically to produce a net force This is why you learned to add vectors together. It wasnt just for your health.

7 Example of finding net force – Tug of war Senior Class Freshman Class 1000 Newtons 800 Newtons Add individual vectors together = resultant Net force =

8 Example of finding net force – Pushing a car Earl Bubba 1000 Newtons 800 Newtons Add individual vectors together = resultant Net Force =

9 But Dr. Mason, those examples were far too easy Lets look at a contrived example: Three kids pulling on a tire. Top down view Witness the artistry ropes tire

10 Kids pulling on tire problem Lets look at the forces from each individual rope Θ 2 = Θ 1 = 45 0 Θ 3 = N 14.14N

11 Kids pulling on tire problem What are the horizontal and vertical components of each force? Θ 1 = N F = 10N F X = 7.07N F Y = 7.07N Find the horizontal and vertical components of force the same way you do for other vectors.

12 Kids pulling on tire problem What are the horizontal and vertical components of each force? Θ 2 = N F = 10N F X = -7.07N F Y = 7.07N

13 Now lets look at all the forces again Forces acting horizontally: – X-component of force 1 = 7.07N – X-component of force 2 = -7.07N – X-component of force 3 = 0 N Forces acting vertically: – Y-component of force 1 = 7.07N – Y-component of force 2 = 7.07N – Y-component of force 3 = N So, all forces in the X-direction add to Zero and all forces in Y-direction add to Zero So net force on tire is Zero Newtons

14 Inertia and Mass Inertia: the tendency for an object to remain at rest or in motion in a straight line. Mass: a quantitative measure of inertia – SI unit for inertia/mass: kilogram (kg) We will discuss mass at length shortly

15 Newtons Second Law This is the most important physics law you will learn. It is way more important than that E=MC 2 business. You ready for it? Here it is: F = MA – Force = mass X acceleration

16 Investigating Newtons Second Law F = ma Or, more specifically, ΣF = ma – Where Σ = the greek letter sigma and means sum of – So the sum of all the forces on an object are equal to that objects mass times its acceleration – As before, the forces on an object add together vectorally. This is how you find the resultant.

17 Examples of Newtons 2 nd law You are pushing a 200 kg sled over a completely frictionless frozen lake. Your applied force is 150 N. What is the sleds acceleration? F = ma -> 150 N = (200 kg)(a) a = (150 N)/(200 kg) = 0.75 m/s 2

18 More Examples of Newtons 2 nd law Now, you are pushing a 200 kg sled over a frozen lake with some friction. Your applied force is still 150 N, but now it is counteracted by a 100 N friction force. What is the sleds acceleration? ΣF = ma So, 150 N – 100 N = ma 50 N = (200 kg)(a) a = (50 N)/(200 kg) = 0.25 m/s 2

19 Mini-competition 1: Question 1 Chuck Norris (who once successfully ordered a Big Mac at Burger King) has a fist of mass 0.7kg. Starting from rest, his fist achieves a velocity of 8m/s in 0.15 sec. What avg force does he generate to achieve this level of performance? 37 N If paper beats rock, rock beats scissors, and scissors beats paper, what beats all 3 at the same time? Chuck Norris.

20 Mini-competition 1: Question 2 Carla Wong serves a 58 gram tennis ball and accelerates it from rest to 45m/s. The impact gives it a constant acceleration over 44cm. What is the magnitude of the force on the ball? 133 N

21 Newtons 3 rd Law Whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body. Whenever you push on something, that something pushes back on you with the same force.

22 Newtons Third law example From the book: – An astronaut (92kg) pushes on a satellite (11,000kg) with a force of 36N. What is the accel of the astronaut? Of the Satellite? F astronaut on satellite = - F satellite on astronaut A astronaut = -F/m astronaut = -36N/92kg = -0.30m/s 2 A satellite = F/m satellite = 36N/11,000kg = m/s 2

23 Mini-competition 2: Question 1 Charles Steigerwald pushes on a wall to the right with a force of 200 N. What is the force that the wall exerts on Charles? -200N

24 Mini-competition 2: Question 2 Kurt Stegmeier (mass 63 kg) and Dr. Mason (mass 110 Kg) are sitting opposite one another in rolling chairs. They push off one another in opposite directions with a force of 100 N. Assuming perfectly frictionless school chairs and floors, what is the magnitude of the acceleration each experiences? Dr. Mason: m/s 2 Kurt: 1.58 m/s 2

25 Gravitational Force Gravitational force is always attractive F g = Gm 1 m 2 /r 2 F g = force of gravity M 1 = mass of first object M 2 = mass of second object r = distance between them (or specifically, between their centers of mass G = 6.67X Nm 2 /kg 2

26 A note on the center of mass Every object has a center of mass For purposes of calculating forces like gravity, you can assume that all the mass of the object is at the center of mass. For a sphere (assuming uniform density), center of mass is the actual center ris like r m1m2 m1m2

27 Gravitational Force example 1 What is magnitude of gravitational force between two particles of masses m 1 = 12kg and m 2 = 25 kg separated by distance r = 1.2m? F = Gm 1 m 2 /r 2 = 6.67X Nm 2 /kg 2 (12kg)(25kg)/(1.2m) 2 = 1.4 X N F g is very small for normal sized objects, which is why you dont feel yourself being pulled sideways every time a car passes by.

28 Gravitational Force example 2 What is the magnitude of F g between the earth and the moon? M earth = × kilograms M moon = 7.36 × kilograms According to the internet (which is never wrong), distance between centers of mass = 384,403 Km, or 384,403,000m F g = 1.98 X N

29 Super-neato explanation of g Recall that F = Gm 1 m 2 /r 2 If we set m 1 = mass of the earth and r = radius of the earth, lets see what happens: r earth = km = X10 6 m F = (6.67X Nm 2 /kg 2 )( × kg)(m 2 )/(6.3781X10 6 m) 2 F = (9.795 m/s 2 )(m 2 ) i.e. all those numbers other than m 2 work out to be 9.8 m/s 2

30 Mini-competition 3: Question 1 Saturn has an equatorial radius of 6 X 10 7 m and a mass of 5.67 X kg. What is the acceleration due to gravity at the equator of Saturn? What is the ratio of Sami Fekadus weight on Saturn to that on Earth? 10.5 m/s 2 Ratio = 1.07

31 Mini-competition 3: Question 2 The mass of the earth is × kg and the mass of the sun is 1.99 X kg. They are separated by a distance of 146 X 10 9 m. What is the magnitude of the gravitational attraction on the earth by the sun? 3.72 X N

32 Gravity and Weight Mass is the amount of material that makes up an object Weight is the gravitational force on that mass W = mg W = weight, m = mass and g = -9.8m/s 2

33 Weight example: Hubble Telescope Hubble telescope has a mass of 11,600kg What is its weight – Resting on the surface of the earth – In orbit at 598km above the surface of the earth? W surface = 1.14 X 10 5 N W orbit : need total distance from earth center = 6.38 X 10 6 m X 10 3 m = 6.98 X 10 6 m W orbit = 0.95 X 10 5 N Weighs less in orbit, as expected

34 More about gravity According to F = Gm 1 m 2 /r 2, force of gravity decreases as square of distance from center of mass

35 Mini-competition 4: Question 1 On the surface of the mysterious planet Jay-Z, the acceleration due to gravity is 14.5 m/s 2. If an object weighing 45 N on earth were transported to this other planet, how much would it weigh? 66.58N

36 Mini-competition 4: Question 2 In the Worlds Strongest Man competition, the past three different winners were Žydrūnas Savickas (mass 175 kg), Mariusz Pudzianowski (mass 142 kg) and Phil Pfister (mass 156 kg). What is their combined weight in Newtons? If one kg weighs 2.2 pounds, what was their combined weight in pounds? N 1040 pounds

37 Other forces: Normal force Normal means perpendicular F N = Normal force = the force that surface exerts on an object sitting on it. If a 10kg block is sitting on a table, it weighs 98N The table, then, exerts a 98N force upwards on the block

38 Normal force examples A 5kg book is resting on a table. What is the normal force exerted on the book by the table? A 15 N block is on a table. I push on the block with an additional 11 N. What is F N on the block? The same 15N block is on the table, but now I pull upwards on it with 11N. What is F N on the block?

39 Normal force and Apparent Weight

40 Explanation of apparent weight Remember, Σ F = ma When standing in an elevator, F N is the normal force (force from floor on you). Your weight = W = mg ΣF y = F N – mg = ma, so F N = mg + ma F N is your apparent weight, mg = your true weight

41 Example: moving elevators A 100 kg man is on an elevator that is accelerating upwards at a rate of 5 m/s 2. What is the mans apparent weight? F N = mg + ma (a is positive if going upwards) F N = (100kg)(9.8m/s 2 ) + (100kg)(5m/s2) F N = (100kg)(14.8m/s 2 ) = 1480 N Now, how about 5m/s 2 downwards? F N = (100kg)(4.8m/s 2 ) = 480 N

42 Friction Friction is a force that resists the motion of two surfaces sliding along one another. Motion must be relative Recall the normal force, F N. The normal force is perpendicular to the surface. By contrast, the frictional force is parallel to the surface. Friction always opposes the motion. Always always always.

43 Coefficient of Friction The coefficient of friction is a dimensionless number (no units! The only one youll get all year!) that is a proportionality constant between the normal force and the frictional force between two surfaces. F f = μ F N The symbol for the coefficient of friction is the greek letter μ (written mu and pronounced myoo) Fun Fact: there is no friction in New Zealand

44 But Dr. Mason, tell us more interesting and awesome facts about friction There are two kinds of friction, static and kinetic – Static friction is friction between two objects that are not moving relative to one another Example: a heavy crate on a carpet floor that when you push on it, you cant move it. – Kinetic friction is friction between two objects that are moving relative to one another Example: same heavy crate that Tim Tebow is pushing on, who can move it.

45 Why the distinction between static and kinetic friction? Because, in general, the two are different magnitudes, even when you have the same two surfaces in contact with one another. For example, to pull two random materials off the internet, the coefficients of static and kinetic friction between glass and nickel are μ S = 0.78 μ K = 0.56 In general, kinetic friction is less than static friction. I.e. its harder to get a stuck object to move than to keep a sliding object moving

46 A freaky note on static friction Picture, if you will, a block resting on the ground. Let us say that the block is very heavy and is hard to get moving. The force of static friction can change depending on how much sideways force you apply to the block. If you apply a small force, the frictional force is small to counteract your push. If you apply a medium-sized force, the frictional force will be medium-sized to counteract your push.

47 Freaky note continued In fact, the frictional force will vary up to a maximum value of F F max = μ S F N After it starts moving, there is a constant force of kinetic friction on the block = μ K F N. What would the implication be if the static frictional force did not vary with applied horizontal force?

48 Example from the book A sled is resting on some snow with μ S = 0.35 Sled + rider have mass = 38 kg. What is the magnitude of horizontal force needed to get the sled moving? F needed = μ S F N = μ S mg = (0.35)(38kg)(9.8m/s 2 ) = 130N

49 A different example from the book A sled is traveling with an initial velocity of 4m/s on a horizontal stretch of snow. The coefficient of kinetic friction μ K = How far does the sled go before coming to a halt? F f = μ K F N = μ K mg Omg, we dont know m! Whatever shall we do?? By Newtons 2 nd law, a x = -F f /m = - μ K mg/m = - μ K g Now can use V f 2 = V i 2 + 2ax to find x X = 16.3m

50 Mini-competition 5: Question 1 A rocket blasts off from rest and attains a speed of 45 m/s in 15 sec. An astronaut on board has a mass of 57 kg. What is his apparent weight during takeoff? 730N

51 Mini-competition 5: Question 2 A box of calculators that students perpetually borrow weighs 45N and sits at rest on a horizontal table. The coefficients of static and kinetic friction are 0.65 and 0.42, respectively. A horizontal force of 36N is applied to the box. Will it move and, if so, what will its acceleration be? Yes, it will 3.72m/s 2.

52 Mini-competition 5: Question 3 Edward is driving a car at constant speed and Jacob is in the passenger seat. A cup of hot coffee is on the dashboard and μ S = 0.3 between it and the dash. What is the maximum acceleration the car can have without spilling the coffee on Jacob and making him go all aggro n stuff? A = F/m = F f /m = μ K mg/m = μ K g = 2.94m/s 2. You know vampires cant go out during the day, right? I mean, like, seriously?

53 Equilibrium Equilibrium means that the sum of all the forces acting on an object = zero. If ΣF = ma and ΣF = 0, then what does a =? a = 0 For an object to be truly in equilibrium, we must look at both horizontal and vertical directions. – ΣF X = 0 – ΣF Y = 0

54 You must achieve….balance…. How to analyze equilibrium situations: Identify the system you wish to analyze – Is it a single object? Are two objects acting together? Are there ropes or pulleys or other things? Draw a free body diagram Choose a set of x- and y-axes. Resolve all forces along these axes. Apply ΣF X = 0 and ΣF Y = 0 Solve for any unknown quantities

55 Example from the book – replacing an engine Engine T1T1 T2T Yoon Jae Sung is replacing the engine to his car. There is a rope with tension T 1 that goes up to a pulley and he pulls with tension T 2 as shown in the diagram. If the engine weighs 3150 N, find the tensions T 1 and T 2. W

56 Example from the book – replacing an engine Engine T1T1 T2T Step 1: identify the system -The system is the engine plus two ropes Step 2: Draw the free body diagram All forces are acting on the ring at the center, so use that point as the center of your free body diagram. W

57 Example from the book – replacing an engine Engine T1T1 T2T W T1T1 T2T2 W Free Body Diagram

58 Example from the book – replacing an engine 10 0 T1T1 T2T2 W Step 3: Choose a set of x- and y-axes. -Done Step 4: Resolve all forces along these axes. Force X-compY-comp T 1 T 1 cos(100 0 ) T 1 sin(100 0 ) T 2 T 2 cos(350 0 ) T 2 sin(350 0 ) W 0- W

59 Example from the book – replacing an engine 10 0 T1T1 T2T2 W Step 5: Apply ΣF X = 0 and ΣF Y = 0 ΣF X = 0 so T 1 cos(100 0 ) + T 2 cos(350 0 ) = 0 cos(100 0 ) = cos(350 0 ) = so T 1 *(-0.173) + T 2 *(0.985) = 0

60 Example from the book – replacing an engine 10 0 T1T1 T2T2 W Step 5: Apply ΣF X = 0 and ΣF Y = 0 ΣF Y = 0 so T 1 sin(100 0 ) + T 2 sin(350 0 ) – W = 0 sin(100 0 ) = sin(350 0 ) = so T 1 *(0.985) + T 2 *(-0.173) – W = 0

61 Example from the book – replacing an engine Step 6: Solve for the unknowns T 1 *(-0.173) + T 2 *(0.985) = 0 T 1 *(0.985) + T 2 *(-0.173) – W = 0 T 1 *(-0.173) + T 2 *(0.985) = 0, so T 1 = T 2 [(0.985/0.173)] Substitute this into second equation T 2 [(0.985/0.173)]*(0.985) + T 2 *(-0.173) – 3150 = 0 T 2 = 579 N, so then T 1 = 3300 N


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