* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Kinetics Key to Success Eight Step Process Friction
Survey
Document related concepts
Equations of motion wikipedia , lookup
Coriolis force wikipedia , lookup
Jerk (physics) wikipedia , lookup
Seismometer wikipedia , lookup
Nuclear force wikipedia , lookup
Fundamental interaction wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Fictitious force wikipedia , lookup
Mass versus weight wikipedia , lookup
Centrifugal force wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Transcript
Kinetics Key to Success Kinetics is the study of forces that cause accelerated motion. ⇒ Newton’s 2nd Law governs: ∑F = ma Rectangular Coordinates: ∑ Fx ∑ Fy = may Normal and Tangential: ∑F ∑F at = v& an = v2 ρ 1. Impending motion known to exist. Draw FBD with friction force in opposite direction of impending motion. Solve for unknown force or coefficient of friction. 3) 4) 5) 6) 7) n = m an t 1) 2) = m at The Free Body Diagram = The Kinetic Diagram Remember: ^ e Friction t e^n ^ x i Eight Step Process y ^ j = m ax 8) Decide what needs to be isolated. (may be the hardest part). Draw the isolated Free Body & Kinetic Diagrams (complete with all external boundaries) and set them equal to each other. Choose a Coordinate System (C.S.). a) Add all EXTERNALLY APPLIED forces & moments acting ON the Free Body Diagram: a1) Given forces and moments including weight. a2) Support reactions (where the body is cut from the rest of the world). b) Add all mass*acceleration terms to the Kinetic Diagram. Add all necessary dimensions. Enforce Newton’s 2nd Law: a) If necessary, set up any required Kinematic equations. b) Solve ALL equations for ALL unknowns. Check work and answers for units, directions, proper notation, S.D., reasonableness, etc. Fmax - about to slip (impending motion) F F = μs N F = P F < Fmax & F < P 45o No Motion P Motion 2. Determine whether body is at rest or is sliding. Friction force F required to maintain equilibrium is obtained from FBD. If: If: If: F < μs N No sliding F = μs N Impending motion F > μs N Body is sliding. F = μ k N Solve using F = ma 1 Non-Uniform Circular Motion v = v eˆt v2 a = v& eˆt + a = a = ρ time 2 ^ e t Motion along a circular path with constant speed path r 2 n ⎛ v2 ⎞ v& + ⎜⎜ ⎟⎟ ⎝ρ⎠ time 1 ^ e t 2 2 Time rate of change of the direction of the velocity ∑ Ft = m at = mv& v2 ∑ Fn = m an = m ρ Time rate of change of the magnitude of the velocity Note that the acceleration and the force are inward. They are a centripetal acceleration and force. A force acting on an object as it moves through a fluid (liquid or gas). It is opposite the direction of motion Similar to sliding friction Linear FD = bv a = Quadratic = e^n ^ e an eˆn ∑ Ft = m at = 0 t Time rate of change of the direction of the velocity v ∑ Fn = m an = m 2 ρ Note that the acceleration and the force are inward. They are a centripetal acceleration and force. mg = b vter = 2mg ρAC D m1m2 r212 Scalar form F =G Vector form F12 = − G Where 1 FD = ρAC D v 2 2 ρ r eˆn = at eˆt + an eˆn time 1 vter v2 eˆn r v2 e^n Newton’s Law of Universal Gravitation Drag & Terminal Velocity a = v& eˆt + t path v = v eˆt e^n a +a 2 t v = constant time 2 ^ e Motion along a circular path with constant speed e^n eˆn = at eˆt + an eˆn Uniform Circular Motion m1m2 r̂21 r212 G = universal gravitational constant = 6.67 x 10-11 N⋅m2/kg m1 = mass of body 1 m2 = mass of body 2 r21 = position vector from body 2 to body 1 r21 = magnitude of the position vector from body 2 to body 1 r̂21 = position vector from body 2 to body 1 2 Satellite Motion Kepler’s Laws: The objective is to “balance” the force of gravity with the normal (as in direction) acceleration. Fr = G mS mE v2 = m S r2 r m ⇒ G E = v2 r Must be on Engineering paper Front side ONLY In YOUR handwriting Put your name at top Anything you want to write You are to turn it in with the exam NO other books, notes, etc. are allowed. During equal time intervals the radius vector from the Sun to a planet sweeps out equal areas. If T is the time that it takes for a planet to make one full revolution around the Sun, and if S is half the major axis of the ellipse (S reduces to the radius of the planet’s orbit if that orbit is circular), then: where C is a constant T2 = C whose value is the 3 same for all planets. S m1mSun v2 = m1 1 r1 r12 v1 = 2πr1 T1 ⇒ C= 4π 2 GmSun Exam 2 Covers Module 2 Will be a combination of short, medium, and long problems You can bring one sheet of notes (i.e., a “cheat sheet”) Planets move in planar elliptical paths with the Sun at one focus of the ellipse. ΣFn = G Exam 2 – Wednesday, Oct. 17 You must show ALL work on ALL problems Coordinate system on all problems - When needed and not provided - FBD = KD on all problems that need them - Equations - Units - - - Make sure they are easily read, everything is labeled and the are complete Provide generic versions of the equations Final answers must have correct and consistent units Watch unit conversions and unit consistency - Significant Figures (SF) - Vector symbols and answers when required - - - Final answers must have the correct SF Use over bars or over arrows (hats for unit vectors) Vector answers must be in either I j notation or magnitude and direction format. Some questions may specify which format. etc. MORE SLIDES TO FOLLOW!!! 3 ConcepTest 5.1a Gravity and Weight I ConcepTest 5.1b Gravity and Weight II What can you say 1) Fg is greater on the feather What can you say 1) about the force of 2) Fg is greater on the stone about the acceleration 2) it is greater on the stone 3) Fg is zero on both due to vacuum of gravity acting on the 3) it is zero on both due to vacuum 4) Fg is equal on both always stone and the feather? gravity Fg acting on a stone and a feather? it is greater on the feather 4) it is equal on both always 5) it is zero on both always 5) Fg is zero on both always The acceleration is given by F/m so here the mass divides out. Since we The force of gravity (weight) depends know that the force of gravity (weight) on the mass of the object!! The stone is mg, then we end up with acceleration has more mass, therefore more weight. g for both objects. Follow-up: Which one hits the bottom first? ConcepTest 5.2 On the Moon An astronaut on Earth kicks a bowling ball and hurts his foot. A year later, the same astronaut kicks a bowling ConcepTest 5.3a Tension I 1) more 2) less 3) the same You tie a rope to a tree and you 1) 0 N pull on the rope with a force of 2) 50 N 100 N. What is the tension in ball on the Moon with the 3) 100 N 4) 150 N the rope? 5) 200 N same force. His foot hurts... Ouch! The masses of both the bowling ball The tension in the rope is the force that the rope and the astronaut remain the same, so “feels” across any section of it (or that you would his foot feels the same resistance and feel if you replaced a piece of the rope). Since you hurts the same as before. are pulling with a force of 100 N, that is the tension in the rope. Follow-up: What is different about the bowling ball on the Moon? ConcepTest 5.3b Tension II ConcepTest 5.3c Tension III Two tug-of-war opponents each 1) 0 N You and a friend can pull with a force of 100 N on 2) 50 N each pull with a force of 3) 100 N 20 N. If you want to rip 4) 150 N a rope in half, what is 5) 200 N the best way? opposite ends of a rope. What is the tension in the rope? 1) you and your friend each pull on opposite ends of the rope 2) tie the rope to a tree, and you both pull from the same end 3) it doesn’t matter -- both of the above are equivalent 4) get a large dog to bite the rope This is literally the identical situation to the Take advantage of the fact that the tree can previous question. The tension is not 200 N !! pull with almost any force (until it falls down, Whether the other end of the rope is pulled by a that is!). You and your friend should team up person, or pulled by a tree, the tension in the rope on one end, and let the tree make the effort on is still 100 N !! the other end. Page 1 1 ConcepTest 5.4 Three Blocks ConcepTest 5.5 Over the Edge Three blocks of mass 3m, 2m, and 1) T1 > T2 > T3 In which case does block m experience 1) case 1 m are connected by strings and 2) T1 < T2 < T3 a larger acceleration? In (1) there is a pulled with constant acceleration a. 10 kg mass hanging from a rope and 3) T1 = T2 = T3 What is the relationship between the tension in each of the strings? 5) tensions are random constant downward force of 98 N. 4) depends on value of m Assume massless ropes. 5) case 2 In (2) the tension is 98 N T1 pulls the whole set a the tension is less than must be the largest. 98 N because the block T2 pulls the last two T3 3m T2 2m masses, but T3 only m T1 m m due to the hand. In (1) of blocks along, so it pulls the last mass. 3) both cases are the same falling. In (2) a hand is providing a 4) all tensions are zero 2) acceleration is zero a a 10kg F = 98 N is accelerating down. Only if the block were at Case (1) Case (2) rest would the tension Follow-up: What is T1 in terms of m and a? be equal to 98 N. ConcepTest 5.6a Going Up I ConcepTest 5.6b Going Up II A block of mass m rests on the floor of 1) N > mg A block of mass m rests on the 1) N > mg an elevator that is moving upward at 2) N = mg floor of an elevator that is 2) N = mg 3) N < mg (but not zero) accelerating upward. What is 3) N < mg (but not zero) constant speed. What is the relationship between the force due to the relationship between the gravity and the normal force on the 4) N = 0 block? 5) depends on the size of the elevator 4) N = 0 force due to gravity and the 5) depends on the size of the elevator normal force on the block? The block is accelerating upward, so The block is moving at constant speed, so v a>0 mg so N must be greater than mg in order just like the block at rest on a table. m to give the net upward force! Follow-up: What is the normal force if the elevator is in free fall downward? ConcepTest 5.7 Normal Force Σ F = N – mg = ma > 0 ∴ N > mg ConcepTest 5.8 On an Incline 1) case 1 Consider two identical blocks, 1) case A 2) case 2 one resting on a flat surface, 2) case B and the other resting on an 3) it’s the same for both incline. For which case is the 4) depends on the magnitude of the force F normal force greater? 3) both the same (N = mg) 4) both the same (0 < N < mg) 5) both the same (N = 0) 5) depends on the ice surface In Case A, we know that N = W. Case 1 In Case 1, the force F is pushing down In Case B, due to the angle of (in addition to mg), so the normal force the incline, N < W. In fact, we needs to be larger. In Case 2, the force F can see that N = W cos(θ). is pulling up, against gravity, so the m forces on it are N (up) and mg (down), on it are N (up) and mg (down), so N = mg, Below you see two cases: a physics student pulling or pushing a sled with a force F which is applied at an angle θ. In which case is the normal force greater? N it must have a net upward force. The it must have no net force on it. The forces y N f x Case 2 θ Wy normal force is lessened. θ W Page 2 2 ConcepTest 5.9 Friction ConcepTest 5.10 Antilock Brakes on a frictionless truck bed. 1) the force from the rushing air pushed it off When the truck accelerates 2) the force of friction pushed it off forward, the box slides off 3) no net force acted on the box the back of the truck 4) truck went into reverse by accident A box sits in a pickup truck because: Antilock brakes keep the car wheels from locking and skidding during a sudden stop. Why does this help slow the car down? 1) μk > μs so sliding friction is better 2) μk > μs so static friction is better 3) μs > μk so sliding friction is better 4) μs > μk so static friction is better 5) none of the above 5) none of the above Generally, the reason that the box in the truck bed would move Static friction is greater than sliding friction, so with the truck is due to friction between the box and the bed. by keeping the wheels from skidding, the static friction force will help slow the car down more If there is no friction, there is no force to push the box along, efficiently than the sliding friction that occurs and it remains at rest. The truck accelerated away, essentially during a skid. leaving the box behind!! ConcepTest 5.11 Going Sledding Your little sister wants you to give her a ride on her sled. On level ground, what is the ConcepTest 5.12 Will it Budge? 1) pushing her from behind 3) both are equivalent easiest way to 4) it is impossible to move the sled accomplish this? 5) tell her to get out and walk In Case 1, the force F is pushing down 3) moves up 4) moves down 5) the box does not move maximum of μsN = 40 N. The force is larger. In Case 2, the force F Static friction (μs = 0.4 ) tension in the rope is only 30 N. 1 is pulling up, against gravity, so the enough to overcome friction. the frictional force is proportional to 2 the normal force. Follow-up: What happens if the tension is 35 N? What about 45 N? ConcepTest 5.13a Sliding Down I ConcepTest 5.13b Sliding Down II A mass m is placed on an inclined plane (μ > 0) and slides down the plane with constant speed. If a similar block (same μ) of mass 2m were placed on the same incline, it would: 1) component of the gravity force parallel to the plane increased 2) coeff. of static friction decreased 3) normal force exerted by the board decreased 4) both #1 and #3 5) all of #1, #2 and #3 1) not move at all 2) slide a bit, slow down, then stop 3) accelerate down the incline 4) slide down at constant speed 5) slide up at constant speed As the angle increases, the component The component of gravity acting down of weight parallel to the plane increases N f the plane is double for 2m. However, and the component perpendicular to the the normal force (and hence the friction Normal plane decreases (and so does the Normal force) is also double (the same factor!). force). Since friction depends on Normal This means the two forces still cancel Net Force force, we see that the friction force gets smaller and the force pulling the box T m So the pulling force is not big normal force is lessened. Recall that z 2) moves to the right The static friction force has a (in addition to mg), so the normal A box sits on a flat board. You lift one end of the board, making an angle with the floor. As you increase the angle, the box will eventually begin to slide down. Why? 1) moves to the left A box of weight 100 N is at rest on a floor where μs = 0.5. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move? 2) pulling her from the front to give a net force of zero. Weight Wy θ Wx W θ down the plane gets bigger. Page 3 3 ConcepTest 5.14 Tetherball In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point? ConcepTest 5.15a Around the Curve I 1) Toward the top of the pole You are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn. From your perspective in the car, what do you feel is happening to you? 2) Toward the ground 3) Along the horizontal component of the tension force 4) Along the vertical component of the tension force 5) Tangential to the circle The vertical component of the T T (4) You are thrown to the ceiling (5) You are thrown to the floor line. From your perspective in the car, it feels like you are being thrown to the right, hitting the points toward the center of the passenger door. W ConcepTest 5.15b Around the Curve II During that sharp left turn, you found yourself hitting the passenger door. What is the correct description of what is actually happening? (3) You are thrown to the left to continue moving in a straight W provides the centripetal force that circle. (2) You feel no particular change The passenger has the tendency tension balances the weight. The horizontal component of tension (1) You are thrown to the right ConcepTest 5.15c Around the Curve III (1) centrifugal force is pushing you into the door You drive your dad’s car too fast around a curve and the car starts to skid. What is the correct description of this situation? (2) the door is exerting a leftward force on you (3) both of the above (4) neither of the above (1) car’s engine is not strong enough to keep the car from being pushed out (2) friction between tires and road is not strong enough to keep car in a circle (3) car is too heavy to make the turn (4) a deer caused you to skid (5) none of the above The friction force between tires and The passenger has the tendency road provides the centripetal force to continue moving in a straight that keeps the car moving in a circle. line. There is a centripetal force, If this force is too small, the car provided by the door, that forces continues in a straight line! the passenger into a circular path. Follow-up: What could be done to the road or car to prevent skidding? ConcepTest 5.16 Missing Link ConcepTest 5.17 Ball and String 1) T2 = 1/4 T1 A ping pong ball is shot into a Two equal-mass rocks tied to strings are circular tube that is lying flat whirled in horizontal circles. The radius of (horizontal) on a tabletop. When circle 2 is twice that of circle 1. If the period 3) T2 = T1 of motion is the same for both rocks, what 4) T2 = 2 T1 is the tension in cord 2 compared to cord 1? 5) T2 = 4 T1 the ping pong ball leaves the track, which path will it follow? 2) T2 = 1/2 T1 The centripetal force in this case is given by the z Once the ball leaves the tube, there is no longer tension, so T = mv2/r. For the same period, we find a force to keep it going in a circle. Therefore, it that v2 = 2v1 (and this term is squared). However, for simply continues in a straight line, as Newton’s the denominator, we see that r2 = 2r1 which gives us First Law requires! the relation T2 = 2T1. Follow-up: What physical force provides the centripetal force? Page 4 4 ConcepTest 5.18 Barrel of Fun A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? ConcepTest 5.19a Going in Circles I You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by 1 3 2 4 5 1) N remains equal to mg 2) N is smaller than mg your seat is equal to your weight mg. 3) N is larger than mg How does N change at the top of the 4) None of the above Ferris wheel when you are in motion? The normal force of the wall on the rider provides the centripetal force You are in circular motion, so there needed to keep her going around has to be a centripetal force pointing in a circle. The downward force of inward. At the top, the only two gravity is balanced by the upward forces are mg (down) and N (up), so frictional force on her, so she does N must be smaller than mg. not slip vertically. Follow-up: Where is N larger than mg? Follow-up: What happens if the rotation of the ride slows down? ConcepTest 5.19b Going in Circles II ConcepTest 5.19c Going in Circles III A skier goes over a small round hill 1) Fc = N + mg You swing a ball at the end of string 1) Fc = T – mg with radius R. Since she is in circular 2) Fc = mg – N in a vertical circle. Since the ball is 2) Fc = T + N – mg motion, there has to be a centripetal 3) Fc = T + N – mg force. At the top of the hill, what is 4) Fc = N Fc of the skier equal to? Fc points toward the center of the circle, i.e., downward in this case. The weight vector points down and the normal force (exerted by the hill) points up. The magnitude of the net force, therefore, is: Fc = mg – N in circular motion there has to be a centripetal force. At the top of the ball’s path, what is Fc equal to? 4) Fc = T 5) Fc = mg 5) Fc = mg v Fc points toward the center of the circle, i.e. downward in this case. The mg 3) Fc = T + mg N weight vector points down and the v mg T tension (exerted by the string) also R points down. The magnitude of the net force, therefore, is: Follow-up: What happens when the skier goes into a small dip? Fc = T + mg R Follow-up: What is Fc at the bottom of the ball’s path? Page 5 5