* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Electromagnetism wikipedia, lookup

Equivalence principle wikipedia, lookup

Pioneer anomaly wikipedia, lookup

Coriolis force wikipedia, lookup

Lorentz force wikipedia, lookup

Newton's law of universal gravitation wikipedia, lookup

Modified Newtonian dynamics wikipedia, lookup

Centrifugal force wikipedia, lookup

Fictitious force wikipedia, lookup

Artificial gravity wikipedia, lookup

Transcript

Newton’s Second Law Chapter 6 The Second Law Force = mass X acceleration SF = ma Sum of all the forces acting on a body Vector quantity SF = 0 or SF = ma -Still object -Accelerating object -Obj. at constant velocity The Second Law Situation One: Non-moving Object • Still has forces Force of the material of the rock Force of gravity http://alfa.ist.utl.pt/~vguerra/Other/Rodin/thinker.jpg The Second Law Situation Two: Moving Object: Constant Velocity SF = 0 Fpedalling = Fair + Ffriction Fair Fpedalling Ffriction http://2003.tour-de-france.cz/images/foto/05-07-2003/armstrong4.jpg The Second Law Situation Two: Moving Object: Accelerating SF = ma ma = Fpedalling – Fair - Ffriction Fair Fpedalling Ffriction http://2003.tour-de-france.cz/images/foto/05-07-2003/armstrong4.jpg The Second Law Unit of Force = the Newton SF=ma SF = (kg)(m/s2) 1 N = 1 kg-m/s2 (MKS) 1 Newton can accelerate a 1 kg object from rest to 1 m/s in 1 s. Equilibrium 1. No motion 2. Constant velocity BOTH INDICATE NO ACCELERATION SF=0 Three ropes are tied together for a wacky tug-ofwar. One person pulls west with 100 N of force, another south with 200 N of force. Calculate the magnitude and direction of the third force. ? 100 N 200 N A car with weight 15,000 N is being towed up a 20o slope (smooth) at a constant velocity. The tow rope is rated at 6000 N. Will it break? Accelerating Systems • SF=ma • Must add up all forces on the object What Force is needed to accelerate a 5 kg bowling ball from 0 to 20 m/s over a time period of 2 seconds? Calculate the net force required to stop a 1500 kg car from a speed of 100 km/h within a distance of 55 m. 100 km/h = 28 m/s v2 = vo2 + 2a(x-xo) a = (v2 - vo2)/2(x-xo) a = 02 – (28 m/s)2/2(55m) = -7.1 m/s2 A 1500 kg car is pulled by a tow truck. The tension in the rope is 2500 N and the 200 N frictional force opposes the motion. The car starts from rest. a. Calculate the net force on the car b. Calculate the car’s speed after 5.0 s A 500.0 gram model rocket (weight = 4.90 N) is launched straight up from rest by an engine that burns for 5 seconds at 20.0 N. a. Calculate the net force on the rocket b. Calculate the acceleration of the rocket c. Calculate the height and velocity of the rocket after 5 s d. Calculate the maximum height of the rocket even after the engine has burned out. Calculate the sum of the two forces acting on the boat shown below. (53.3 N, +11.0o) Mass vs. Weight Mass – The amount of matter in an object/INTRINSIC PROPERTY – Independent of gravity – Measured in kilograms Weight – Force that results from gravity pulling on an object – Weight = mg (g = 9.8 m/s2) Mass vs. Weight • Weight = mg is really a re-write of F=ma. – Weight is a force – g is the acceleration (a) of gravity • Metric unit of weight is a Newton • English unit is a pound A 60.0 kg person weighs 1554 N on Jupiter. What is the acceleration of gravity on Jupiter? Elevator at Constant Velocity a= 0 SF = FN – mg ma = FN – mg 0 = FN – mg FN = mg Suppose Chewbacca has a mass of 102 kg: FN = mg = (102kg)(9.8m/s2) FN = 1000 N FN mg a is zero Elevator Accelerating Upward a = 4.9 m/s2 SF = FN – mg ma = FN – mg FN = ma + mg FN = m(a + g) FN=(102kg)(4.9m/s2+9.8 m/s2) FN = 1500 N FN mg a is upward Elevator Accelerating Downward a = -4.9 m/s2 SF = FN – mg ma = FN – mg FN = ma + mg FN = m(a + g) FN=(102kg)(-4.9m/s2+9.8 m/s2) FN = 500 N FN mg a is down At what acceleration will he feel weightless? FN = 0 SF = FN – mg ma = FN – mg ma = 0 – mg ma = -mg a = -9.8 m/s2 Apparent weightlessness occurs if a > g FN mg A 10.o kg present is sitting on a table. Calculate the weight and the normal force. FN Fg = W Suppose someone leans on the box, adding an additional 40.0 N of force. Calculate the normal force. Now your friend lifts up with a string (but does not lift the box off the table). Calculate the normal force. What happen when the person pulls upward with a force of 100 N? SF = FN+ Fp – mg SF = 0 +100.0N – 98N = 2.0N Fp = 100.0 N ma = 2N a = 2N/10.0 kg = 0.2 m/s2 Fg = mg = 98.0 N Free Body Diagrams: Ex. 3 A person pulls on the box (10.0 kg) at an angle as shown below. Calculate the acceleration of the box and the normal force. (78.0 N) Fp = 40.0 N 30o FN mg Friction • Always opposes the direction of motion. • Proportional to the Normal Force (more massive objects have more friction) FN Ffr FN mg Fa mg Friction Static -opposes motion before it moves (ms) • Generally greater than kinetic friction • Fmax = Force needed to get an objct moving Fmax = msFN Kinetic - opposes motion while it moves (mk) • Generally less than static friction Ffr = mkFN Friction and Rolling Wheels Rolling uses static friction – A new part of the wheel/tire is coming in contact with the road every instant B A Braking uses kinetic friction Point A gets drug across the surface A A 50.0 kg wooden box is pushed across a wooden floor (mk=0.20) at a steady speed of 2.0 m/s. a. How much force does she exert? (98 N) b. If she stops pushing, calculate the acceleration. (-1.96 m/s2) c. Calculate how far the box slides until it stops. (1.00 m) A 100 kg box is on the back of a truck (ms = 0.40). The box is 50 cm X 50 cm X 50 cm. a. Calculate the maximum acceleration of the truck before the box starts to slip. Your little sister wants a ride on her sled. Should you push or pull her? Inclines What trigonometric function does this resemble? Inclines FN mg q Inclines FN q mgcosq q mg mgsinq Inclines FN Ffr mgsinq mgcosq q A 50.0 kg file cabinet is in the back of a dump truck (ms = 0.800). a. Calculate the magnitude of the static friction on the cabinet when the bed of the truck is tilted at 20.0o (170 N) b. Calculate the angle at which the cabinet will start to slide. (39o) Given the following drawing: a. Calculate the acceleration of the skier. (snow has a mk of 0.10) (4.0 m/s2) b. Calculate her speed after 4.0 s? (16 m/s) First we need to resolve the force of gravity into x and y components: FGy = mgcos30o FGx = mgsin30o The pull down the hill is: FGx = mgsin30o The pull up the hill is: Ffr= mkFN Ffr= (0.10)(mgcos30o) SF = pull down – pull up SF = mgsin30o– (0.10)(mgcos30o) ma = mgsin30o– (0.10)(mgcos30o) ma = mgsin30o– (0.10)(mgcos30o) a = gsin30o– (0.10)(gcos30o) a = 4.0 m/s2 (note that this is independent of the skier’s mass) To find the speed after 4 seconds: v = vo + at v = 0 + (4.0 m/s2)(4.0 s) = 16 m/s Suppose the snow is slushy and the skier moves at a constant speed. Calculate mk SF = pull down – pull up ma = mgsin30o– (mk)(mgcos30o) ma = mgsin30o– (mk)(mgcos30o) a = gsin30o– (mk)(gcos30o) Since the speed is constant, acceleration =0 0 = gsin30o– (mk)(gcos30o) (mk)(gcos30o) = gsin30o mk= gsin30o gcos30o = sin30o = tan30o =0.577 cos30o Drag D ≈ ¼Av2 D = drag force A = Area V = velocity Fails for • Very small particles (dust) • Very fast (airplanes) • Water and dense fluids Finding Acceleration with Drag Derive the formula for the acceleration of a freefalling object including the drag force. Terminal Speed a. Find the formula for terminal speed (a=0) for a freefalling body b. Calculate the terminal velocity of a person who is 1.8 m tall, 0.40 m wide, and 75 kg. (64 m/s) A 1500 kg car is travelling at 30 m/s when the driver slams on the brakes (mk = 0.800). Calculate the stopping distance: a. On a level road. (57.0 m) b. Up a 10.0o incline (48.0 m) c. Down a 10.0o incline (75.0 m) A dogsled has a mass of 200 kg. The sled reaches cruising speed, 5.0 m/s in 15 m. Two ropes are attached to the sled at 10.0o, one on each side connected to the dogs. (mk = 0.060) a. Calculate the acceleration of the sled. (0.833 m/s) b. Calculate T1 and T2 during the acceleration period. (140 N) Formula Wrap-Up SF=ma Weight = mg (g = 9.8 m/s2) Fmax = msFN Ffr = mkFN