Sample Problem 3/4The design model for a new ship has a mass of 10 kg and is tested in an exper-imental towing tank to determine its resistance to motion through the water atvarious speeds. The test results are plotted on the accompanying graph, and theresistance R may be closely approximated by the dashed parabolic curve shown. Ifthe model is released when it has a speed of 2 m/s, determine the time t requiredfor it to reduce its speed to 1 m/s and the corresponding travel distance x.Solution. We approximate the resistance-velocity relation by R  kv2andfind k by substituting R  8 N and v  2 m/s into the equation, which gives k  8/22 2 Thus, R  2v2.The only horizontal force on the model is R, so thatWe separate the variables and integrate to obtainThus, when v  v0/2  1 m/s, the time is t 2.5 s. Ans.The distance traveled during the 2.5 seconds is obtained by integrating v dx/dt. Thus, v  10/(5  2t) so thatAns.Sample Problem 3/5The collar of mass m slides up the vertical shaft under the action of a forceF of constant magnitude but variable direction. If   kt where k is a constantand if the collar starts from rest with   0, determine the magnitude F of theforce which will result in the collar coming to rest as  reaches /2. The coeffi-cient of kinetic friction between the collar and shaft is k.Solution. After drawing the free-body diagram, we apply the equation of mo-tion in the y-direction to getwhere equilibrium in the horizontal direction requires N  F sin . Substituting  kt and integrating first between general limits givewhich becomesFor   /2 the time becomes t  /2k, and v  0 so thatAns.Fk [1  k(0  1)]  mg2k  0 and F  mg2(1  k)Fk [sin kt  k(cos kt  1)]  mgt  mv t0 (F cos kt  k F sin kt  mg) dt  m v0 dvF cos   k N  mg  m dvdt[ΣFy  may] x0 dx  2.50 105  2t dt x  102 ln (5  2t)2.50  3.47 m5(11  12 ) t0 dt  5 v2 dvv2 t  51v  12 sR  max or 2v2  10 dvdt[ΣFx  max]N s2/m2.130 Chapter 3 Kinetics of Particles12024680v, m/sR, NRv0 = 2 m/svxB = WWmµkFθNθFmgµkNHelpful Hints Be careful to observe the minus signfor R. Suggestion: Express the distance xafter release in terms of the velocityv and see if you agree with the re-sulting relation x  5 ln (v0/v).Helpful Hints If  were expressed as a function ofthe vertical displacement y insteadof the time t, the acceleration wouldbecome a function of the displace-ment and we would use vdv ady. We see that the results do not de-pend on k, the rate at which theforce changes direction.c03.qxd 6/15/06 12:31 PM Page 130EXAMPLE 13.413.4 E QUATIONS OF M OTION: RECTANGULAR C OORDINATES 115A smooth 2-kg collar C ,shown in Fig. 13–9a ,is attached to a springhaving a stiffness and an unstretched length of 0.75 m. Ifthe collar is released from rest at A ,determine its acceleration and thenormal force of the rod on the collar at the instant SOLUTIONFree-Body Diagram. The free-body diagram of the collar when it islocated at the arbitrary position y is shown in Fig. 13–9b .Note that theweight is Furthermore,the collar is assumedto be accelerating so that “ a ”acts downward in the positive y direction.There are four unknowns,namely,a,and Equations of Motion.(1)(2)From Eq. 2 it is seen that the acceleration depends on the magnitudeand direction of the spring force.Solution for and a is possibleonce and are known.The magnitude of the spring force is a function of the stretch s of thespring; i.e., Here the unstretched length is Fig. 13–9a ;therefore,Since then(3)From Fig.13–9a ,the angle is related to y by trigonometry.(4)Substituting into Eqs.3 and 4 yields andSubstituting these results into Eqs.1 and 2, we obtainAns.Ans.NOTE: This is not acase of constant acceleration, since the spring forcechanges both its magnitude and direction as the collar moves downward.a = 9.21 m> s2TNC= 0.900 Nu = 53.1°.Fs= 1.50 Ny = 1 mtan u =y0.75uFs= ks = 3A4y2+ 1 0.7522- 0.75Bk = 3 N> m,s = CB - AB =4y2+ 1 0.7522- 0.75.AB = 0.75 m,Fs= ks.uFsNC19.62 - Fs sin u = 2 a+T© Fy= may;- NC+ Fs cos u = 0:+© Fx= max;u .Fs,NC,W = 2 1 9.812 = 19.62 N.y = 1 m.k = 3 N> my0.75 mACk $ 3 N/ m(a)Buxya(b)NC19.62 NFsuFig. 13–9Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from thepublisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.13.5 E QUATIONS OF M OTION: NORMAL AND T ANGENTIAL C OORDINATES 129Determine the banking angle for the race track so that the wheels ofthe racing cars shown in Fig. 13–12a will not have to depend uponfriction to prevent any car from sliding up or down the track. Assumethe cars have negligible size,a mass m ,and travel around the curve ofradius with a speed v .ru(b)bnanNCW $ mguFig. 13–12u(a)EXAMPLE 13.6SOLUTIONBefore looking at the following solution, give some thought as to whyit should be solved using t, n, b coordinates.Free-Body Diagram. As shown in Fig. 13–12b ,and as stated in theproblem, no frictional force acts on the car.Here represents theresultant of the ground on all four wheels.Since can be calculated,the unknowns are and Equations of Motion. Using the n, b axes shown,(1)(2)Eliminating and m from these equations by dividing Eq. 1 by Eq. 2, we obtainAns.NOTE: The result is independent of the mass of the car.Also,a forcesummation in the tangential direction is of no consequence to thesolution. If it were considered, then since the carmoves with constant speed.A further analysis of this problem isdiscussed in Prob.21–48.at= dv> dt = 0,u = tan- 1av2g rb t an u =v2g rNCNC cos u - mg = 0+c© Fb= 0;NC sin u = mv2r:+© Fn= man;u .NCanNCUnpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from thepublisher prior to any