* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download document
Survey
Document related concepts
Frictional contact mechanics wikipedia , lookup
Classical mechanics wikipedia , lookup
Coriolis force wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Equations of motion wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Jerk (physics) wikipedia , lookup
Seismometer wikipedia , lookup
Centrifugal force wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Fictitious force wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Transcript
Lecture 3 Chapter 6 Force and Motion II Friction More Circular motion Apparent Weight Drag Forces Numerical integration Derivatives/Integration (Lecture 1) Tension between Blocks (Lecture 2) Misconceptions Friction You are standing still, then begin to walk. What was the external forced that caused you to accelerate? Hint: It is very hard to start walking if you are standing on ice. What force causes a car to accelerate when a traffic light turns green? Friction Fixed block N mg F fs Static Friction: Push with a force F and block does not move because fs = F. The force of friction varies from 0 up to some maximum. The maximum value equals fs = msN, where N is the normal force. Above we would have fs = msmg. The coefficient of static friction ranges from 0 to 1.2 Kinetic Friction: If we increase F until the block starts to move, the friction force decreases to fk= mkN and remains constant throughout the motion. Frictional Forces Friction is an attractive force between two surfaces that is a result of the vector sum of many electrical forces between the surface atoms of the two different bodies. Only about 10-4 of the surface atoms actually contribute. Model of dry friction 3 or 4 asperites support top block. Temporarily weld together The Friction and Lubrication of Solids F. P. Bowden and D. Tabor, Oxford University Press 1964 Models of friction See Chabay and Sherwood Matter and Ineractions Volume 1 ISBN 0-471-35491-0 Problem Solving with Newton’s 2nd Law involving friction • Vector sum of external forces in x direction = max • Vector sum of external forces in y direction = may • If no acceleration, then set sum equal to 0 Mass on Inclined plane Question: How high can we tilt the book before the coin slides down the block? Apply Newton's 2nd law in the x and y directions and using fs = ms N x direction fs mgsin 0 ms N mgsin 0 y direction N mg cos 0 N mg cos ms mg sin h tan mg cos d Mass on Inclined plane Question: How high can we tilt the book before the coin slides down the block? Apply Newton's 2nd law in the x and y directions and using fs = ms N Take x direction along the plane fs mgsin 0 ms N mgsin 0 F Mass On Inclined Plane x direction fk mgsin F 0 mk N mgsin F 0 y direction F N mg cos 0 N mg cos F mg sin m k mg cos Free Body Diagram of an accelerating system: Atwood’s machine with friction. Find a and T. N f T fk m k N Frictionless pulley Mg T f Ma T Ma m k Mg T fk T ----- N Mg T mg ma Ma m k Mg mg ma T T mg mg mk Mg a Mm T (1 mk ) -y M mg Mm Problem 13 Chapter 6 edition 6 and 7 Question: What is the minimum magnitude force required to start the crate moving? +y m 68kg f 15 deg rees N m s 0.5 f Tcos f fs mg x components T cos f fs 0 T +x y components T sin f N mg 0 N mg T sin f T cos f m s N m s (mg T sin f ), Solve for T m s mg (0.5)(68)(9.8) T 304N (cos f m s sin f ) cos(15) 0.5sin(15) Problem 13 continued Question: What is the initial acceleration if mk0.35? m 68kg f 15 deg rees +y N m s 0.35 f Tcos f fs T +x mg Newton's 2nd law x components T cos f fk ma y components T sin f N mg 0 fk m k N m k (mg T sin f ) ma T cos f m k (mg T sin f ) 304 cos(15) 0.35sin(15) T (cos f m k sin f ) mk g (0.35)(9.8) m 68 a 1.3m / s 2 a Inertial Drag Force and Terminal Velocity Drag force: Whenever you have a body like a ball moving through a medium that behaves like a fluid, there will be a drag force opposing the motion. D 1 C Av 2 2 Imagine a falling ball slowed down due to elastic collisions with air molecules. Simply pushing the air out of the way. Hand waving argument dp d(mv) dv dm F m v dt dt dt dt dm Ady dm dy A Av dt dt F v Av Av 2 ball v . A dy air molecules Inertial drag Terminal speeds in air Using Newton’s 2nd law, 1 ma mg C Av 2 2 where m is the mass of the falling ball 1 mg C Av0 2 0 2 Solve for v0 2mg v0 CA Stokes-Napier Law TERMINAL SPEEDS IN AIR Object ` Feather Snowflake BB Mouse Tennis ball Baseball Sky diver Cannonball Speed (m/s) Speed (mph) 0.4 0.9 1 9 13 31 42 60 -120 250 Show demo of falling feather in vacuum 2.2 20 29 66 86 134 -268 560 How to solve this equation? Two ways 1 F Mg C Av2 2 F Mg bv2 One way is to use a spread sheet in Exel. Use Newtons 2nd Law Initial component of momentum: Initial force on ball: Using 2nd Law Find new p p1 Mv1 F Mg bv12 dp p p2 p1 F dt t t p2 p1 Ft p2 p1 Ft Substitute in for p1 and F Newtons 2nd Law p2 p1 Ft p2 p1 (Mg bv12 )t p2 b 2 v2 v1 [g ( )v1 ]t M M v1 v2 x2 x1 v2 t x1 ( )t 2 Go to Excel Spread Sheet delta_t= g= m_1= b_1= v_init= 0.05 10 0.75 0.25 0 time gravity mass drag coefficient initial downward velocity velocity 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0 0.5 0.995833 1.479305 1.942833 2.379923 2.785522 3.156203 3.490176 3.787154 4.048112 distance 0 0.0125 0.049896 0.111774 0.197328 0.305397 0.434533 0.583076 0.749235 0.931169 1.12705 b _1 vi vi 1 g ( )vi 12 t m _1 vi 1 vi xi xi 1 ( )t 2 631 Website: Lecture 3 Materials =C16+(g-(b_1/m_1)*C16*C16)*delta_t =D16+1/2*(C16+C17)*delta_t Velocity vs. Time 10 9 Velocity (m/s) 8 7 6 Drag 5 No Drag 4 reduce mass by 4 3 2 1 0 0 0.5 1 Time (s) 1.5 2 Displacement vrs Time Displacement (m) 30 Displacement w/ drag 20 Displacement w/out drag mass reduced by 4 10 0 0 0.5 1 Time (s) 1.5 2 We can also solve the equation to get the velocity as a function of time before it reaches terminal velocity. Let b = 1/2CA dv m mg bv 2 dt dv m /b mg/b v 2 dt let v 0 = mg b dv 2 m /b v0 v 2 dt 1 dv b /mdt 2 2 v0 v Solving equation continued 1 dv b /mdt 2 2 v0 v 1 1 1 1 ( 0 0 ) 2 2 v v v v 2v 0 v0 v dv dv 2v 0b dt v0 v v0 v m v 0 dv v0 v v 0 dv 2v 0b v0 v m t dt 0 v 0 dv v0 v v dv 2v 0b v0 v m t dt 0 This can be integrated, and fixing the constant of integration by the requirement that the velocity be zero at t = 0 , which is the case for free fall we find: 2v b 0 t 1 e m v v 0 2v 0 b t m 1 e Now show comparison of this solution with numerical integration with Excel. Comparison True Terminal Velocity Function 60 Velocity (m/s) 50 40 TRUE Veloc ity Squared 30 20 10 0 0 5 10 15 20 Time (s) The curve modeled by velocity squared for terminal velocity Differs from the true equation due to a large delta t. When delta t becomes small enough the two curves are Indistinguishable. Water Resistance and Drag Forces Whenever you have a body moving through a liquid there will be a drag force opposing the motion. Here the drag force is proportional to - kv. Viscous drag. Stokes Law: terminal velocity is proportional to mass A 1000 km boat in the water shuts off its engine at 90 km/hr. Find the time required to slow down to 45 km/hr due to a water drag force equal to -70v, where v is the speed of the boat. Let k = 70. dv kv dt mdv kvdt dv k dt v m v dv k t v m dt v0 0 m ln v v ln v ln v0 v ma = - kv v/v0 = 45/90 =1/2 t = m/k ln 2=1000/70 ln 2 = 9.9 s 0 ln v k t v0 m v mk t e v0 Uniform circular motion T Ma 2 v TM R VERTICAL CIRCULAR MOTION Down is negative, Up is positive ac ac mg N N At the top: mv 2 N mg m(ac ) r mv 2 N mg r Minimum v for N = 0: (apparent weightlessness) 2 mg mv mg r v rg VERTICAL CIRCULAR MOTION Down is negative, Up is positive ac ac mg N At the bottom: N mv N Mg mac r2 2 Apparent weight = N = + mv /r + mg mg What do we mean by Fictitious Forces Ff = - ma (the fictitious force always acts in the opposite direction of acceleration) T ma T mg T mg ma ma tan mg a g tan Example of fictitious force (Ff = - ma) In a vertically accelerated reference frame, eg. an elevator, what is your apparent weight? Apparent weight = N Upwards is positive Downwards is negative Upward acceleration N mg ma N mg ma Downward acceleration N mg ma N mg ma Free fall a = g N = 0 Weightless condition Tug-of-war demo illustrates how a small sideways force can produce a large horizontal force Problem 13-10 Suppose two guys in the tug of war are at 4.5 meters apart and I pull the rope out 0.15 meters. Then f4degrees T 0.15 tan f 2.25 f 4o T 2.25 m T f F 2 F sin f 2T F T 2sin f If F 100lbs thenT 700lbs Therefore, F F T 7F o 2sin(4 ) (2)(0.07) The smaller the angle the larger the magnification ConcepTest 4.6 Force and Two Masses A force F acts on mass m1 giving acceleration a1. The same force acts on a different mass m2 giving acceleration a2 = 2a1. If m1 and m2 are glued together and the same force F acts on this combination, what is the resulting acceleration? 1) 3/4 a1 2) 3/2 a1 3) 1/2 a1 4) 4/3 a1 5) 2/3 a1 F F F m1 a1 m2 m2 m1 a2 = 2a1 a 3 ConcepTest 4.6 Force and Two Masses A force F acts on mass m1 giving acceleration a1. The same force acts on a different mass m2 giving acceleration a2 = 2a1. If m1 and m2 are glued together and the same force F acts on this combination, what is the resulting acceleration? 1) 3/4 a1 2) 3/2 a1 3) 1/2 a1 4) 4/3 a1 5) 2/3 a1 F m1 a1 F = m1 a1 a2 = 2a1 F m2 F = m2 a2 = (1/2 m1 )(2a1 ) Mass m2 must be (1/2)m1 because its acceleration was 2a1 with the same force. Adding the two masses together gives (3/2)m1, leading to an F m2 m1 acceleration of (2/3)a1 for the same a 3 F = (3/2)m1 a3 => a3 = (2/3) a1 applied force. ConcepTest 4.7 Climbing the Rope When you climb up a rope, the first 1) this slows your initial velocity which is already upward thing you do is pull down on the 2) you don’t go up, you’re too heavy rope. How do you manage to go 3) you’re not really pulling down – it just seems that way up the rope by doing that?? 4) the rope actually pulls you up 5) you are pulling the ceiling down ConcepTest 4.7 Climbing the Rope When you climb up a rope, the first 1) this slows your initial velocity which is already upward thing you do is pull down on the 2) you don’t go up, you’re too heavy rope. How do you manage to go 3) you’re not really pulling down – it just seems that way up the rope by doing that?? 4) the rope actually pulls you up 5) you are pulling the ceiling down When you pull down on the rope, the rope pulls up on you!! It is actually this upward force by the rope that makes you move up! This is the “reaction” force (by the rope on you) to the force that you exerted on the rope. And voilá, this is Newton’s 3rd Law. ConcepTest 4.8a Bowling vs. Ping-Pong I In outer space, a bowling ball and a ping-pong ball attract each other due to gravitational forces. How do the magnitudes of these attractive forces compare? 1) The bowling ball exerts a greater force on the ping-pong ball 2) The ping-pong ball exerts a greater force on the bowling ball 3) The forces are equal 4) The forces are zero because they cancel out 5) There are actually no forces at all F12 F21 ConcepTest 4.8a Bowling vs. Ping-Pong I In outer space, a bowling ball and a ping-pong ball attract each other due to gravitational forces. How do the magnitudes of these 1) The bowling ball exerts a greater force on the ping-pong ball 2) The ping-pong ball exerts a greater force on the bowling ball 3) The forces are equal attractive forces compare? 4) The forces are zero because they cancel out 5) There are actually no forces at all The forces are equal and opposite by Newton’s 3rd Law! F12 F21 ConcepTest 4.10a Contact Force I If you push with force F on either the heavy box (m1) or the light box (m2), in which of the two cases below is the contact force between the two boxes larger? 1) case A 2) case B 3) same in both cases A m2 F m1 B m2 m1 F ConcepTest 4.10a Contact Force I If you push with force F on either the heavy box (m1) or the light box (m2), in which of the two cases below is the contact force between the two boxes larger? 1) case A 2) case B 3) same in both cases The acceleration of both masses together A is the same in either case. But the contact force is the only force that accelerates m1 m2 F m1 in case A (or m2 in case B). Since m1 is the B larger mass, it requires the larger contact force to achieve the same acceleration. Follow-up: What is the accel. of each mass? m2 m1 F ConcepTest 5.3b Tension II Two tug-of-war opponents each pull with a force of 100 N on opposite ends of a rope. What is the tension in the rope? 1) 0 N 2) 50 N 3) 100 N 4) 150 N 5) 200 N ConcepTest 5.3b Tension II Two tug-of-war opponents each pull 1) 0 N with a force of 100 N on opposite 2) 50 N ends of a rope. What is the tension 3) 100 N in the rope? 4) 150 N 5) 200 N This is literally the identical situation to the previous question. The tension is not 200 N !! Whether the other end of the rope is pulled by a person, or pulled by a tree, the tension in the rope is still 100 N !! ConcepTest 5.4 Three Blocks Three blocks of mass 3m, 2m, and m 1) T1 > T2 > T3 are connected by strings and pulled 2) T1 < T2 < T3 with constant acceleration a. What is 3) T1 = T2 = T3 the relationship between the tension in 4) all tensions are zero each of the strings? 5) tensions are random a 3m T3 2m T2 m T1 ConcepTest 5.4 Three Blocks Three blocks of mass 3m, 2m, and m 1) T1 > T2 > T3 are connected by strings and pulled 2) T1 < T2 < T3 with constant acceleration a. What is 3) T1 = T2 = T3 the relationship between the tension in 4) all tensions are zero each of the strings? 5) tensions are random T1 pulls the whole set of blocks along, so it a must be the largest. T2 pulls the last two 3m T3 2m T2 m T1 masses, but T3 only pulls the last mass. Follow-up: What is T1 in terms of m and a? ConcepTest 5.5 Over the Edge In which case does block m experience a 1) case 1 larger acceleration? In (1) there is a 10 kg 2) acceleration is zero mass hanging from a rope and falling. In 3) both cases are the same (2) a hand is providing a constant downward force of 98 N. Assume massless 4) depends on value of m ropes. 5) case 2 m m 10kg a a F = 98 N Case (1) Case (2) ConcepTest 5.5 Over the Edge In which case does block m experience a 1) case 1 larger acceleration? In (1) there is a 10 kg 2) acceleration is zero mass hanging from a rope and falling. In 3) both cases are the same (2) a hand is providing a constant downward force of 98 N. Assume massless 4) depends on value of m ropes. 5) case 2 In (2) the tension is 98 N due to the hand. In (1) the tension is less than m m 10kg a 98 N because the block a F = 98 N is accelerating down. Only if the block were at rest would the tension be equal to 98 N. Case (1) Case (2) ConcepTest 5.12 Will it Budge? A box of weight 100 N is at rest on a floor where ms = 0.5. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move? 1) moves to the left 2) moves to the right 3) moves up 4) moves down 5) the box does not move Static friction (ms = 0.4 ) m T ConcepTest 5.12 Will it Budge? A box of weight 100 N is at rest on a floor where ms = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move? 1) moves to the left 2) moves to the right 3) moves up 4) moves down 5) the box does not move The static friction force has a maximum of msN = 40 N. The tension in the rope is only 30 N. Static friction (ms = 0.4 ) m T So the pulling force is not big enough to overcome friction. Follow-up: What happens if the tension is 35 N? What about 45 N? ConcepTest 5.19c Going in Circles III You swing a ball at the end of string in a 1) Fc = T – mg vertical circle. Since the ball is in 2) Fc = T + N – mg circular motion there has to be a 3) Fc = T + mg centripetal force. At the top of the ball’s path, what is Fc equal to? 4) Fc = T 5) Fc = mg top v R ConcepTest 5.19c Going in Circles III You swing a ball at the end of string in a 1) Fc = T – mg vertical circle. Since the ball is in 2) Fc = T + N – mg circular motion there has to be a centripetal force. At the top of the ball’s 3) Fc = T + mg path, what is Fc equal to? 4) Fc = T 5) Fc = mg Fc points toward the center of the circle, i.e. downward in this case. The v mg T weight vector points down and the tension (exerted by the string) also points down. The magnitude of the net force, therefore, is: Fc = T + mg Follow-up: What is Fc at the bottom of the ball’s path? R