z link full downoad https://getbooksolutions.com/solution-manual-forengineering-mechanics-dynamics-13th-edition-byrussell-c-hibbeler/ F y F 13–1 The 6-lb particle is subjected to the action of its weight and forces F1 = 52i + 6j - 2tk6 lb, F2 = 5t i - 4tj - 1k6 lb, and F3 = - 2ti6 lb, where t is in seconds Determine the distance the ball is from the origin s after being released from rest x SOLUTION ©F = ma; (2i + 6j - 2tk) + (t i - 4tj - 1k) - 2ti - 6k = ¢32 ≤(axi + ay j + azk) Equating components: 6 ¢ 32.2 ≤ax = t - 2t + ¢ 32.2 ≤ay = - 4t + ¢ 32.2 ≤az = - 2t - Since dv = a dt, integrating from n = 0, t = 0, yields t 6 2 ¢ 32.2 ≤vx = - t + 2t ¢32.2 ≤ s = x 12 - ¢ 32.2 ≤vy = - 2t + 6t ¢32.2 ≤ +t ¢ 32.2 ≤vz = - t ¢ s y = -3 + 3t Since ds = v dt, integrating from s = 0, t = yields t instructors sx = 14.31 ft, sy = 35.78 ft sz World learning = 89.44of ft United the is not on use s= (14.31)2 + (35.78)2 permitted States When t = s then, for + ( - 89.44)2 = 97.4 ft by Thus, student the protected of the assessing is solely work work provided and of This is their des troy sale will integrity this the and courses part any of or Web) 32.2 ≤sz = - - Dissemination copyright Wide 2t laws teaching t - 7t work and Ans F1 © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher 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, Pearson Education, Inc., Upper Saddle River, NJ 07458 13–2 v = 10 ft/s The 10-lb block has an initial velocity of 10 ft>s on the smooth plane If a force F = 12.5t2 lb, where t is in seconds, acts on the block for s, determine the final velocity of the block and the distance the block travels during this time F = (2.5t) lb SOLUTION 10 + 2.5t = ¢ 32.2 ≤a : ©Fx = max; a = 8.05t dv = a dt n L10 t dv = L0 8.05t dt v = 4.025t + 10 When t = s, laws v = 46.2 ft>s teaching Web) Dissemination copyright Wide A s ds = v dt s or t L0 ds = L0 (4.025t + 10) dt States s = 1.3417t + 10t United use When t = s, for instructors World permitted of learning the is not on and the student (including work by s = 66.2 ft protected is of the assessing solely work work provided and of integri ty this This is Ans the and courses part any of their des troy sale will © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher 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, Pearson Education, Inc., Upper Saddle River, NJ 07458 13–3 P If the coefficient of kinetic friction between the 50-kg crate and the ground is mk = 0.3, determine the distance the crate travels and its velocity when t = s The crate starts from rest, and P = 200 N 308 SOLUTION Free-Body Diagram: The kinetic friction Ff = mkN is directed to the left to oppose the motion of the crate which is to the right, Fig a Equations of Motion: Here, ay = Thus, + c ©Fy = 0; N - 50(9.81) + 200 sin 30° = N = 390.5 N + : ©Fx = max; 200 cos 30° - 0.3(390.5) = 50a a = 1.121 m>s Kinematics: Since the acceleration a of the crate is constant, + laws :B A v = v0 + act teaching v = + 1.121(3) = 3.36 m>s s = s0 + v0t + : A act B instructors of the learning Uniteduse s= Web) copyrightAns Wide Dissemination States World permitted and + or on by + + (1.121) A3 B = 5.04 m the student work (including protected of the is solely work assessing this not is and Ans for workprovided and of integrity This is the and courses part of their destroy any sale will © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher 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, Pearson Education, Inc., Upper Saddle River, NJ 07458 *13–4 P If the 50-kg crate starts from rest and achieves a velocity of v = m>s when it travels a distance of m to the right, determine the magnitude of force P acting on the crate The coefficient of kinetic friction between the crate and the ground is mk = 0.3 308 SOLUTION Kinematics: The acceleration a of the crate will be determined first since its motion is known : + ( ) = v + - v0 2ac(s s0) = + 2a(5 - 0) a = 1.60 m>s : Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.3N is required to be directed to the left to oppose the motion of the crate which is to the right, Fig a Equations of Motion: + c ©Fy = may; N + P sin 30° - 50(9.81) = 50(0) laws teaching N = 490.5 - 0.5P or Web) copyright Wide Dissemination Using the results of N and a, + : ©Fx States = max; P cos 30° - 0.3(490.5 - 0.5P) United P = 224 N World permitted not of learning the is on andAns instructors = 50(1.60) use by the student (including work protected of the is solely work assessing this for workprovided and of integrity This is the and courses part of their destroy any sale will © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher 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, Pearson Education, Inc., Upper Saddle River, NJ 07458 13–5 The water-park ride consists of an 800-lb sled which slides from rest down the incline and then into the pool If the frictional resistance on the incline is Fr = 30 lb, and in the pool for a short distance Fr = 80 lb, determine how fast the sled is traveling when s = ft 100 ft SOLUTION + b aFx = max; 800 a 32.2 800 sin 45° - 30 = 100 ft s a = 21.561 ft>s 2 v1 = v0 + 2ac(s - s0) v1 = + 2(21.561)(100 2 - 0)) v1 = = 78.093 ft>s 800 + ; aFx = max; -80 = 32.2 a a = -3.22 ft>s = (78.093) v2 v2 = v1 laws Web) teaching Dissemination or copyright + 2( -3.22)(5 - 0) + 2ac(s2 - s1) permitted instructors States v2 Wide of United = 77.9 ft>s World learning on the use not Aisns and the student for (including work by protected of the assessing is solely work provided and of This is work integrity this the and courses part any of their des troy sale will © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher 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, Pearson Education, Inc., Upper Saddle River, NJ 07458 13–6 If P = 400 N and the coefficient of kinetic friction between the 50-kg crate and the inclined plane is mk = 0.25, determine the velocity of the crate after it travels m up the plane The crate starts from rest P 30° SOLUTION 30° Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.25N is required to be directed down the plane to oppose the motion of the crate which is assumed to be directed up the plane The acceleration a of the crate is also assumed to be directed up the plane, Fig a Equations of Motion: Here, ay¿ = Thus, ©Fy¿ = may¿; N + 400 sin 30° - 50(9.81) cos 30° = 50(0) N = 224.79 N laws teaching Using the result of N, or Dissemination copyright Wide a = 0.8993 m>s ©Fx¿ Web) 400 cos 30° - 50(9.81) sin 30° - 0.25(224.79) = 50a = may¿; States United instructors World permitted of learning the is Kinematics: Since the acceleration a of the crate is constant, use v = v0 + 2ac(s - s0) and student by the v not on protected = + 2(0.8993)(6 - 0) for (including of the work assessing v = 3.29 m>s is solely work work provided and of integrity this This is Ans the and courses part any of their des troy sale will © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher 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, Pearson Education, Inc., Upper Saddle River, NJ 07458 © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher 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, Pearson Education, Inc., Upper Saddle River, NJ 07458 13–126 A probe has a circular orbit around a planet of radius R and mass M If the radius of the orbit is nR and the explorer is traveling with a constant speed v0, determine the angle u at which it lands on the surface of the planet B when its speed is reduced to kv0, where k at point A B u R A SOLUTION nR When the probe is orbiting the planet in a circular orbit of radius rO = nR, its speed is given by GM v = GM r B O = B nR The probe will enter the elliptical trajectory with its apoapsis at point A if its speed is GM decreased to va = kvO = kB nR at this point When it lands on the surface of the planet, r = rB = R O 1 r GM r a1 - = r v P P R a = rP P GM 2 rP vP - = nRakA nR b = k Since h = rava GM rPvP = h copyright k2nGMRlearning rP vP = rP Also, 2GM = rP rPvP 2GM -1 nR = is rv of P P 2nGMR vP = rp(rp + nR) Solving Eqs.(2) and (3), k n rp = 2-k R Substituting the result of rp and vp into Eq (1), 2-k R = a k nR -1 u = cos k na 1-k Here u was measured from periapsis.When measured from apoapsis, as in the figure then k n-1 -1 u = p - cos a © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher 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, Pearson Education, Inc., Upper Saddle River, NJ 07458 13–127 Upon completion of the moon exploration mission, the command module, which was originally in a circular orbit as shown, is given a boost so that it escapes from the moon’s gravitational field Determine the necessary increase in velocity so that the command module follows a parabolic trajectory The mass of the moon is 0.01230 Me A Mm SOLUTION When the command module r0 = 3(10 ) m, its velocity is v= c is moving GM r B 66.73(10 around - 12 24 )(0.0123)(5.976)(10 ) m = B the circular orbit of radius 3(10 ) = 1278.67 m>s The escape velocity of the command module entering into the parabolic trajectory is v= e 2GM 2(66.73)(10 - 12 B r0 24 )(0.0123)(5.976)(10 ) m =B 3(10 ) = 1808.31 m>s laws teaching Web) Dissemination copyright Wide Thus, the required increase in the command module is ¢v = v e or c = 1808.31 - 1278.67 = 529.64 m >s -v = 530 m> permi tted Ans instructors States United of World learningthe is not on use and the student for (including work by protected of the assessing is solely work provided and of This is work integrity this the and courses part any of their des troy sale will © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher 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, Pearson Education, Inc., Upper Saddle River, NJ 07458 *13–128 The rocket is traveling in a free-flight elliptical orbit about the earth such that e = 0.76 and its perigee is Mm as shown Determine its speed when it is at point B Also determine the sudden decrease in speed the rocket must experience at A in order to travel in a circular orbit about the earth A B SOLUTION Mm GMe h = r0 v r0 a - r0 v0 b [Eq 13–21] and Central-Force Motion: Here C = [Eq 13–20] Substitute these values into Eq 13–17 gives ch e B r v 22 A - GM 0 rv 2 r0 r0v0 0 e = GMe = Rearrange Eq.(1) gives = GMe - GMe (1) GMe + e = r0 v0 Rearrange Eq.(2), we have laws( 2) or teaching Dissemination Web) copyright B v0 = instructors permitted World States (3) r0 11 + Wide e2 GMe learning Substitute Eq.(2) into Eq 13–27, = a= r0 12GMe >r0 v0 - r r0 United of use ,w student for the work is not and (includi ng (4) by the 2A B -1 protected of the assessing 1+ e is solely work work provided and of Rearrange Eq.(4), we have 1+e + 0.76 This is 1-e - 0.76 and coursesany part the their = a integrity this b C 9110 b r0 = a 2D = destroy 66.0110 2m of m into Eq.(3)saleyieldswill Substitute r0 = rp = 9A 10 B (1 + 0.76)(66.73)(10 v = p D - 12 24 )(5.976) (10 ) = 8830.82 m>s 9(10 6) Applying Eq 13–20, we have 9110 rp v a = a b np = B 66.0110 R18830.822 = 1204.2 m>s = 1.20 km>s Ans If the rocket travels in a circular free-flight trajectory, its speed is given by Eq 13–25 GM - 12 66.73110 24 215.9762110 e 9110 v e = D r0 = D = 6656.48 m>s The speed for which the rocket must be decreased in order to have a circular orbit is ¢v = v p - v c = 8830.82 - 6656.48 = 2174.34 m>s = 2.17 km>s Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher 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, Pearson Education, Inc., Upper Saddle River, NJ 07458 13–129 A rocket is in circular orbit about the earth at an altitude above the earth’s surface of h = Mm Determine the minimum increment in speed it must have in order to escape the earth’s gravitational field h Mm SOLUTION Circular orbit: GM 66.73(10 - 12 24 )5.976(10 ) e vC = B r0 3 = B 4000(10 ) + 6378(10 ) = 6198.8 m>s Parabolic orbit: 2GMe ve = B r0 2(66.73)(10 =B 4000(10 ) - 12 24 )5.976(10 ) + 6378(10 ) = 8766.4 m>s ¢v = ve - vC = 8766.4 - 6198.8 = 2567.6 m>s laws Ans ¢v = 2.57 km>s or teaching Web) Dissemination copyright Wide instructors permitted States World United use for of learning the is not on and the student (including work by protected of the assessing is solely work provided and of This is work integrity this the and courses part any of their des troy sale will © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher 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, Pearson Education, Inc., Upper Saddle River, NJ 07458 13–130 The satellite is in an elliptical orbit having an eccentricity of e 15 Mm/h = 0.15 If its velocity at perigee is vP = 15 Mm>h, determine its velocity at apogee A and the period of the satellite P SOLUTION m Here, vP = c 15(10 ) h d a 3600 1h sb A = 4166.67 m>s h = rPvP h = rP (4166.67) = 4166.67rp (1) and GMe C = rP ¢1 - rP vP ≤ -12 24 66.73(10 )(5.976)(10 ) C = rp B1 - R rp(4166.67 ) laws( R C = rP B1 rP 22.97(10 ) e= 2) teaching GMe Ch DisseminationWide copyright 22.97(10 ) instructors States r P r B1 - R(4166.67 rP) P 0.15 = 66.73(10 -12 Web) permitted World United by use 24 )(5.976)(10 ) for rP = 26.415(106) m protected of learningthe is not student the on and work of the assessing Using the result of rp is work rP rA = This is 2GMe solely and work and of thisintegrity provided the part destroy cours es their any -1 rP vP of 26.415(10 ) = 2(66.73)(10 -12 sale will 24 )(5.976)(10 ) - 26.415(10 )(4166.67 ) = 35.738(10 ) m Since h = rP vP = 26.415(10 )(4166.67 ) = 110.06(10 ) m >s is constant, rA vA = h 35.738(10 )vA = 110.06(10 ) vA = 3079.71 m>s = 3.08 km>s Ans Using the results of h, rA, and rP, p T =6 A rP + rA B r r P A p 6 6 = 110.06(10 ) C 26.415(10 ) + 35.738(10 )D 226.415(10 )(35.738)(10 ) = 54 508.43 s = 15.1 hr Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher 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, Pearson Education, Inc., Upper Saddle River, NJ 07458 13–131 A rocket is in a free-flight elliptical orbit about the earth such that the eccentricity of its orbit is e and its perigee is r0 Determine the minimum increment of speed it should have in order to escape the earth’s gravitational field when it is at this point along its orbit SOLUTION To escape the earth’s gravitational field, the rocket has to make a parabolic trajectory Parabolic Trajectory: 2GMe ye = A r0 Elliptical Orbit: Ch e = GMe where C = ¢ GMe r0 e=a GMe r0 ¢ - r0y0 ≤ and h = r0 y0 GMe ≤ laws ry or - 1b Dissemination Web) r0 y0 copyright (r0 y0) 1- e= =e+ GMe (e + 1) United y0 = r0 B Ar - A student =Ar r a 22 - 21 + GMe (e + 1) 2GMe World use GMe ¢y = instructors not permitted learning of the is on States Wide GMe r0 y0 teaching GMe pro tecte d byfor the assessing is solely work and work (includingebthe Ans of work provided and of integrity this This is the and courses part any of their des troy sale will © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher 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, Pearson Education, Inc., Upper Saddle River, NJ 07458 *13–132 The rocket shown is originally in a circular orbit Mm above the surface of the earth.It is required that it travel in another circular orbit having an altitude of 14 Mm.To this,the rocket is given a short pulse of power at A so that it travels in free flight along the gray elliptical path from the first orbit to the second orbit.Determine the necessary speed it must have at A just after the power pulse, and at the time required to get to the outer orbit along the path AA ¿ What adjustment in speed must be made at A¿ to maintain the second circular orbit? A A' O Mm 14 Mm SOLUTION r0 Central-Force Motion: Substitute Eq 13–27, 6 = (14 + 6.378)(10 ) = 20.378(10 ) m and r0 = (2GM>r0v0 ) - , with = rp = (6 + 6.378)(10 ) = 12.378(10 ) m, we have 2(66.73)(10 - 12 ¢ 24 )[5.976(10 )] 12.378(10 )vp 20.378(10 ) = ≤ - laws teaching 12.378(10 ) Dissemination or copyright Wide instructors permitted States World of learning the is United on not vp = 6331.27 m>s r Applying Eq 13–20 we have 12.378(10 ) p use and student v = a ¢ra ≤ v (including work for = B 20.378(106) R (6331.27) = 3845.74 m> sby p the protected is Eq 13–20 gives h = rp vp = 12.378(10 )(6331.27) p Eq 13–31, we have = = This and and of work the 78.368(10 ) m >s.Thus, applying integrity this courses part the is destroy any [(12.378 + 20.378)(106 )]their212.378(20.378)(10 of p solely provided ra)2rp of assessing work h T = (rp + ) sale will 78.368(10 ) = 20854.54 s The time required for the rocket to go from A to A¿ (half the orbit) is given by T t = = 10427.38 s = 2.90 hr Ans In order for the satellite to stay in the second circular orbit, it must achieve a speed of (Eq 13–25) GM vc = A e r = 66.73(10 A Web) - 12 24 )(5.976)(10 ) 20.378(10 ) = 4423.69 m>s = 4.42 km>s Ans The speed for which the rocket must be increased in order to enter the second circular orbit at A¿ is ¢v = vc - va = 4423.69 - 3845.74 = 578 m s Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher 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, Pearson Education, Inc., Upper Saddle River, NJ 07458 ... crate’s acceleration in t = s if the coefficient of static friction is ms = 0.4, the coefficient of kinetic friction is 20 mk = 0.3, and the towing force is P = (90t ) N, where t is in seconds SOLUTION. .. determine the crate’s initial acceleration if the coefficient of static friction is ms = 0.5 and the coefficient of kinetic friction 20 is mk = 0.3 SOLUTION Equations of Equilibrium: If the crate is... magnitude is increased until the crate begins to slide Determine the crate’s initial acceleration if the coefficient of static friction is ms = 0.6 and the coefficient of kinetic friction is mk