* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Document
Survey
Document related concepts
Jerk (physics) wikipedia , lookup
Classical mechanics wikipedia , lookup
Frictional contact mechanics wikipedia , lookup
Fictitious force wikipedia , lookup
Equations of motion wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Mass versus weight wikipedia , lookup
Hunting oscillation wikipedia , lookup
Centrifugal force wikipedia , lookup
Seismometer wikipedia , lookup
Classical central-force problem wikipedia , lookup
Centripetal force wikipedia , lookup
Transcript
example F-F12 = m1 a 56 - m1 a = F12 56 -12 x1.75 =35 N 22 May 2017 Norah Ali Al-moneef 1 If objects in motion tend to stay in motion, why don’t moving objects keep moving forever? Things don’t keep moving forever because there’s almost always an unbalanced force acting upon it. A book sliding across a table slows down and stops because of the force of friction. Why Doesn’t Gravity Make the Box Fall? Force of Floor acting on Box Force from floor on box cancels gravity. If the floor vanished, the box would begin to fall. Force of Earth acting on Box (weight) 22 May 2017 Norah Ali Al-moneef 3 Force on box by person Force on floor by box Force on person by box Force on box by floor It’s the sum of all the forces that determines the acceleration. Every force has an equal & opposite partner. What’s missing in this picture? A pair of forces acting between person and floor. 22 May 2017 Norah Ali Al-moneef 4 Friction Mechanism Corrugations in the surfaces grind when things slide. Lubricants fill in the gaps and let things slide more easily. Don’t all forces then cancel? • How does anything ever move (accelerate) if every force has an opposing pair? • The important thing is the net force on the object of interest Net Force on box 22 May 2017 Norah Ali Al-moneef 5 Friction Forces When two surfaces are in contact, friction forces oppose relative motion or impending motion. F Friction forces are parallel to the surfaces in contact and oppose motion or impending motion. Static Friction: No relative motion. 22 May 2017 Kinetic Friction: Relative motion. Norah Ali Al-moneef 6 Example When you push a book against a wall, the static friction between the wall and the book can prevent it from falling. If you press harder, the friction force will be: A. Larger than before B. The same C. Smaller than before. For the book not to fall down, fS = W Nbook,hand Pushing harder (increasing Nbook,hand) increases Nbook,wall and therefore fS MAX increases, but not the actual value of fS that we had, which needs to continue to be exactly W ! 22 May 2017 Norah Ali Al-moneef fS,book,wall Nbook,wal l Wbook,Earth 7 22 May 2017 Norah Ali Al-moneef 8 Friction and the Normal Force 8N 4N 12 N N N2 N N 4N 6N The force required to overcome static or kinetic friction is proportional to the normal force,N fs = msN 22 May 2017 fk = mkN Norah Ali Al-moneef 9 Friction forces are independent of area. 4N 4N If the total mass pulled is constant, the same force (4 N) is required to overcome friction even with twice the area of contact. For this to be true, it is essential that ALL other variables be rigidly controlled. 22 May 2017 Norah Ali Al-moneef 10 The Static Friction Force When an attempt is made to move an object on a surface, static friction slowly increases to a MAXIMUM value. N fs n F W In this module, when we use the following equation, we refer only to the maximum value of static friction and simply write: fs = msN 22 May 2017 Norah Ali Al-moneef 11 Constant or Impending Motion For motion that is impending and for motion at constant speed, the resultant force is zero and SF = 0. (Equilibrium) F fs Rest F– fs = 0 fk F Constant Speed F – fk = 0 Here the weight and normal forces are balanced and do not affect motion. 22 May 2017 Norah Ali Al-moneef 12 Friction and Acceleration When F is greater than the maximum fs the resultant force produces acceleration. a fk F Constant Speed fk = mkN Note that the kinetic friction force remains constant even as the velocity increases. 22 May 2017 Norah Ali Al-moneef 13 Friction force opposes motion by acting opposite the object’s velocity and direction of travel. “When 2 things are in contact with each other, there will be friction acting between them” 22 May 2017 Norah Ali Al-moneef 14 Coefficient of Friction • µ is determined from experiment. • Determine how much force is required, f, to just get an object moving on a level surface. • Weigh the object to determine the contact force, N. (Level surface N = weight = mg) • Divide f/N. This is the coefficient of friction and characterizes the slickness of the surface. Two Kind of Coefficients • Static Coefficient of Friction, µs-for objects at rest. • Kinetic Coefficient of Friction, µk-for objects in motion. • µk < µs ( Wood on wood, µs = 0.06, µk = 0.04 Rubber of dry concrete, µs = 1.2, µk = 0.9) • Two factors govern the magnitude of the force or maximum static friction or kinetic friction in any situation: the coefficient of friction, represented by the small Greek letter mu (m), and the normal (perpendicular) reaction force (N). F = mN • 22 May 2017 Norah Ali Al-moneef 15 22 May 2017 Norah Ali Al-moneef 16 22 May 2017 Norah Ali Al-moneef 17 22 May 2017 Norah Ali Al-moneef 18 Static Friction... • Opposes motion except here a = 0 is the constant x: Fnet fS = 0 y: N = mg While the block is static: fS Fnet (unlike kinetic friction) fs is NOT fixed in magnitude N y Fnet x fS 22 May 2017 mg Norah Ali Al-moneef 19 Static Friction... • The maximum possible force that the friction between two objects can provide is fMAX = mSN, where ms is the “coefficient of static friction”. – So fS mS N. – As one increases F, fS gets bigger until fS = mSN and the object “breaks loose” and starts to move. N y F x fS 22 May 2017 mg Norah Ali Al-moneef 20 Static Friction... • mS is discovered by increasing F until the block starts to slide: x: FMAX mSN = 0 y: N = mg mS FMAX / mg N FMAX mSmg 22 May 2017 Active Figure y x mg Norah Ali Al-moneef 21 There is some maximum static frictional force, fsmax. Once the applied force exceeds it, the book moves f max S mS N Magnitudes not vectors • ms is the coefficient of static friction, it is a dimensionless number, different for each surface-object pair (wood-wood, wood-metal); also depends on surface preparation • ms does not depend on the mass or surface area of the object •If no applied vertical force 22 May 2017 Norah Ali Al-moneef f Smax m S mg 22 Forces of Friction • Static friction, fs • Kinetic friction, fk 22 May 2017 Norah Ali Al-moneef 23 22 May 2017 Norah Ali Al-moneef 24 The following empirical laws hold true about friction: - Friction force, f, is proportional to normal force, n. f s ms N f k mk N - ms and mk: coefficients of static and kinetic friction, respectively - Direction of frictional force is opposite to direction of relative motion - Values of ms and mk depend on nature of surface. - ms and mk don’t depend on the area of contact. - ms and mk don’t depend on speed. - ms, max is usually a bit larger than mk. - Range from about 0.003 (mk for synovial joints in humans) to 1 (ms for rubber on concrete). See table 5.2 in book. 22 May 2017 Norah Ali Al-moneef 25 Kinetic friction When the relative motion between the two objects is not zero. It “slows down” the sliding motion. Experimentally, it is observed that the kinetic friction force between two surfaces: • Is parallel to the surface (and thus perpendicular to the normal) • Its magnitude does not depend on the speed (except when v = 0) or on the area of the contact surfaces. • Its magnitude is proportional to the magnitude of the normal force between the two surfaces: normal force between the two surfaces: Use the correct one!!! (N is NOT always mg) fk = μ k N μk = coefficient of kinetic friction (depends on the materials) 22 May 2017 Norah Ali Al-moneef 26 EXAMPLE : If mk = 0.3 and ms = 0.5, what horizontal pull F is required to just start a 250-N block moving? 1. Draw sketch and free-body diagram as shown. 2. List givens and label what is to be found N mk = 0.3; ms = 0.5; W = 250 N Find: F = ? to just start F fs F– fs = 0 + W 3. Next we find fs from: fs = msN = 0.5 (250 N) 4.For this case: F– fs = 0 F = fs = 0.5 (250 N) 22 May 2017 Norah Ali Al-moneef 27 EXAMPLE ms = 0.5, W = 250 N. Find F to overcome fs (max). Now we know N = 250 N. 6. Next we find fs from: fs = msN = 0.5 (250 N) 7. For this case: fs N F + F– fs = 0 F = fs = 0.5 (250 N) 250 N ms = 0.5 F= 125 N This force (125 N) is needed to just start motion. Next we consider F needed for constant speed. 22 May 2017 Norah Ali Al-moneef 28 If mk = 0.3 and ms = 0.5, what horizontal pull F is required to move with constant speed? (Overcoming kinetic friction) SFy = m ay = 0 mk = 0.3 N F fk N= W fk = mkN = mk W N -W=0 + mg Now: SFx = 0; F - fk = 0 F = fk = mkW F = 75.0 N F = ( 0.3)(250 N) 22 May 2017 Norah Ali Al-moneef 29 The direction of the friction force, F, is always opposite of the direction of possible motion Possible motion W y F x F y F 0; N W No motion: Impending motion: Motion: 22 May 2017 N F F F x 0; P F Fmax ms N x 0; P F Fmax ms N x 0; P Fmax ms N ; F Fk mk N Norah Ali Al-moneef 30 The direction of the friction force, F, is always opposite of the direction of possible motion Possible motion F y Fy W Fx x F y F 0; N W Py No motion: Impending motion: Motion: 22 May 2017 N F F F x 0; Px F Fmax ms N x 0; Px F Fmax ms N 0; P Fmax ms N ; F Fk mk N x Norah Ali Al-moneef x 31 y x W Wy Wx F F y 0; N Wy N Small P Fx 0; Wx P F Fmax ms N No motion: Impending motion: Fx 0; Wx P F Fmax ms N Motion: Fx 0; Wx P Fmax ms N ; F Fk mk N 22 May 2017 Norah Ali Al-moneef 32 y x W Wy Wx F F y 0; N Wy N No motion: Impending motion: Motion: 22 May 2017 F F F x 0; Wx F Fmax ms N x 0; Wx F Fmax ms N x 0; Wx Fmax ms N ; F Fk mk N Norah Ali Al-moneef 33 y x W Wy Wx F F y 0; N Wy N Large P Fx 0; P Wx F Fmax ms N No motion: Impending motion: Fx 0; P Wx F Fmax ms N Motion: Fx 0; P Wx Fmax ms N ; F Fk mk N 22 May 2017 Norah Ali Al-moneef 34 A 10-kg box is being pulled across the table to the right at a constant speed with a force of 50N. a) Calculate the Force of Friction b) Calculate the Force Normal FN Ff Fa Fa F f 50 N mg N (10)(9.8) 98 N mg 22 May 2017 Norah Ali Al-moneef 35 Suppose the same box is now pulled at an angle of 30 degrees above the horizontal. a) Calculate the Force of Friction Fax Fa cos 50 cos 30 43.3N b) Calculate the Force Normal F f Fax 43.3N FN Ff Fay 30 Fax mg 22 May 2017 Fa N mg! N Fay mg N mg Fay (10)(9.8) 50 sin 30 N 73N Norah Ali Al-moneef 36 example A hockey puck is given an initial speed of 20.0 m/s. It slides 115 m before coming to rest. (a) Determine the coefficient of kinetic friction between the puck and the ice. Applying Newton's Second Law, ΣFx = -Ffk = max ΣFy = N - mg = 0 Since Ffk = μkN ΣFx = -μkN = -μkmg = max ax = -μkg vxf2 = vxi2 + 2ax(xf - xi) 0 = vxi2 - 2μkgxf μk = 0.177 22 May 2017 Norah Ali Al-moneef 37 Applying Newton's Second Law, ΣFx = -Ffk = max ΣFy = N - mg = 0 Since Ffk = μkN ΣFx = -μkN = -μkmg = max ax = -μkg vxf2 = vxi2 + 2ax(xf - xi) 0 = vxi2 - 2μkgxf Solving for μk , μk = 0.177 22 May 2017 Norah Ali Al-moneef 38 example A 1200 kg car moving at 100 km/h coasts to a stop in 25 seconds. What is the value of the static coefficient of friction? m = 1200 kg vo = 100km/h = 27.8 m/s t = 25 s µ=F/N where F = ma & N=mg = ma/mg =(m(|v – vo|)/t)/mg vo f From v = vo + at, a = (v-vo)/t =(27.8m/s/25 s )/ 9.8 m/s2) µ = 0.113 22 May 2017 Norah Ali Al-moneef 39 Example What is the weight of an object that is being pulled at a constant velocity by a force of 25 N if the coefficient of sliding friction between the object and the surface is 0.75? ∑ Fy = 0 N - W =0 ∑ Fx = ma (a =0 ∑ Fx = 0 Fk= F = 25 N Fk= μk N N = Fk / μk W = N = 25 / 0.75 = 33.3 N W = 33.3 N 22 May 2017 N=W constant velocity ) Norah Ali Al-moneef 40 example A 1200 kg car moving at 100 km/h coasts to a stop in 25 seconds. What is the value of the static coefficient of friction? m = 1200 kg vo = 100km/h = 27.8 m/s t = 25 s µ=f/N, where f = ma & N=mg = ma/mg =(m(|v – vo|)/t)/mg vo f From v = vo + at, a = (v-vo)/t =(27.8m/s/25 s )/ 9.8 m/s2) 22 May 2017 µ = 0.113 Norah Ali Al-moneef 41 example Suppose 35 kg crate was not moving at a constant speed, but rather accelerating at 0.70 m/s2. Calculate the applied force. The coefficient of kinetic friction is still 0.30. N Fa Ff mg 22 May 2017 Example A box slides down an incline with angle θ = 45°. The coefficient of kinetic friction between the box and the plane is μk = 0.2. Find the acceleration of the box. ∑ Fx = ma ma fk –Wx= - ma ∑ Fx =0 N – Wy = 0 mg sinθ – μk mg cosθ = ma N – mg cosθ = 0 N fk mg sinθ – μk N = a = g (sinθ – μk cosθ ) = 5.5 m/s2 Wx Wy θ W 22 May 2017 Norah Ali Al-moneef θ 43 Example A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 25.0° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier’s mass is 55.0 kg , and the coefficient of kinetic friction between the skis and the snow is 0.120. Find the magnitude of the force that the tow bar exerts on the skier. Given: m = 55.0 kg, mk = 0.120, = 25.0° since velocity is constant, ax=0; also ay=0 since skier remains on slope equilibrium F 0 Fx 0, Fy 0 22 May 2017 Norah Ali Al-moneef 44 F y 0 N mg cos 0 N mg cos F x 0 f k F mg sin 0 F f k mg sin m k N mg sin m k mg cos mg sin mg ( m k cos sin ) 286 N 22 May 2017 Norah Ali Al-moneef 45 Example A 5-kg block sits on a 30 degree incline. It is attached to string that is thread over a pulley mounted at the top of the incline. A 7.5-kg block hangs from the string. • a) Calculate the tension in the string if the acceleration of the system is 1.2 m/s/s • b) Calculate the coefficient of kinetic friction. T FN m2gcos30 30 T Ff m2g m1 30 m2gsin30 m1g 22 May 2017 FNET ma m1 g T m1a m1 g m1a T (7.5)(9.8) (7.5)(1.2) T T 64.5 N 22 May 2017 Norah Ali Al-moneef 47 Example: The figure shows a wooden block B at rest on another blocks are connected by astringe which goes round a fixed smooth pulley .the kinetic friction between block A and the table is 14 N and that between blocks B and A is 10 N . What is the horizontal force P which is required to move block A with uniform velocity? 22 May 2017 Norah Ali Al-moneef 48 Example A metal block of mass 4Kg is on a plane inclined at an angle of 300 to the horizontal. The block slides down the plane from rest and travels a distance of 1.0 m in 2.0 s. (assume g = 9.81ms-2) a) What is the friction between the block and the inclined plane when the block is sliding ? b) What is the coefficient of kinetic friction ? 22 May 2017 Norah Ali Al-moneef 49 22 May 2017 Norah Ali Al-moneef 50 µ µ 22 May 2017 Norah Ali Al-moneef 51 22 May 2017 Norah Ali Al-moneef 52 22 May 2017 Norah Ali Al-moneef 53 22 May 2017 Norah Ali Al-moneef 54 22 May 2017 Norah Ali Al-moneef 55 If the box moves with constant velocity 22 May 2017 Norah Ali Al-moneef 56 22 May 2017 Norah Ali Al-moneef 57 example A person pushes a 30-kg shopping cart up a 10 degree incline with a force of 85 N. Calculate the coefficient of friction if the cart is pushed at a constant speed. Fa mg cos N Ff mg mg sin 22 May 2017 Example Block A, with mass 4.0 kg, sits on a frictionless table. Block B, with mass 2.0 kg, hangs from a rope connected through a pulley to block A. What is the acceleration of block A? + NA = 39.2 N T aA ≡ a T WA = 39.2 N 22 May 2017 aB ≡ a Norah Ali Al-moneef WB = 19.6 N + 59 For mass A (4 kg), (+T) = (4 kg)(+a) For mass B (2 kg), (+T)+(-19.6 N) = (2 kg)(-a) We now have two equations in two unknowns (a and T). We must solve these simultaneously. Substitute the first into the second, (4 kg)(+a) + (-19.6 N) = (2 kg)(-a) Solving: a 22 May 2017 = (+19.6 N)/(6 kg) = 3.27 m/s2 Norah Ali Al-moneef 60 22 May 2017 Norah Ali Al-moneef 61 22 May 2017 Norah Ali Al-moneef 62 Example :Pulling against friction. A 10.0-kg box is pulled along a horizontal surface by a force of 40.0 N applied at a 30.0° angle above horizontal. The coefficient of kinetic friction is 0.30. Calculate the acceleration. : Lack of vertical motion gives us the normal force (remembering to include the y component of FP), which isFy= 78 N. The frictional force is= μk N = 23.4 N, and the horizontal component of FP is Fx=34.6 N, Fx- Fk =ma 34.6- 23.4 = 10 a a = 11.2/ 10 = 1.1 m/s2 22 May 2017 Norah Ali Al-moneef 63 Example : To push or to pull a sled? Your little sister wants a ride on her sled. If you are on flat ground, will you exert less force if you push her or pull her? Assume the same angle θ in each case. : Pulling decreases the normal force, while pushing increases it. Better pull 22 May 2017 Norah Ali Al-moneef 64 Example : Two boxes and a pulley. Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box A and the table is 0.20. We ignore the mass of the cord and pulley and any friction in the pulley, which means we can assume that a force applied to one end of the cord will have the same magnitude at the other end. We wish to find the acceleration, a, of the system, which will have the same magnitude for both boxes assuming the cord doesn’t stretch. As box B moves down, box A moves to the right. For box A, the normal force equals the weight; the magnitude of the frictional force, but not of the tension in the cord. Fk =μk N = μk mg = 0.20 x 5 x 9.8 Box B has no horizontal forces; the vertical forces on it are its weight and the tension in the cord .Solving the two free-body equations for a gives 1.4 m/s2 (and the tension as 17 N). 22 May 2017 Norah Ali Al-moneef 65 Example : The skier. This skier is descending a 30° slope, at constant speed. What can you say about the coefficient of kinetic friction? FG = mg is the force of gravity (weight) on the skier. Since the speed is constant, there is no net force in any direction. Fk = mg sin 300 μkN = mg sin 300 μkmg cos 300 = mg sin 300 μkcos 300 = sin 300 / cos 300 μk = 0.5 / 0.886 = 0.58 22 May 2017 Norah Ali Al-moneef 66 example In the figure, block B has a mass of 8.0 kg and moves up the incline, where the coefficient of kinetic friction is 0.4. Mass A is 20. kg. Assume an ideal pulley with no mass or friction. a ) Determine the friction force acting on block B. b ) Determine the acceleration of each block. c ) Determine the tension in the connecting rope a b c a) Since B is moving up the incline, the force of friction points down the incline with magnitude fk = μkN Since N = mBg cosθ, fk = μkmBg cosθ Substituting, fk = (0.4)(8.0 kg)(9.8 m/s2)cos 30 fk = 27 N 22 May 2017 Norah Ali Al-moneef 67 The forces acting on block B along incline are T – mB g sinθ - fk = mB a Since fk = μk mB g cos θ, T – mB g sinθ – μkmB g cos θ = mB a (1) The forces acting on block A are T - mA g = - mA a Solving for T, T = mA g – mA a (2) Substituting (2) into (1) : mA g – mA a – mB g sinθ – μkmB g cos θ = mB a Moving a's to same side, mAg - mBg sinθ - μkmBg cosθ = mBa + mAa Factoring out a, mAg - mBg sinθ - μkmBg cosθ = a(mB + mA) Solving for a, a = (mAg - mBg sinθ - μkmBg cosθ)/(mB + mA) Substituting values : a = [(20)(9.8) - (8.0)(9.8)(0.500) - (0.4)(8.0)(9.8)(0.866)]/(28) a = 4.6 m/s2 22 May 2017 Norah Ali Al-moneef 68 C ) Since the forces acting on block A are T - mAg = - mAa Solving for T, T = m Ag - m Aa Substituting values, T = (20) (9.8) - (20) (4.6) T = 104 N 22 May 2017 Norah Ali Al-moneef 69 Example : A ramp, a pulley, and two boxes. Box A, of mass 10.0 kg, rests on a surface inclined at 37° to the horizontal. It is connected by a lightweight cord, which passes over a massless and frictionless pulley, to a second box B, which hangs freely as shown. (a) If the coefficient of static friction is 0.40, determine what range of values for mass B will keep the system at rest. (b) If the coefficient of kinetic friction is 0.30, and mB = 10.0 kg, determine the acceleration of the system. Pick axes along and perpendicular to the slope. Box A has no movement perpendicular to the slope; box B has no horizontal movement. The accelerations of both boxes, and the tension in the cord, are the same. Solve the resulting two equations for these two unknowns. (a)The mass of B has to be between 2.8 and 9.2 kg (so A doesn’t slide either up or down). (b) 220.78 m/s2 May 2017 Norah Ali Al-moneef 70 example Two blocks of masses m1 and m2, with m1 > m2, are placed in contact with each other on a frictionless, horizontal surface, as shown below. A constant horizontal force F is applied to m1 as shown. a ) Find the magnitude of the acceleration of the system. b) Determine the magnitude of the contact force between the two blocks. a a ) Fx(net) = (m1 + m2) ax so, ax = F/(m1 + m2) b )The contact force will be called P Fx(net) = P12 = m2ax since (from part a), ax = F/(m1 + m2) P12 = [m2/(m1 + m2)] F 22 May 2017 Norah Ali Al-moneef 71 Example 3- 29 A block is at rest on an inclined plane the coefficient of static fraction is µs .what is s the the maximum possible angle of inclination θ ( max ) of the surface for which block will remain at rest Fs = w sin θ N = w cos θ tanθ 22 May 2017 Norah Ali Al-moneef 72 22 May 2017 Norah Ali Al-moneef 73 example 22 May 2017 Norah Ali Al-moneef 74 22 May 2017 Norah Ali Al-moneef 75 Summary of Chapter 3 • Newton’s first law: If the net force on an object is zero, it will remain either at rest or moving in a straight line at constant speed. • Newton’s second law: • Newton’s third law: • Weight is the gravitational force on an object W= m g. • Free-body diagrams are essential for problem-solving. Do one object at a time, make sure you have all the forces, pick a coordinate system and find the force components, and apply Newton’s second law along each axis. •Kinetic friction: Fk = μkN • Static friction: Fs ≤ μsN . • Fk < Fs 22 May 2017 μk < μs Norah Ali Al-moneef 76 Newton’s First Law is also called the Law of Inertia Inertia: the tendency of an object to resist changes in its state of motion The First Law states that all objects have inertia. The more mass an object has, the more inertia it has (and the harder it is to change its motion). • Inertia is a term used to measure the ability of an object to resist a change in its state of motion. • An object with a lot of inertia takes a lot of force to start or stop; an object with a small amount of inertia requires a small amount of force to start or stop. • The word “inertia” comes from the Latin word inertus, which can be translated to mean “lazy.” 22 May 2017 Norah Ali Al-moneef 77 • Identifying Newton’s third law pairs – Each force has the same magnitude – Each force acts along the same line but in opposite directions – Each force acts at the same time – Each force acts on a different object – Each force is of the same type 22 May 2017 Norah Ali Al-moneef 78