Ch. 5.pdf - Chapter 5 Using Newton’s Laws Friction...

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Unformatted text preview: Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces Forces of Friction . When an object is in motion on a surface, there will be a resistance to the motion, called the force of friction - Fm, (not a constant), n1a:»:imun1=_ui5 F,, . F,,f,,= pk F,, (constant) no motion “—F"|-'—'sliding-—-I- mi P—n grad-Lari ':. 61"er Quad ( ‘0“) Some Coefficients of Friction TRBLE H Coefficients ef brittle“i Coefficient ef {Inefficient of Surfaces Slutle Friction. fl. Kinetic I-‘rletiuu. pk Weed en wee-cl {L4 i1: lee eu iee [Ll ilflfi Metal en metal {lubricated} {1.15 M)? Steel en steel {unlubrieateu} EL? be Rubber e11 erg.r eenerete Hi {18 Rubber en wet eunerete Ill? [LS Rubber urn ether eulid surteees 1—4 1 Tenant eu Tbfleu in air em em Teflun un steel in air 0.0-4 ilfl-t Lubricated ball bearings {eel {:1th Sync-vial joints [in human limbs} ilfll [Li]! *Vetues are epprmtimele and intended only as e guide. Friction in Newton’ 5 Laws Problems - Friction is a force, so it simply is included in the Fir: (we may use f for simplicity) in Newton’s Laws - The direction of the force of friction is against motion - The determination of magnitude of the force of friction needs the normal force, F" (or n). a Two blacks eennected by a rape bf negligible mass are being dragged by a horizontal farce F. Suppose that F = 100.0 N, m1 = 10.0 kg, mr2 = 20.0 kg, and the caefficient bf kinetic frictien between each black and the surface is 0.200. - (a) Draw a free-bnd'yr diagram fur each black. — (b) Determine the tensien T and the magnitude ef the aceeleratinn of the system. _. m; {4111 IN” Edflmk: WW Aaif'mi-hn 9‘15”“1 :0 .EL “I '-"- 01.3 6; FL» F- “: mu. ““3 1 ft. T E in FF VF.“ 15 1 J; CD T- “Wits—.min. ‘— «Q03: flfilJ mini WM (9 FIT— lama"; W‘s“- iea— Sq 1. a: (3.. Askew): E T ‘3‘: \Jl'fi's [HI-Pm: L) T: (’Q.fi+I‘JHt :L qu.fi+in)m : 1.1m; Q 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 puliey, to a second box B, which hangs freelyr as shown. If the coefficient of kinetic friction is 0.30, and m3 = 10.0 kg, determine the acceleration of the system. Flaw-mates FT— MHJSing - fim“5wfi=fl“h Mas-ma me- «wastes: we)... A 3.0-kg hiock sits on top of a 5.0-kg block which is on a horizontal surface. The 5.0-kg block is pulled to the right with a force F=lflflN as shown in the figure . The coefficient of static friction between all surfaces is 0.60 and the kinetic coefficient is 0.50. What is the acceleration of each block? a as: IJ~= ““5 “Pi-d “I T E -.'. Jim -----‘..'l|I Uniform Circular Motion—Kinematics Uniform circular motion: motion in a circle of constant radius at constant speed Instantaneous yelocity is always tangent to the circle. Looking at the change in velocity in the limit that the time interyal becomes infinitesimally small, we see that I‘ll.- .1!" -_'_—_'_-'—-u- 1": RB .- r‘ This acceleration is ca _=_ _-_:. centripetal, or radial, :: _‘ 1 _ k“; “93V. and it points toward th center of ’5‘" "" ...—-—-——-L" -~ _...--—- the circle. N: i“ 0x EV HG if Arr-W4 The rater of an ultracentrifuge rotates at 60,000 rpm (reunlutiene per minute cr 5280 reefs). A particle at the tap cf a test tube is 5.00 cm from the ratatien axis. Calculate its centripetal acceleratien, in “g’s.” Dynamics of Uniform Circular Motion For an object to be in uniform circular motion, there must be a net force acting on it. We already know the acceleration, so can in'Imetliatei'l»F write the force: EFR = milk = I?!" We can see that the force must be inward by thinking about a ball on a string. Stringe only:r pull; they never push. There is no centrifugal force painting outward; what happens is that the natural tendency cf the object to mauve in a straight line must he evercerne. If the centripetal farce vanishes, the abject flies off at a tangent to the circle. .i" DUESNT HAPPEN O Highwayr curves are marked with a suggested speed. If this speed is based an what wculd be safe in wet weather, estimate the radius of curvature fer a curve marked 45 milesfh {er 2i} mfs}. The caefficient of static fricticn of rubber en wet cencrete is {3.3, the ceeFficient cf kinetic frictien ef rubber an wet cancrete is 0.5. r GA:- (F;Mfi> L “11““ Asmfi: P1 ”F P“ 1 :2 mi“; .L .. L21: Fm .1145 flfl‘l-t) ”31:44.55 ’H‘Elm i. 4M1; fig” 3-. “”59”“3Jtsu'. a 3 )(L ngmu +b41-Fi:-i55"=i"'i.53:t L ‘ " “its“ Hw-rr‘a‘os)‘ I A reller caaster car has a mass ef E, see kg when fully,i leaded with passengers. - a? If the vehicle has a speed af .0 mfs at aint A, what is the fence exerte b the track an the car at this pain ? ... {b} What is the maximum speed_ the vehicle can have at B and 5t!“ remain an the track? at: V1 2. i it 1': Hams...) n: U1 "(37) “5 T" 5"" 4*2'1-8512) up : qgflaaI-N mu: J. ‘9 in: {Cm-L MAI {pr-‘4 I} an" A I P31} J ”I r- _ u‘ ——- 11L .______ __ f V- F3 V: ifif—w An amusement park ride censists cf a retating circular piatferm 8.00 m in diameter from which 10.0-kg seats are suspended at the end of 2.50-m massless chains. When the system retates, the chains make an angle 6' = 30.fl° with the vertical. (a) iWhat is the speed of each seat? (In) Draw a free-hady diagram of a 40.0— kg child riding in a seat and find the tension in the chain. hauls) ‘— KM*h|£}a- 'CififlsTfilqfifls-j-fl' t3) W1 [m5 .mgsihhmg)figgnfi A model airplane of mass 0.?50 kg flies in a horizontal circle at the end of a 60.fl-m control wire with a speed of 35.0 me. Compute the tension in the wire assuming that it makes a constant angle of 20.0“ with the horizon al. The forces exerted on the air ne are the pull of the control wire, e gravitational force, and odynamic lift, which acts at 20.0“ inwa from the 1vertical as s wn in Figure. Motion with Resistive Forces I Motion can be through a medium — Either a liquid or a gas I The medium exerts a resistive force, FD, which depends on the medium and is opposite to the direction of motion. I FD nearly always increa s with increasing Speed fl FD = @v” . n=1; Good approximation for slew motions or small objects - 11:2; Good approximation for large objects FD Proportional To v, I Analyzing the motion results in “T = 31'” = "3 I b Ham. 1' is; called time Eflflfi'i’flflf FD Proportional To v o=fl| Initially, v = o and dvfdt = g H ° 1 As 1‘ increases, R increases and a decreases 0 * I; The acceleration approaches 0 when FD 9 mg At this point, v approaches 0 ‘ i the terminal speed of the object 0 ‘ ii 1! FD=4n o ‘ ' A small, spherical bead ef mass 4.00 g is released from rest at I: [J in a battle ef liquid. The terminal speed is ebserved te be vT = 2.00 emfs. Find — (a) the value of the eenstant b, — (b) the time at whieh the bead reaehes 0.632 vT — (e) the value nf the resistive fnree when the bead reaches terminal speed. 2F=mg— 2m 1 E fl=g_[DpA FD Proportional To v2 - For objects moving at high speeds through air, the resistive force is approximately equal to the square of the spe FD = to flow? .0 is a dimensi empirical Iled t ' 'ent p is t sity of air A i the eros -seotional area of the o ‘ect u is the speed of t object antity that drag £39sz = mo rhea-rind Him —-mhm Haas-trod Drag commits FD Prop- ional To 1:52, Termi s Speed 1- The terminal speed ' occur 1when the FD acceleratinn goes to zern - Solving the equatipn gi‘u’ES 'I't-I'II'IiIInE HIM-HI fur l’nrinum {H'Ilil'l‘IJ-«i Mil-Eing 'I'hl'uugl'l Air flhject. Hm {kg} Malian-1m {m3} mum; 5ij diver 1'5 me an Baseball [radius 3.? cm] 0.115 4.? K 113-3 43 Gnifhall {radius 2.1 cm} (1.1145 1.4 3:: III—5 44 Haitmne {radius 0.5:} cm} 4.3 an: 111'" 19 x 10*5 I4 Raindrnp {radius flit} cm} 3.4 F: “1‘5 1.3 H. 1111-5 9J1 ...
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