Forces and Newton's Laws of Motion - Loudoun County Public ...

Forces and Newton's Laws of Motion - Loudoun County Public ...

Forces and Newtons Laws of Motion The most important chapter of physics EVAR 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) 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. 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. 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. 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. Example of finding net force Tug of war Freshman Class Senior Class 800 Newtons 1000 Newtons

Add individual vectors together resultant = Net force = 200 N @ 00 Example of finding net force Pushing a car Bubba Earl

800 Newtons Add individual vectors together resultant = 1000 Newtons Net Force =

1800 Newtons @ 00 But Dr. Mason, those examples were far too easy Lets look at a contrived example: Three kids pulling on a tire. ropes Witness the artistry Top down view

tire Kids pulling on tire problem Lets look at the forces from each individual rope 10N 10N 2 = 1350 3 = 2700

14.14N 1 = 450 Kids pulling on tire problem 10N 1 = 450 FY = 7.07N

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

10N 2 = 1350 FX = -7.07N FY = 7.07N What are the horizontal and vertical components of each force? 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 = -14.14 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 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 Newtons Second Law This is the most important physics law you will

learn. It is way more important than that E=MC2 business. You ready for it? Here it is: F = MA Force = mass X acceleration Investigating Newtons Second Law F = ma Or, more specifically, F = maF = ma Where F = ma = 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. Examples of Newtons 2 law nd 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/s2 More Examples of Newtons 2nd 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?

F = maF = ma So, 150 N 100 N = ma 50 N = (200 kg)(a) a = (50 N)/(200 kg) = 0.25 m/s2 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. 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 Newtons 3 Law rd 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. 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? Fastronaut on satellite = - Fsatellite on astronaut Aastronaut = -F/mastronaut = -36N/92kg = -0.30m/s2 Asatellite = F/msatellite = 36N/11,000kg = 0.0033m/s2

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 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: 0.909 m/s2 Kurt: 1.58 m/s2 Gravitational Force

Gravitational force is always attractive Fg = Gm1m2/r2 Fg = force of gravity M1 = mass of first object M2 = mass of second object r = distance between them (or specifically, between their centers of mass G = 6.67X10-11 Nm2/kg2 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 r m1 is like m2

r m1 m2 Gravitational Force example 1 What is magnitude of gravitational force between two particles of masses m1 = 12kg and m2 = 25 kg separated by distance r = 1.2m? F = Gm1m2/r2 = 6.67X10-11 Nm2/kg2 (12kg)(25kg)/(1.2m)2

= 1.4 X 10-8 N Fg is very small for normal sized objects, which is why you dont feel yourself being pulled sideways every time a car passes by. Gravitational Force example 2 What is the magnitude of Fg between the earth and the moon? Mearth = 5.9742 1024 kilograms Mmoon = 7.36 1022 kilograms According to the internet (which is never wrong), distance between centers of mass =

384,403 Km, or 384,403,000m Fg = 1.98 X 1020 N Super-neato explanation of g Recall that F = Gm1m2/r2 If we set m1 = mass of the earth and r = radius of the earth, lets see what happens: rearth = 6 378.1 km = 6.3781X106m F = (6.67X10-11Nm2/kg2)(5.9742 1024 kg)(m2)/(6.3781X106m)2 F = (9.795 m/s2)(m2) i.e. all those numbers other than m2 work out

to be 9.8 m/s2 Mini-competition 3: Question 1 Saturn has an equatorial radius of 6 X 107 m and a mass of 5.67 X 1026 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/s2 Ratio = 1.07 Mini-competition 3: Question 2

The mass of the earth is 5.9742 1024 kg and the mass of the sun is 1.99 X 10 30 kg. They are separated by a distance of 146 X 109 m. What is the magnitude of the gravitational attraction on the earth by the sun? 3.72 X 1022 N 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/s2 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? Wsurface = 1.14 X 105 N Worbit: need total distance from earth center = 6.38 X 106 m + 598 X 103 m = 6.98 X 106m Worbit = 0.95 X 105 N

Weighs less in orbit, as expected More about gravity According to F = Gm1m2/r2 , force of gravity decreases as square of distance from center of mass www.rpdp.net/sciencetips_v2/P12B4.htm Mini-competition 4: Question 1 On the surface of the mysterious planet Jay-Z, the acceleration due to gravity is 14.5 m/s2. If

an object weighing 45 N on earth were transported to this other planet, how much would it weigh? 66.58N Mini-competition 4: Question 2 In the Worlds Strongest Man competition, the past three different winners were ydrnas 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? 4635.4N 1040 pounds Other forces: Normal force Normal means perpendicular FN = 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

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 FN on the block? The same 15N block is on the table, but now I pull upwards on it with 11N. What is FN on the block? Normal force and Apparent Weight

Explanation of apparent weight Remember, F = ma When standing in an elevator, FN is the normal force (force from floor on you). Your weight = W = mg F = maFy = FN mg = ma, so FN = mg + ma FN is your apparent weight, mg = your true weight Example: moving elevators A 100 kg man is on an elevator that is

accelerating upwards at a rate of 5 m/s2. What is the mans apparent weight? FN = mg + ma (a is positive if going upwards) FN = (100kg)(9.8m/s2) + (100kg)(5m/s2) FN = (100kg)(14.8m/s2) = 1480 N Now, how about 5m/s2 downwards? FN = (100kg)(4.8m/s2) = 480 N Friction Friction is a force that resists the motion of two surfaces sliding along one another. Motion must be relative

Recall the normal force, FN. 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. 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. Ff = FN The symbol for the coefficient of friction is the greek letter (written mu and pronounced myoo) Fun Fact: there is no friction in New Zealand 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. 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 S = 0.78 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

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.

Freaky note continued In fact, the frictional force will vary up to a maximum value of FFmax = S FN After it starts moving, there is a constant force of kinetic friction on the block = K FN. What would the implication be if the static frictional force did not vary with applied horizontal force? Example from the book A sled is resting on some snow with S = 0.35 Sled + rider have mass = 38 kg.

What is the magnitude of horizontal force needed to get the sled moving? Fneeded = S FN = S mg = (0.35)(38kg)(9.8m/s2) = 130N 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 K = 0.05. How far does the sled go before coming to a halt? Ff = K FN = Kmg Omg, we dont know m! Whatever shall we do??

By Newtons 2nd law, ax = -Ff/m = - Kmg/m = - Kg Now can use Vf2 = Vi2 + 2ax to find x X = 16.3m 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 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 move @ 3.72m/s2. 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 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 = Ff/m = Kmg/m = Kg = 2.94m/s2. You know vampires cant go out during the day, right? I mean, like, seriously? Equilibrium Equilibrium means that the sum of all the forces acting on an object = zero.

If F = maF = ma and F = maF = 0, then what does a =? a=0 For an object to be truly in equilibrium, we must look at both horizontal and vertical directions. F = maFX = 0 F = maFY = 0 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 F = maFX = 0 and F = maFY = 0 Solve for any unknown quantities Example from the book replacing an engine 100

T1 Yoon Jae Sung is replacing the engine to his car. There is a rope with tension T1 that goes up to a pulley and he pulls with tension T2 as shown in the diagram. If the engine weighs 3150 N, find the tensions T1 and T2. 100 Engine

W T2 Example from the book replacing an engine 100 T1 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. 100 Engine W T2

Example from the book replacing an engine 100 T1 100 T1 T2 W 100

Engine W 100 T2 Free Body Diagram Example from the book replacing an engine

100 T1 Step 3: Choose a set of x- and y-axes. -Done Step 4: Resolve all forces along these axes. 100 T2

W Force X-comp Y-comp T1 T1cos(1000) T1sin(1000) T2 T2cos(3500) T2sin(3500) W 0 -W Example from the book replacing an engine

Step 5: Apply F = maFX = 0 and F = maFY = 0 100 T1 F = maFX = 0 so T1cos(1000) + T2cos(3500) = 0 cos(1000) = -0.173 cos(3500) = 0.985 100 T2

W so T1*(-0.173) + T2*(0.985) = 0 Example from the book replacing an engine Step 5: Apply F = maFX = 0 and F = maFY = 0 100 T1 F = maFY = 0

so T1sin(1000) + T2sin(3500) W = 0 sin(1000) = 0.985 sin(3500) = -0.173 100 T2 W so T1*(0.985) + T2*(-0.173) W = 0 Example from the book replacing an engine

Step 6: Solve for the unknowns T1*(-0.173) + T2*(0.985) = 0 T1*(0.985) + T2*(-0.173) W = 0 T1*(-0.173) + T2*(0.985) = 0, so T1 = T2[(0.985/0.173)] Substitute this into second equation T2[(0.985/0.173)]*(0.985) + T2*(-0.173) 3150 = 0 T2 = 579 N, so then T1 = 3300 N

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