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Transcript
NEWTON’S LAWS • Force push or a pull • Contact Force physical contact • Field Forces No contact: gravity, magnetic force, etc. INERTIA INERTIA •Newton’s 1st Law •An object wants to keep on doing what it is already doing •In order to change, it needs a net force not equal to zero å F= Sum of all forces NEWTON’S ND 2 LAW: ACCELERATION •When you do have a net force… •object will accelerate in direction of the net force •This acceleration is proportional to the net force and inversely proportional to its mass acceleration µ force 1 acceleration µ mass EXPLORING ND 2 LAW •Put the two together: Fnet a= mtot F = ma •mass units: kg •acceleration units: m/s2 kg × m •This makes force units: 2 s •1 kg•m/s2 is called a Newton (N) MASS IS NOT WEIGHT •Weight is a common force •Weight is the force exerted by gravity on an objects mass •Weight can change if gravity changes Mass: 70 kg Mass: 70 kg Weight: 700 Newtons Weight: 112 Newtons (157 pounds) (25 pounds) Mass is Same!! MASS WEIGHT •Mass is a measure of how much matter makes up an object • Weight is a Force • Depends on object itself – NEVER CHANGES • Mass Is a measure of inertia •Mass is measured in: grams, Kilograms • Depends on object’s location • Weight is measured in: Newtons (N) or pounds (lbs) Mass vs. Weight You can convert mass into weight using Newton’s Second Law F = ma On Earth, a = 9.8m/s2 Example 1: An object has a mass of 10.0 kg, find its weight Example 2: A 2500 kg car is pushed with a 250 N force, what is the acceleration acting on the car due to the force? If the force is doubled, what will happen to the acceleration? Example 3: An artillery shell has a mass of 55 kg. The projectile is fired from the piece and has a velocity of 770 m/s when it leaves the barrel. The gun barrel is 1.5 m long. Assuming the force and therefore the acceleration is constant while the projectile is in the barrel, what is the force that acted on the projectile? Equilibrium vs. Static Equilibrium • If an object is in equilibrium it is not acceleration (no change in motion) • If an object is in Static equilibrium it is at rest (not moving) • An object is in equilibrium if the sum of the forces acting on the object is zero SF = 0 FREE BODY DIAGRAMS 1. Identify all forces acting upon an object 2. Show direction and relative size of forces 3. Vectors originate from the center of mass Which boxes are in equilibrium? Only # 3 Which boxes are definitely in static equilibrium? NONE! b/c it’s possible that they are all moving 1 2 3 FREE BODY DIAGRAMS Scenario 1: 10 kg block going upwards 10 kg SFy = W W FREE BODY DIAGRAMS Scenario 2: 10 kg block pulled across a frictionless floor by a string Y Does it fall through the floor? FN Just like projectiles, we treat x and y separately 10 kg W Normal Force • perpendicular () to the surface • equal to the force acting on the opposite side of surface SFy = FN -W = 0 FREE BODY DIAGRAMS Scenario 2: 10 kg block pulled across a frictionless floor by a string X N 3 Forces T 10 kg W net force only in x direction S Fy = 0 S Fx = FT FREE BODY DIAGRAMS A 1000kg car moving right using its brakes to Scenario 3: slow down on a level freeway. Consider friction. Does it fall through Just like projectiles, the floor? FN we treat x and y separately Y 10 kg W S Fy = FN -W = 0 FREE BODY DIAGRAMS A 1000kg car moving right, using its Scenario 3: brakes to slow down on a level freeway. Consider friction. X N 1000 kg f W WHAT’S UP WITH friction?! X f does not actually act on the COM, but at the surface. N 1000 kg f • Note: f is a W force. • it always acts against the motion FREE BODY DIAGRAMS A 1000kg car moving right, using its Scenario 3: brakes to slow down on a level freeway. Consider friction. X N 4 Forces Fapplied 1000 kg f net force only in x direction W S Fy = 0 S Fx = -Fapplied - f FREE BODY DIAGRAMS 10 kg block sliding up a ramp at 30° with Scenario 4: friction 10 kg FREE BODY DIAGRAMS 10 kg block sliding up a ramp at 30° with Scenario 4: friction N There are 3 forces 10 kg fk W FREE BODY DIAGRAMS 10 kg block sliding up a ramp at 30° with Scenario 4: friction N There are 3 forces Choose x and y to line up with the direction of motion Weight has both an x and y component 10 kg fk W FREE BODY DIAGRAMS q The angle between the weight and Fy is the same as the angle between the ramp and the ground W Wy= W cosq W = W sin q x q FREE BODY DIAGRAMS Scenario 4: 10 kg block on a frictionless ramp at 30° N The y forces cancel (block does not accelerate in the y direction) SFy = N -Wy = 0 the only Net Force is: SFx= -W sinq - fk 10 kg fk W Fwy Fwx Practice! Drawing a FBD for # 5 from HW 3 35 HW 3: #5 A 34.5 kg block rests on the ramp as shown in the drawing. What is the tension in the line that is connected to the top of ramp? 35 Language Objective Students will be able to explain to each other what force causes a ball to accelerate down a hill Force Applied Variable F1,… F2 Direction Direction of push/pull Direction of push Air resistance / Drag Fair Opposite to motion Kinetic Friction fk Opposite to motion Static Friction fs Opposite to motion Weight / gravity W or Fw Normal N or FN 90° to surface Tension T or FT Along the rope/string Straight down Practice! Complete the Free Body Exercises in your HW packet ND 2 LAW: EXAMPLE The Batman, with a mass of 70-kg, rappels down a rope from his bat-copter with a downward acceleration of 3.5 m/s2. What vertical force does the rope exert on Batman? ND 2 LAW: EXAMPLE The Batman, with a mass of 70-kg, rappels down a rope from his bat-copter with a downward acceleration of 3.5 m/s2. What vertical force does the rope exert on Batman? Start all force questions with a diagram showing the forces T Pick the positive direction (make direction of motion positive) m1 = 70 kg anet = 3.5m / s Identify all givens with symbols W 2 ND 2 LAW: EXAMPLE The Batman, with a mass of 70-kg, rappels down a rope from his bat-copter with a downward acceleration of 3.5 m/s2. What vertical force does the rope exert on Batman? Given: m = 70 kg anet = 3.5m / s 2 W = mg = (70kg)(9.8m /s ) W = 686 N Want: 2 T m1 = 70 kg anet = 3.5m / s T W 2 ND 2 LAW: EXAMPLE T = 441N Calculations: Fnet = mtot anet SF = W -T = mtot anet T = W - mtot anet W T = (686 N) - (70kg)(3.5m / s ) 2 T = 441N Practice! Try # 7 from HW 1 AIR RESISTANCE •Most falling objects don’t accelerate at 9.8 m/s2! •Because air resistance makes objects accelerate less 9.8 m/s2! •Two things determine the magnitude of air resistance •surface area (shape) •speed TERMINAL VELOCITY •Terminal velocity is the fastest an object can fall due to air resistance! •When FAir resistance = W an object is falling at terminal velocity and not accelerating TERMINAL VELOCITY •When FAir resistance = W the object’s a = zero 10 kg block in freefall traveling at terminal Scenario 1: velocity Fair 10 kg SFy = 0 Say = 0 W TIME TO PRACTICE Start your homework Turn to pg. 405 Atwood Machine Video http://www.youtube.com/watch ?v=i2bGTC27OJU NEWTON’S RD 3 LAW •Equal and Opposite •For every force that one object exerts on a second object, the second object exerts an equal (in size) and opposite (in direction) reaction force on the first object. NEWTON’S RD 3 LAW SUBTLETIES •Forces only come in pairs (you cannot create a single force) •Does action cancel out reaction? •NO! Each force acts on a different object. •Each force has different effect on their object! Action Fore: A bat applies a force to a baseball. Action force effect: •The baseball accelerations a LOT Reaction force: •The baseball applies a force to a bat Reaction force effect: •The baseball bat (de)accelerates a little When I am zooming down the highway I crush small bugs that hit my car Which exerts the larger force? a. the bug on my car b. My car on the bug •Neither! The forces are equal (but opposite) How does a rocket work in space if there is no air to push it? Lets watch WALL-E •What happens when the pod reaches “cruising speed”? •How does WALL-E move in space when there is no air or ground for him to push against? TIME TO PRACTICE Check Yourself #11-15 Answers 11. 12. 13. 14. 15. A B B B A http://www.youtube.com/watch?v=8bTdM mNZm2M 1. Why does the moon orbit the earth? 2. Does the moon put a gravitational pull on the earth? 3. How does that force compare to the force the earth puts on the moon? FRICTION •Friction is everywhere there is motion on a surface or within a fluid •Friction is a resistive force (a force that opposes motion) •Friction evidence? •Heat •Noise (air resistance) FRICTION •Let’s examine sandpaper 40 grit 400 grit FRICTION •Even “smooth” sandpaper is rough 40 grit 400 grit FRICTION •Slide the surfaces of two pieces together •Atoms at the peaks bond with atoms at the peaks of the other •It requires force to break these bonds •opposing force is friction PERPENDICULAR FORCE •Rub your hands together •What happens to friction if you press harder? •So, f µ FNormal NATURE OF SURFACES •Rub your hands together again •What would happen to friction if you added some oil to your hands? •What would happen to friction if you added some dry glue? •This condition of a surface is given as a quantity known as the coefficient of friction (m) FORCE OF FRICTION •Friction between solid surfaces depends on: 1. The perpendicular (Normal) force between the surfaces in contact 2. The nature of the surfaces in contact •together, these create a net frictional force NATURE OF SURFACES •Each situation has two coefficients of friction •μs = static friction •when an object is being pushed, but hasn’t moved yet Surfaces steel on steel glass on glass μs μk 0.74 0.94 0.57 0.40 tire on dry road 1.0 tire on icy road 0.30 bone joints 0.010 0.80 0.015 0.0030 •μk = kinetic friction •the friction once the object starts moving NATURE OF SURFACES • μs is always bigger • This means … It takes more force to make an object start moving than it takes to keep an object moving Surfaces steel on steel glass on glass μs μk 0.74 0.94 0.57 0.40 tire on dry road 1.0 tire on icy road 0.30 bone joints 0.010 0.80 0.015 0.0030 FRICTIONAL FORCE •Friction is the product of the normal force and the coefficient of friction f = m FN •when μ = 0, surface contact is frictionless FRICTION EXAMPLE A 500kg car is going 30 m/s. How long will it take to come to a stop on a dry road compared to an icy one? Givens m = 500 kg vo= 30 m/s vf = 0 Unknown t=? adry = ? aice = ? Equation t= v f - v0 a FRICTION EXAMPLE A 500kg car is going 30 m/s. How long will it take to come to a stop on a dry road compared to an icy one? 1. Draw a FBD! 2. Identify the + direction 3. Write the Net Force Eqs N SFx = - f SFy = N -W = 0 f W FRICTION EXAMPLE A 500kg car is going 30 m/s. How long will it take to come to a stop on a dry road compared to an icy one? 4. Solve for the net acceleration SFx = - f = mtot a net -f = a net mtot N f W FRICTION EXAMPLE A 500kg car is going 30 m/s. How long will it take to come to a stop on a dry road compared to an icy one? m fdry = mk FN = (0.8)(500kg)(9.8 2 ) = 3920N sm ficy = mk FN = (0.015)(500kg)(9.8 2 ) = 73.5N s The difference in friction is why it takes so long to stop on ice N f W FRICTION EXAMPLE A 500kg car is going 30 m/s. How long will it take to come to a stop on a dry road compared to an icy one? 4. Solve for the net acceleration for each road - fdry = a dry mtot -3920N m = a dry= -7.84 2 s 500kg N f W FRICTION EXAMPLE A 500kg car is going 30 m/s. How long will it take to come to a stop on a dry road compared to an icy one? 4. Solve for the net acceleration for each road - fice = a ice mtot -73.5N m = a ice = -0.14 2 s 500kg N f W FRICTION EXAMPLE A 500kg car is going 30 m/s. How long will it take to come to a stop on a dry road compared to an icy one? Equation Givens m = 500 kg v f - v0 t= vo= 30 m/s a vf = 0 Unknown t=? tdry = 214s m/s2 adry = -7.84 aice = -0.14 m/s2 tice = 3.83s Solving Force Problems 1. 2. 3. 4. Draw a FBD Identify direction of movement Write the expression for Fnet = F1+F2-F3… Set Fnet expression equal to mtotanet F1+F2-F3…= mtotanet