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STU DENT SOLUTIONS MAN UAL J. Richard Christman Professor Emeritus U.,S. Coast Guard Academy FUNDAMENTALS OF PHYSICS Eighth Edition David Halliday (Jnivers iQ of P itts burgh Robert Resnick Rens s elaer Polytechnic Institute Jearl Walker Cleveland State Univers iQ John Wiley & Sons, Inc.
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Cover Image: @ Eric Heller/Photo ResearchersBicentennial Logo Design: Richard J. PacificoCopyright @ 2008 John Wiley & Sons, Inc. All rights reserved.No part of this publication may be reproduced, stored in a retrieval system or transmittedin any form or by any means, electronic, mechantcal, photocopying, recording, scanning,or otherwise, except as perrnitted under Sections 107 or 108 of the I 97 6 lJnited StatesCopyright Act, without either the prior written permission of the Publisher, orauthorizationthrough payment of the appropriate per-copy fee to the CopyrightClearance Center, Inc ., 222 Rosewood Drive, Danvers, MA 01923, or on the web atwww.copyright.com. Requests to the Publisher for permission should be addressed to thePermissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ07030-5714, (201) 148-6011, fax (201) 748-6008, or online athttp //www. wilelz. c om/ go/p ermi : ss ions.To order books or for customer service, please call 1-800-CALL-WILEY (225-5945).rsBN- 13 978- 0-47 r-779s8-2Printed in the United States of Amerca10 9 8 7 6 s 4 3 2 |Printed and bound by Bind-Rite Graphics.
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PREFACEThis solutions manual is designed for use with the textbook Fundamentals of Physics, eighthedition, by David Halliday, Robert Resnick, and Jearl Walker. Its primary puqpose is to showstudents by example how to solve various types of problems given at the ends of chapters in thetext.Most of the solutions start from definitions or fundamental relationships and the final equationis derived. This technique highlights the fundamentals and at the same time gives students theopportunity to review the mathematical steps required to obtain a solution. The mere pluggingof numbers into equations derived in the text is avoided for the most part. We hope students willlearn to examine any assumptions that are made in setting up and solving each problem.Problems in this manual were selected by Jearl Walker. Their solutions are the responsibility ofthe author alone.The author is extremely grateful to Geraldine Osnato, who oversaw this project, and to hercapable assistant Aly Rentrop. For their help and encouragement, special thanks go to the goodpeople of Wiley who saw this manual through production. The author is especially thankful forthe dedicated work of Karen Christman, who carefully read and coffected an earlier version ofthis manual. He is also grateful for the encouragement and strong support of his wife, MaryEllen Christman.J. Richard ChristmanProfessor EmeritusIJ.S. Coast Guard AcademyNew London, CT 06320
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Chapter L 1yd - (0.9144mX106 pmlm) - 9.144 x 105 pm.3use the given conversion factors.(a) The distance d, in rods is (4.0 furlongsX2Ol . 168 m/furlong) d - 4.0 furlongs - 5.0292mf rod(b) The distance in chains is (40 furlongsX20l 168 m/furlong) d - 4.Lfurrongs - : 4|chains . 20.L7 mlchain-I(a) The circumference of a sphere of radius R is given by 2r R. Substitute R - (6.37 x106mX10-tk^lm)should obtain 4.00 x 104km.(b) The surface area of a sphere is given by 4trR2, so the surface area of Earth is 4n(6.37 x103 k*)(c) The volume of a sphere is given by (4nlrR3, so the volume of Earth is G"13X6.37 x103 k*)3 _ 1.08 x 1012 km3.t7None of the clocks advance by exactly 24h in a 24-h period but this is not the most importantcriterion for judging their quality for measuring time intervals. What is important is that theclock advance by the same amount in each 24-h period. The clock reading can then easily beadjusted to give the correct interval. If the clock reading jumps around from one 24-h period toanother, it cannot be coffected since it would impossible to tell what the coffection should be.The followittg table gives the coffections (in seconds) that must be applied to the reading oneach clock for each 24-h period. The entries were determined by subtracting the clock readingat the end of the interval from the clock reading at the beginning. Chapter I
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CLOCK Sun. Mon. Tues. Wed. Thurs. Fri. -Mon. -Tues. -Wed. -Thurs. -Fri. -Sat A -16 -16 -15 -17 -15 -15 B -3 +5 -10 +5 +6 -7 C -58 -58 -s8 -s8 -58 -58 D +67 +67 +67 +67 +67 +67 E +70 +55 +2 +20 +10 +10Clocks C and D are the most consistent. For each clock the same coffection must be appliedfor each period. The coffection for clock C is less than the coffection for clock D, so we judgeclock C to be the best and clock D to be the next best. The coffection that must be applied toclock A is in the range from 15 s to 17s. For clock B it is the range from -5 s to *10 s, for clockE it is in the range from -70 s to -2s. After C and D, A has the smallest range of correction,B has the next smallest range, and E has the greatest range. From best the worst, the ranking ofthe clocks is C, D, A, B, E.21(a) Convert grams to kilograms and cubic centimeters to cubic meters: 1 g _ I x 10- kg and1cm3 - (1 x To-2 m)3 rs-(1 e)(#) (,%) t :rx1o3kg(b) Divide the mass (in kilograms) of the water by the time (in seconds) taken to drain it. Themass is the product of the volume of water and its density: M - (5700m3X1 x 103 kg/nt)5.70 x 106kg. The time is t - (10.0hX3600s/h) - 3.60 x 104 s, so the mass flow rate n is M- 5.70 x 106kg- R - t 3.60 x 104 s 1 58 kg/s "3s(a) The amount of fuel she believes she needs is (750mi)l(0mif gal): 18.8gal. This is actuallythe number of IJ.K. gallons she needs although she believes it is the number of IJ.S. gallons.(b) The ratio of the U.K. gallon to the U.S. gallon is (4.545963 IL)l(3.785 3060L) : 1.201.The number of U.S. gallons she actually needs is (18.8 IJ.K. galx 1.201 U.S gallu.K. gal) : 22.5 IJ.S. gal .39The volume of a cord of wood is V - (8 ftx 4 tt)(4 ft) : I28ft3. IJse l ftAppendix D) to obtain V - I28ft3X0 .3048 ft)3 - 3.62m3. Thus 1.0 m3 of wood coffespondsto (l 13.62) cord - 0 .28 cord. ^l2 Chapter I
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4T(a) The difference in the total amount between 73 freight tons and 73 displacement tons is (8 barrel bulk/freight ton)(73 freight ton) - (7 barrel bulk/displacement ton)(73 displacementton) : 73 barrel bulk .Now l banel bulk - 0.141 5m3 - (0.!41 5m3)(28.378U.S. bushel): 4.01 5U.S. bushel ,SO 73barceI bulk - (73 barrel bulk)(4.01 5 U.S. bushellbarcel bulk) - 293 U.S. bushel .(b) The difference in the total amount between 73 register tons and 73 displacement tons is (z}barrel bulk/register ton)(7 3 register ton) - (7 barrel bulk/displacement ton)(73 displacementton) : 949 barrel bulk.Thus g4:gbarrel bulk - (949barrel bulk)(4.01 5U.S. bushellbarcel bulk):3810U.S. bushel .45 m-1x10-3km, 0.12 ATJ lmin.57(a) We want to convert parsecs to astronomical units. The distancebetween two points on a circle of radius r is d - 2r stn 0 12, where0 is the angle subtended by the radtal lines to the points. See thefigure to the right. Thus r - d,lz sin 0 12 and t /-" l pc - ^",., - 2.06 x 105 AU, ., " 012 2 sin( I" 12)where !" - - (2.78 x 10-4) was used. Finally (I13600)" 1 AU - (1 AU) 1Q.06 x 105 AU lpc) : 4.9 x 10-u p, .(b) A light year is (1.86 x 10s mrlsxl.0 yX3 65.3 daly)Q|hld)(3600 s/h) - 5 .87 x 1012 mtand l Au : 929 x J l.5g x 1o- ly. 5.87xlot2milly- },9^6 Tt,; LJ) Chapter I
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Chapter 21(a) The average velocity during any time interval is the displacement during that interval dividedby the duration of the interval: ?)av - Lr I Lt, where Lr is the displacement and Lt is the timeinterval. In this case the interval is divided into two parts. During the first part the displacementis Lr 1 : 40 km and the time interval is : (40km) Ltr (30 k*/h) -1.33h.During the second part the displacement is Lrz: 40km and the time intenral is (4gq) Ltz-(60 l*ttr - o 67h Both displacements arc in the same direction, so the total displacement is Lr40km +40km:80km. The total time interval is Lt: Lt1 * Ltz- 1.33h+0.67h:2.00h.The average velocity is (80km) : nl avg 40k*/h. (2.0 h)(b) The average speed is the total distance traveled divided by the time. In this case the totaldistance is the magnitude of the total displacement, so the average speed is 40 km fh.(c) Assume the automobile passes the origin at r g0time t : 0. Then its coordinate as a function of Gm)time is as shown as the solid lines on the graph 60 -- --Q- - --J -- -r -to the right. The average velocity is the slope of 40the dotted line. 20 0.5 1.0 1.5 2.0 t (h)-5Substitute,inturn, t- I,2,3, and4sintotheexpression r(t):3t-4*+#,where r isinmeters and t is in seconds: (a) r(Is): (3^lsxl s) - (mls2xl r)2 + (l^11311 s;3 - 0 (b) r(zs) : (3mlsx2 s) - (4mls(2s)2 + Qmls;12 s) - -2m (c) r(3s): (3^lsx3 s) - (4mls2x3 s)2 + Omlr311: r13 -0 (d) r(4s): (3mlsx4s) - (4mls2x4s)2 + Q^lr3;1+s;3(e) The displacement during an interval is the coordinate at the end of the interval minusthe coordinate at the beginning. For the interval from t -- 0 to t - 4 s, the displacement isLr - r(4s) - tr(0) - I2m- 0 - +12m. The displacement is in the positive r direction.4 Chapter 2
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(0 The average velocity during an interval is defined as the displacement over the intervaldivided by the duration of the interval: uavs- L, I Lt. For the interval from t- 2s to t-4sthe displacement is Lr - r(4 s) r(2 s) - I2m (-2m) - I4m and the time intenral isLt: 4s - 2s : 2s. Thus Lr uavg:E- l4m -7m/s.(d) The solid curye on the graph to the right ^ r r2.oshows the coordrnate r as a function of time. (m)The slope of the dotted line is the average 9.0velocitybetween t- 2.0s and t- 4.0s. 6.0 3.0 0.0 -3.019If ur is the velocity at the beginning of a time interval (at time t)and u2 is the velocity at theend (at tz), then the average acceleration in the interval is given by eavTake h : 0, u1 : 18 m/s, t2 - 2.4 s, and u2 auus -30 m/s - 18 m/s - t -zo ^lr 2.4s .The negative sign indicates that the acceleration is opposite to the original direction of travel.25(a) Solve u- us* at for t: t - (u - uo) f a. Substitute u-0.1(3.0 x 108 mls):3.0 x I07 ml s,u0 : 0, and e,:9.8(b) Evaluate r ^lr. The result is t - 3.06 x 106 s. This is !,zmonths.4.6 x 1013 m.27Solveu2 -u3+2a(r-r0 fora. TakeffO:0. Then a-(r-ril|2tr. Useu0:1.50x 105 mls,u - 5.70 x 106m/s, and r - 1.0cm - 0.010m. The result is (5.70 x 106m/s)2 - (1.50 x 105m/s)2 a_ - 1.62x l0r, ^/r, . 2(0.010m)33(a) Take frs- 0, and solve r- ust * *ot for a: a_ 2(r ust)f t2. Substitute r- 24.0m,Ug - 56.0k*/h - 15.55 m/s, and t - 2.00 s. The result is 2lzq.0m- (15.55m1sx2.00s)] 1 rr ^ r^Z &- Chapter 2 5
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The negative sign indicates that the acceleration is opposite the direction of motion of the car.The car is slowing down.(b) Evaluate u(30.3 km lh).45(a) Take the A axis to be positive in the upward direction and take t - 0 and A - 0 at the pointfrom which the wrench was dropped. If h is the height from which it was dropped, then theground is at A : -h. Solve u2 - uzo + 2gh for h: u2 ul h- 2sSubstitute uo : 0, 1) : -24mls, and g - g.8 m/s2: (24mls)2 :29.4m. h- 2(9.8 m/s2)(b) Solve 1) : uo - gt for t: (to -u) 24mls t- -2.45s. 9.8 ^l12(c) t (s) t (s) 2 2 0 0 a-10 u -10 (-)-zo (m/s) _20 -30 -30The acceleration is constant until the wrench t (s)hits the ground: a- -g.8 m/s2. Its graph is 0as shown on the right. a-5 (-is2) -10 _1547(a) At the highest point the velocity of the ball is instantaneously zero. Take the A axis to beupward, set u in u2 and solve for u$ us- EA. Substitute g6 Chapter 2
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A:50mtoget 2(9.8^ls2x50m) - 31 m/s.(b) It will be in the atr until A-0 again. Solve A - uot t7t for t. Since A the twosolutions aret:0 and t : 2uo I g . Rej ect the first and accept the second: t-_ .L 2uo 2(31 m/s) - -6.4s. I 9.8 ^lt(c) a60 u40 (m) (n/s) 40 20 20 0 _20 68 t (s) -40The acceleration is constant while the ball is in t (s)flight: a, 0on the right. a-5 (*/r) -10 _1549(a) Take the A axis to be upward and place the origin on the ground, under the balloon. Sincethe package is dropped, its initial velocity is the same as the velocity of the balloon, +lTm/s.The initial coordinate of the package is ao: 80 m; when it hits the ground its coordinate is zero.Solve a : Uo + uot - *gt for t: t-uo+ L r :J.-TD-c I 9.8 ^lt ;-i- ll V (9.8 mls2)2 g.g m/s2where the positive solution was used. A negative value for t coffesponds to a time before thepackage was dropped.(b) Use t):7)0 - gt: l}mls - (9.8 mls2x5.4s) - -.41 m/s. Its speed is 4Imf s.51The speed of the boat is given by u6 _ d lt, where d is the distance of the boat from the bridgewhen the key is dropped (IZm) andt is the time the key takes in falling. To calculate t, put the Chapter 2
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origin of the coordinate system at the point where the key is dropped and take the y axis to bepositive in the upward direction. Take the time to be zero at the instant the key is dropped. Youwant to compute the time t when A - -45 m. Since the initial velocity of the key is zera, thecoordinate of the key is given by A - -+g*. Thus 2(-45 m) -3.03s.This means 12m U6: - 4.0 m/s . 3.03 s--I5First find the velocity of the ball just before it hits the ground. During contact with the groundits average acceleration is given by Lu aavg , Nwhere Lu is the change in its velocity durittg contact and Lt is the time of contact.To find the velocity just before contact take the A axis to be positive in the upward directionand put the origin at the point where the ball is dropped. Take the time t to be zero when it isdropped. The ball hits the ground when A - -15.0m. Its velocity then is found fromu2- -ga,SO - -l -ze.g^ls2x-15.0m) : -17.rmls.The negative sign is used since the ball is traveling downward at the time of contact.The average acceleration during contact with the ground is rt: - (- 17 .I m/s) *avs0 zo.o x 1o-3 s 857 ^lt2The positive sign indicates it is upward.89The velocity at time t is given by u: f adt: f S.tdt:2.5t2+C, where e is aconstantofintegration. use the condition that u- +ITmlsatt- 2.0stoobtain C:u-2.5t2 - ITmls-2.5(2.0 s)2 - 7.0 m/s. The velocity att - 4.0 s * 7.0^lsr2.5t2 - 7.0^ls*2.5(4.0 r)2 : 47 m/s.9l(a) First convert the final velocity to meters per second: 1) : (60 k*/hx1000 m/km)1Q600 s/h) - 16.7mf s. The average acceleration is ult - (16.7^ls)/(5.4s):3.1 ^lr.(b) Since the initial velocity is zero,the distance traveled is r - Lot - Ltl .L ^ls2x5 .4 r)2 - 45 m.8 Chapter 2
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(c) Solve n - *ot for t. The result is 2(0.25 103 m) - 13 s.97The driving time before the change in speed limit was t6 - Lr luo, where Lr is the distance and u6 is the original speed limit. The driving time after the change is to - A^r lro, where I)q isthe new speed limit. The time saved is t6- te- Lr(*;) | I )-r2h. -(700km)(0.62ramilkm)( milh LL ) ss 65 mi lh)This is about t h and TZmrn.99Let t be the time to reach the highest point and us be the initial velocity. The velocity at thehighest point is zero, so 0 - us - gt and us- gt. Thus H- uot i7t: gtz ig*where the substitution was made for ns. Let H2 be the second height. It is given by H2 :LgQt)2 - 2gt2 4H. The balls must be thrown to four times the original height.I07(a) Suppose the iceboat has coordinate Ut at time h and coordinate Az at time t2. If a is theacceleration of the iceboat and tre is its velocity att - 0, thenAr: ?rltt+Lot? andy2: u0t2**t7.Solve these simultaneously for a and us. The results are 2(azh - aftz) t1t2(t2 - tt) ug:ffiandTake h - 2.0s and tz: 3.0s. The graph indicates that at: 16m and Uz - 27 m. These valuesyielda, - 2.0^lt2 and us - 6.0m/s.(b) The velocity of the iceboat at t - 3.0s is ,u: uo* at:6.0 mls * (2.0m1s2x3.0s) - L}mf s.(c) The coordinate at the end of 3.0 s is az - 27 m. The coordinate at the end of 6.0 s isUz: uotz+ |"*3- (6.0^lsx6.0s) + l(2.0m1s2x6.0r) - 72m. The distance traveled during thesecond 3.0-s interval is Uz - Uz:72m - 27 m- 45m. Chapter 2
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Chapter 31 a: : :(a)use lmt to obtain e, (-25.0 m)2 + (+40.0 m)2 47 .2m.(b) The tangent of the angle between the vector and the positive r axis is aa 4oo m tano - clr- -25.0 mThe inverse tangent is -58.0o or -58.0" + 180 - I22". The first angle has a positive cosineand a negative sine. It is not correct. The second angle has a negative cosine and a positive sine.It is correct for a vector with a negative r component and a positive A component.3The r component is given by ar - (7 .3 m) cos 250o -- -2.5 m and the A component is givenby a,athe components can also be computed using ar - -Q.3 m) cos 70" and e,a - -Q,3 m) sin 70o .It is also 20" from the negative A axis, so you might also use a,n - -Q.3 ) sin 20" and a,a :-(7.3 m) cos 20". These expressions give the same results.7(a) The magnitude of the displacement is the distancefrom one corner to the diametrically opposite corner:d - / (3.00 m)2 + (3 .70 m)2 + (4.30 m)2see this, look at the diagram of the room, with thedisplacement vector shown. The length of the diag-onal across the floor, under the displacement vector,is given by the Pythagorean theorem: L- tM,where (. is the length and u is the width of the room.Now this diagonal and the room height form a right tri-angle with the displacement vector as the hypotenuse,so the length of the displacement vector is given by d-{L2+h2-(b), (c), and (d) The displacement vector is along the straight line from the beginning to the endpoint of the trip. Since a straight line is the shortest distance between two points the length ofthe path cannot be less than the magnitude of the displacement. It can be greater, however. Thefly might, for example, crawl along the edges of the room. Its displacement would be the samebut the path length would be (. + ut * h" The path length is the same as the magnitude of thedisplacement if the fly flies along the displacement vector.10 Chapter 3
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(e) Take the r axis to be out of the page, the y axis to be to the right, and the z axis to be upward.Then the r component of the displacement is u) : 3.70 m, the A component of the displacementis 4.30m, and the z component is 3.00m. Thus i- (3.70m)i +(4.30m)j+(3.00m)t. You maywrite an equally correct answer by interchanging the length, width, and height.(f) Suppose the path of the fly is as shown by thedotted lines on the upper diagram. Pretend there isa hinge where the front wall of the room joins thefloor and lay the wall down as shown on the lowerdiagram. The shortest walking distance between the l- /-lower left back of the room and the upper right t. /front corner is the dotted straight line shown on the ./ / ..tdiagram. Its length is ( , T umtn w (3.70m * 3.00 m)2 + (4.30 m)2 - 7 .96m . h2(a) Let d,+6. Then r, : an*b* - i- 4.0m- 13m - -g.0m andrr: &y*b, - 3.0m +7.0m -10m. Thus r-: (-9.0m)i*(10m)j.(b) The magnitude of the resultant is r : ^ lr2 + ,2 - (-9.0m)2 + (10m)2 _ 13 m. Y*a(c) The angle 0 between the resultant and the positive r axis is given by tan? - ,a lr* -(10 m)l(-9.0m) - -1.1. 0 is either -48o or I32". The first angle has a positive cosine anda negative sine while the second angle has a negative cosine and positive sine. Since the ncomponent of the resultant is negative and the A component is positive, 0 - I32o .t7(a) and (b) The vector d has a magnitude 10.0m and makes the angle 30o with the positive naxis, so its components are a,n- (10.0m)cos30o - 8.67m and aa: (10.0m)sin30o - 5.00m.The vector 6hur a magnitude of 10.0m and makes an angle of 135o with the positive n axis,so its components are b*- (10.0m)cos 135o - -7.07 m and ba _ (10.0m) sin 135oThe components of the sum are rn: a* * b*- 8.67 m - 7.07 m- 1.60m and ra: ea * ba:5.0m*7.07 m- 12.1 m.(c) The magnitude of iis r - ym: (1.60m)2 + (12.I - I2.2m.(d) The tangent of the angle 0 between i and the positive r axis is given by tan? _ rr lr* -(lz.Lm)l(l.60m) : 7.56. e is either 82.5o or 262.5o. The first angle has a positive cosine anda positive sine and so is the correct answer. Chapter 3 11
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39Since ab cos Q : a*b* * oab, r a"br, a*b*+a,abr+&"b, cos d: abThe magnitudes of the vectors given in the problem are e, : (3.0)2+(3.0)2+(3.0)2b- (2.0)2 + (1.0)2 + (3.0)2 :3.7. The angle between them is found from (3.0X2.0) + (3.0X1.0) + (3.0X3.0) cos Q: _ 0 .926 (5.2)(3.7)and the angle is d:22o .43(a) and (b) The vector d, is along the r axis, so its r component is arcomponent is zero.(c) and (d) The r component of 6 is b* - b cos 0 -- (4.00 m) cos 30.0ocomponent is ba - b sin 0 - (4.00 m) sin 30.0o - 2.00 m.(e) and (f) The r component of c*is cr -- ccos(0 + 90")- (10.0m)cos I20" - -5.00m and theA component is ca: csin(9 + 90o): (10.0m)sin I20o - 8.66m.(g) and (h) In terms of components cr : pa* * qb* and ca : paa + eba. Solve these equationssimultaneously for p and q. The result is P:ffi- bac* - b*ca (3.46 mX8 .66m) - (2.00 mX-5.00 m) --667and q:ffi- arca -(3.00 - aac* mXS .66 m) -43447(a) The scalar product is d . b - abcos d : (10X6.0) cos 60o : 30(b) The magnitude of the vector product is ld*61 - ab sin d: (10X6.0) sin 60o : 52 .51Take the r axis to run west to east and the A axis to run south to north, with the origin at thestarting point. Let f,dest - (90.0 k-) j be the position of the destination and rrthe position of the sailor after the first leg of his journey and iz be the remaining displacement12 Chapter j
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required to complete the journey. The total journey is the vector sum of the two parts, so?dest : rI + iz and n:ri,,, iz - (90 km)i (50.0 km) i. - -The magnitude of the remaining trip is T2 + r?a (50.0 km)2 + (50.0 km)2 _ 103 km . ?.*The tangent of the angle with the positiye r direction is tan d : r2a lrr* - (90.0 km) 160.0 km) -1.80. The angle is either 60.9 or 180+ 60.9o :24Io. Since the sailor must sail northwest toreach his destination the correct angle is 24Io . This is equivalent to 60.9" north of west.7lAccording to the problem statement i+ E - 6.0i+ 1.0j and A- E - -4.0i +7.0j. Add theseto obtain 2A-2.0i+ 8.0j and then A- 1.0i++.0j. The magnitude of Ais (1.0)2 + (4.0)2 Chapter 3 13
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Chapter 47The average velocity is the total displacement divided by the time interval. The total displacementi is the sum of three displacements, each calcul ated as the product of a velocity and a timeinterval. The first has a magnitude of (60.0 km lh)(40.0 min) l$0.0 min lh)direction is east. If we take the r axis to be toward the east and the A axis to be toward thenorth, then this displacement is i1 - (40.0 km) i.The second displacement has a magnitude of (60.0 km lh)(20.0 min) 1rc0.0 min lh) : 20.0 km. Itsdirection is 50.0o east of north, so it may be written iz - (20.0km) sin50.0i + (20.0km)cos 50.0"i - (15.3 km)i* (12.9k*)i.The third displacement has a magnitude of (60.0 km lh)(50.0 min) l(60.0 min lh) _ 50.0 km. Itsdirection is west, so the displacement may be written fi - (-50 km) i. The total displacement is r-: r-r+ i2+ i, - (40.0km)i+ (15.3km)i+ (I2.9km)j - (50k*)i - (5.3 km) i + (12.9 km) j .The total time for the trip is 40 min * 20 mtn + 50 mininterval to obtain an average velocity of du,,e - (2.9k^lh) i + (7 .05k*/h)i. The magnitude ofthe average velocity is lr-uu*l : -7.6km/sand the angle O it makes with the positive r axis satisfies tanQ:!:.^l,n - 2.43 2.9 k-/h .The angle is d : 68o.11(a) The velocity is the derivative of the position vector with respect to time: 6 - *(i. 4*i +,t) : 8,i + rin meters per second for t in seconds.(b) The acceleration is the derivative of the velocity with respect to time: d-* (r,i.o) :8i14 Chapter 4
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in meters per second squared.17(a) The velocity of the particle at any time t is given by d - 6o + dt, where do is the initialvelocity and d is the acceleration. The r component is trr:uyrl a,nt - 3.00 mls - (1.00 mls2)tand the g component rs uy - tr1a* aat - -(0.500 mlsz)t When the particle reaches its maximumr coordinate trr:0. This means 3.00 mls - (1.00 mlst - 0 or t - 3.00s. The A component ofthe velocity atthis time is rra - (-0.500 mls2x3.00s): -1.50m/s. Thus 6 - (-1.50 mls)j.(b) The coordinates of the particle at any time t are r- nyrt + *.o*t and A - uyat + *,oot . Att - 3.00 s their values are r:(3.00 mlsx3.00s) - )ft00 mls2x3.00s)2 - 4.50mand : I _ a -;(o.5oo m/s2x3.oo s)2 -2.25m .Thus r: (4.50 m) i - (2.25m)i.29(a) Take the y axis to be upward and the r axis to be horizontal. Place the origin at thepoint where the diver leaves the platforrn. The cornponents of the divers initial velocity areuyn - 3.00m/s and uya - 0. At t - 0.800 s the horizontal distance of the diver from the platforrnis :x : uy*t - (2.00 mlsx0.800 s) : 1.60 m.(b) The drivers a coordinate is a- -+g*- -+(9.8 mls2x0.800s)2 - -3. 13m. The distanceabove the water surface is 10.0m - 3.13 m - 6.86m.(c) The driver strikes the water when A: -10.0m. The time he strikes is 2(- 10.0 m) - 1.43 sand the horizontal distance from the platform is r:1)0rt - (2.00 mlsxl .43 s): 2.86m.31(a) Since the projectile is released its initial velocity is the same as the velocity of the plane atthe time of release. Take the A axis to be upward and the r axis to be horizontal. Place theorigin at the point of release and take the time to be zero at release. Let r and y (: -730 m) bethe coordinates of the point on the ground where the projectile hits and let t be the time when ithits. Then : 1. A -uot cos 96 1gt ,,where 0o - 53.0o. This equation gives y+ -730 m + |(g.go mls2x5.oo s)2 ug: -T- *gt, / :202m/s. f cos 0g (5.00 s) cos(53.0o)(b) The horizontal distance traveled is r: uotsin0s - (202*1sX5.00 s) sin(53.0") : 806m. Chapter 4 15
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(c) and (d) The n component of the velocity is t)r : t)0 sin 06 - (202^ls) sin(53.0o) : 161 m/sand the A component is ua: -uscos 06 - gt: -(202mls) cos(53o) - (9.80 mls2x5.00 s) : -17Imf s .39Take the A axis to be upward and the r axis to the horizontal. Place the origin at the firingpoint, let the time be zero at firing, and let 0o be the firing angle. If the target is a distance d,away, then its coordinates are r- dandA:0. The kinematic equations are d:uotcos0s and0 - uotsin 0o - Lgt. Elimrnate t and solve for de. The first equation gives t - dlurcosgs. Thisexpression is substituted into the second equation to obtain 2rB sin 0s cos 9s - gd,- 0. [Jse thetrigonometric identity sin 9s cos ds _ | sin(zlil to obtain ufr sin (200 - gd,or QSmfl2)(45] m) sin(200): 4- ufi (460 mls)z - 2.12x 10-3 .The firing angle is 0o: 0.0606o. If the gun is aimed at a point a distance (. above the target,then tan2o: (, 1d, or (, - dtan?s - (45.7 m)tan0.0606 - 0.0484m - 4.84cm.47You want to know how high the ball is from the ground when its horizontal distance from homeplate is 97.5 m. To calculate this quantity you need to know the components of the initial velocityof the ball. [Jse the range information. Put the origin at the point where the ball is hit, take the yaxis to be upward and the r axis to be horizontal . If r (: 107 m) and y (- 0) are the coordinatesof the ball when it lands, then tr : uyrt and 0 - uyat - *St, where t is the time of flight of theball. The second equation gives t - 2uoa I g and this is substituted into the first equation. Use1)0n - uya, which is true since the initial angle is 0o : 45o . The result is r : 2ua I g. Thus gr (9.8 mls2xl07m) :22.9mf u,a - 2 sNow take r and A to be the coordinates when the ball is at the fence. Again ra: uyat-Lgt. The time to reach the fence is given by t - rluo* - (97.5m)lQ2,9mf s) : 4.26s.When this is substituted into the second equation the result is 1. a: utat - ,gt - (22.9^lsx4 .26s) - )e.8^ls2x4 .26s)2:8.63m.Since the ball started I.22m above the ground, it is 8.63 m + 1,.22m : 9.85 m above the groundwhen it gets to the fence and it is 9.85 m - 7 .32m - 2.53 m above the top of the fence. It goesover the fence.l6 Chapter 4
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51Take the A axis to be upward and the r axis to be horizontal. Place the origin at the point wherethe ball is kicked, or the ground, and take the time to be zero at the instant it is kicked. r andA are the coordinates of ball at the goal post. You want to find the kicking angle 0s so thatA- 3.44m when rand Ainto the second the result is gn2 y: rtan0s 2rocos2 gsYou may solve this by trial and effor: systematically try values of 0g until you find the two thatsatisfy the equation. A little manipulation, however, will give you an algebraic solutioll.Use the trigonometric identity 1/ cos 0o: 1 + tan2 0s to obtain 1 ar2 . I -^^2 tft tan2 os - ntanoo + + ;T- Y oThis is a quadratic equation for tanfls. To simplify writing the solutior, let c - ig*lr| -ifq.B0 mls2x50*)t lQ5mls)z - 19.6m. Then the quadratic equation becomes ctanz 0sr tan?o + y + c : 0. It has the solution tan 0o r* + 4(A + c)c 50m * (50 m)2 - 4(3.44 m + 19.6 mX 19.6 m) 2(19.6 m)The two solutions are tan?sand 0son the goal post.s3Let h be the height of a step and u be the width. To hit step n, the ball must fall a distancenh and travel horizontally a distance between (n - l)u and nu. Take the origin of a coordinatesystem to be at the point where the ball leaves the top of the stairway. Take the U axis to bepositive in the upward direction and the r axis to be horizontal. The coordinates of the ball attime t are given by nthe level of step n:The r coordinate then is 2n(0.203 m) n-uor (r.szmls) - (0.30e n)fr,. Chapter 4 t7
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Try values of n until you find one for which r lu is less than n but greater than n 1. Fornnlw - 2. 15. This is also greater than rL. For n- 3, r_0.535m and rlu - 2.64. This is lessthan n and greater than n - 1. The ball hits the third step.67To calcul ate the centripetal acceleration of the stone you need to know its speed while it is beingwhirled around. This the same as its initial speed when it flies off. [Jse the kinematic equationsof projectile motion to find that speed. Take the A axis to be upward and the r axis to behorrzontal. Place the origin at the point where the stone leaves its circular orbit and take the timeto be zeto when this occurs. Then the coordinates of the stone when it is a projectile are givenby r- uot and A_ -+g*. It hits the ground when r- 10m and A- -2.0m. Note that theinitial velocity is horizontal. Solve the second equation for the time: t : {:di. Substitutethis expression into the first equation and solve for uoi ug:r _9 - (10m) 15.7 m/s . 2y 2(-2.0m)The magnitude of the centripetal acceleration is a,: u2 lr - (15.7*1il2 l(1 .5m): 160 ,. ^/73(a) Take the positive r direction to be to the east and the positive A direction to be to the north.The velocity of ship A is given by d t : -t? knots) sin 4s")r i + r (zknots) cos 4s " r i : _rr")LT] I1and the velocity of ship B is given by dn ""|;;:li : -tQ8 knots) sin 40r i - tQ8 knots) cos 40"r i : _ll; ;LT:] I _rrllnffThe velocity of ship A relative to ship B is 6tn:6A-6s - (1 .0 knots) i + 1f 8. knots) j .The magnitude is uln: u2tnr*Io, (1 .0 knots)z + (38.4 knots)2 - 38.4 knots(b)The angle 0 that 6te makes with the positive r axis is UAB A , 38.4 knots -. 0 - tan-1 - tan-I 1.0 knots - 88.5o UAB rThis direction is 1.5o east of north.18 Chapter 4
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(c) The time t for the separation to become d is given by t d lr tu. Since a knot is a nauticalmile, t - (l60nautical miles) lQS. knots): 4.2h.(d) Ship B will be 1.5o west of south, relative to ship A.75Relative to the cur the velocity of the snowflakes has a vertical component of 8.0 m/s anda horizontal component of 50 km/htan 0 - unlu, - (13.9^ls)/(8.0 mls): I.74. The angle is 60o.77Since the raindrops fall vertic aIIy relative to the train, the honzontal component of the velocity ofa raindrop LS u1" 30^ls, the same as the speed of the train. If u, is the vertical component of the -velocity and 0 is the angle between the direction of motion and the vertical, then tan 0 uh/ur. -Thus t)u:uhf tan? - (30 mls)ltan70" - 10.9mf s. The speed of a raindrop is u - G*,?: (30 mls)2 + (10.9m1s)z - 32mf s.91(a) Take the positive A axis to be downward and place the origin at the firing point. Then the Acoordinate of the bullet is given by A - i7t. If t is the time of flight and A is the distance thebullet hits below the target, then 2(0.019 m) : 6.3 x 10-2 s(b) The mvzzle velocity is the initial velocity of the bullet. It is hortzontal. If r is the honzontaldistance to the target, then r : uOt and 30m Ug -y- t 6.3 x 10-2 s - 4.8 x 102 mlst07(a) Use A : uyat - *gt and na - t)ga - gt, where the origin is at the point where the ball is hit,the positive y direction is upward, and uya is the vertical component of the initial velocity. Atthe highest point ua:0, so uya - gt and a: *gt: ifg.8^ls2x3.0r)t - 44m.(b) Set the time to zero when the ball is at its highest point. The vertical component of the initialvelocity is then zero and the balls initral y coordinate ts 44m. The ball reaches the fence at timet- 2.5 s. Then its height above the ground is a: Uo_ Lst :44m- itg.8^ls(2.5 s)t(c) Since the ball takes 5.5 s to travel the horizontal distance of 97 .5 m to the fence, the horizontalcomponent of the initial velocity is 1)0r: Q7.5m)l(5.5 s) - 17.7 mf s. Since the ball took 3.0 sto rise from the ground to its highest point it must take the same time, 3.0s to fall from thehighest point to the ground. Thus it hits the ground 0.50 s after clearing the fence. The pointwhere it hits is (17 .7 s) : 8.9 m. ^1sX0.50 Chapter 4 T9
24.
111 :(a) The position vector of the particle is given by i 6ot + |dtz, where ds is the velocity at timet - 0 and d is the acceleration. The r component of this equation is r : ?Jont* +e*tT. Sinc a uo*0 this becomes r -*.o*t . The solution for t is t- - 2rfa* l zesm)l(4.0 : uyat + *ort - (8 .0 ^ls)The a coordinate then is A s) + *tz.o^1s2x3.81 s)2 : 45 m. ^1sX3.81(b) The r component of the velocity is ur: rr0* + e*t - (4.0 mls2x3.81s) - 15.2mls andthe a component is rra: uya + Lort: 8.0 mls * (2.0m1s2x3.81 s) - 15.6mf s. The speed is?r: l-j (15 .2^ls)2 + (15 .6mls)2 : 22m/s. lr"*r;--12t(a) and (b) Take the r axis to be from west to east and the A axis to be from south to north. Sum the two displacements from A to the resting place. The first is Lfij sin 37o) : (60 km) i+1+S km) j and the second is Liz- -(65 k*)i. The sum is Ar - (60 km) i-(20k*)i. The magnitude of the total displacement is Lrthe tangent of the angle it makes with the east is tan? - (-20 km)/(60 km) - -0 .33. The angleis 18" south of east.(c) and (d) The total time for the trip and rest is 50h+ 35h+ 5.0h:90h, so the magnitude ofthe ayeruge velocity is (63 km) lQ0 h) - 0.70 kmlh. The average velocity is in the same directionas the displacement, l8o south of east.(e) The average speed is the distance traveled divided by the elapsed time. The distance is75 km + 65 km - 140 km, so the average speed is (140 k*) lQ0 h) : 1.5 km/h.(0 and (g) The camel has I20h 90 h - 30 h to get from the resting place to B. If Lin isthe displacement of B from A and A4.r, is the displacement of the resting place from A, thedisplacement of the camel during this time is LrB- Lr-rest: (90km)i-(60km)i- (-20k*)j:(30k-)i+1zok-)i.Themagnitudeofthedisp1acementis-36kmandthe magnitude of the average velocity must be (36 km) lQO h) _ I.zk*/h. The angle 0 that theaverage velocity must make with the east is given by tan Qangle is 34" north of east.20 Chapter 4
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Chapter 5-5Label the two forces Ft and Fr. According to Newtons second law, F, + Fr: md,F, - rnd, - Fr. In unit vector notation Ft - (20.0 N) i and d - -(1 2^lrt;1ritr 30o) i - (I2mlr;1.os 30o)j - -6.0*lrt) i - (10. a^lr) jThus Fz - (z.0ks)(-6.o^lrt)i +(2.0ks)(-1l. mlrt)i - e0.0N)i - (-32N)i - (2rr.Di(b) and (c) The magnitude of F2 is F2: @- (-32N)2 + (-21 X;z - 3gN. Theangle that F2makes with the positive r axis is given by tan? - FzalFr* - (21 N) lQ2N) - 0.656.The angle is either 33" or 33o + 180 - 213o. Since both the r and A components are negativethe coffect result is 2I3". You could also take the angle to be 180o - 213o - -I47o.13In all three cases the scale is not accelerating, which means that the two cords exert forces ofequal magnitude on it. The scale reads the magnitude of either of these forces. In each casethe magnitude of the tension force of the cord attached to the salami must be the same as themagnitude of the weight of the salami. You know this because the salami is not accelerating.Thus the scale reading is ffig,where mtsthe mass of the salami. Its value is (11.0kgX9 .8*lr2) -108N.t9(a) The free-body diagram is shown in Fig. 5 - 16 of the text. Since the acceleration of theblock is zero, the components of the Newtons second law equation yield T - mg sin 0:0 andl7/r, mg cos 0(8.5 kexg .8^lrtl rin 30" : 42 N.(b) Solve the second equation for ,F^/: Fry - mgcos 0 - (8.5kg)(9.8^lrt;ros30o :72N.(c) When the string is cut it no longer exerts a force on the block and the block accelerates . The rcomponent of the second law becomes -mgsin e - me,, so a,: - gsin 0 - -(9.8^lr;ritt30o-4.9mf s2. The negative sign indicates the acceleration is down the plane.25According to Newtons second law F - ma, where F is the magnitude of the force, a" is themagnitude of the acceleration, and m is the mass. The acceleration can be found using theequations for constant-acceleration motion. Solve u : lJ1 * at for a: e, : u lt The final velocity Chapter 5 2l
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is u - (1600 korlhx1000m/km)1p600 s/h) : 444mls, so o (444mls)/(1.8 s) : : 247 and ^lr ^lst) -the magnitude of the force is Ir - (500 kgQaT I .2 x 105 N.29The acceleration of the electron is vertical and for all practical puqposes the only force actingon it is the electric force. The force of gravity is much smaller. Take the fr axis to be in thedirection of the initial velocity and the A axis to be in the direction of the electrical force. Placethe origin at the initial position of the electron. Since the force and acceleration are constanttheappropriateequations arer_ uot and a- Lot: ifgl*)*, where F- ma, wasusedtosubstitute for the acceleration a. The time taken by the electron to travel a distance n (: 30 mm)horizontally is t - r luo and its deflection in the direction of the force is 4.5 x 1 g-t6 y ,a2 1F 30x 10-3 m a: ;- /.m (#):;( 9.11 x 10-3t kg )( 1.2 x 107 m/s )35The free-body diagram is shown at the right. F* is the normal forceof the plane on the block and mj ts the force of gravity on the block.Take the positive r axis to be down the plane, in the direction ofthe acceleration, and the positive A axis to be in the direction of thenormal force. The r component of Newtons second law is thenmg sin 0 - ma) so the acceleration is a : g stn 0 .(a) Place the origin at the bottom of the plane. The equations formotion alongthe r axis are r: uot+)atz and u: n0*at. The blockstops when u : 0.According to the second equation, this is at the time t - -us f a. The coordinate when it stops is 2--1 3) *)"(+) fr : tJ.( -u0 6 A 2 a, 2 gstn1 2 t_ (-3.s0 mls)2 : -t.t8m | tq .B^lrt; Ltn ritt 32.0" ] (b) The time is t Ug Ug -3.50 m/s 0.674s. g srn? (9.g mlrt;,in 32.0"(c) Now set r: 0 and solve tr : uot + atz for t. The result is , 2uo 2uo 2(-3.50 m/s) t::: a g stn? (9.8 mlrt; ritt 32.0o - 1.35 s.The velocity is u - uol_ at: uo* gtsrnl - -3.50 mls + (9.8 mls2xl.35 s)sin 32o - 3.50 mls,22 Chapter 5
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as expected since there is no friction. The velocity is down the plane.45The free-body dtagrams for the links are drawn below. The force affrows are not to scale. 4onr{ F:onz{ ll-rrrJl--rr-rll fLon:{ fton+{ FA lltrl t o Ior I ltl ltl mgy mgililr on msll F3 on+ msll F+ on s Link I Link 2 Link 3 Link 4 Link 5(a) The links are numbered from bottom to top. The forces on the bottom link are the force ofgravity md, downward, and the force F, onr of hnk2, upward. Take the positive direction to beupward. Then Newtons second law for this link is Fz on 1 - mg : ma. Thus Fz onl : m(a+ g) - (0.100kgX2.50 mlr+ 9.8 mlr) - I.23N.(b) The forces on the second link arc the force of gravity md, downward, the force Fr on zof tink I, downward, and the force Fz onz of link 3, upward. According to Newtons thirdlaw Ft onz has the same magnitude as Fzonr. Newtons second law for the second link isF3 onz - fl on2 - mg : mQ,, SO Fz onz- m(&+ g)+ Fr onz- (0.100kg)(2.50 mlr+9.8 mls2)+ 1.23N - 2.46N,where Newtons third law was used to substitute the value of F2on I for Ft onz.(c) The forces on the third link are the force of gravity mj, downward, the force Fz on z of link2, downward, and the force F+ on z of link 4, upward. Newtons second law for this link isF+onZ- F2on3 -mg:TnA, SO F+on3 - m(a+ g)+ Fzon3 - (0.100NX2.50 mlr+9.8 mls2)+ 2.46N - 3.69N,where Newtons third law was used to substitute the value of F3 on2 for F2 on3.(d) The forces on the fourth link are the force of gravity mj, downward, the force Ft on + oflink 3, downward, and the force Fs on + of link 5, upward. Newtons second law for this link isFS on + - F3 on 4 - mg - mA, SO Fs on+ - m(a+ g)+ Ft on4 - (0.100kgX2.50 mlr+ 9.8 mls2) + 3.6gN - 4.92N,where Newtons third law was used to substitute the value of Faon 3 for F3 on4.(e) The forces on the top link arc the force of gravrty mrt, downward, the force Fq on s oflink 4, downward, and the applied force F, upward. Newtons second law for the top link isF - F+ on S - mg - nLa, SO F -m(a+ g)+ Fqon5 - (0.100k9(2.50 mlrt+9.8 mls2)+ 4.92N- 6.15N,where Newtons third law as used to substitute the value of Fs on 4 for F+ on s. Chapter 5 23
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(0 Each link has the same mass and the same acceleration, so the same net force acts on eachof them: Fnet - ma - (0.100 kg(z.50 mls2) - 0.25 N.s3(a) The free-body dragrams are shown to the right. F is the appliedforce and f is the force of block 1 on block 2. Note that F isapplie,C only to block 1 and that block 2 exerts the force - i onblock 1. Newtons third law has thereby been taken into account.Newtons second law for block 1 is F f - TrL1a, where o, isthe acceleration. The second law for block 2 is f -- TrL2e,. Sincethe blocks move together they have the same acceleration and thesame symbol is used in both equations. Use the second equationto obtain an expression for a: a - f l*r. Substitute into the firstequation to get F - f : TTLI f l*r. Solve for f : n Fmz Q.2NXl.2kg) f - m1 * TTL2 z.3kg + r.zk|- - 1 1 N llrr(b) If F is applied to blo ck2 instead of block 1, the force of contactis i- Fml Q.2NX2.3 kg) a_ TrLl I mz_ 2.3 kg + l.2kg t .t(c) The acceleration of the blocks is the same in the two cases. Since the contact force f isthe only horizontal force on one of the blocks it must be just right to give that block the same ---.1N.acceleration as the block to which F is applied. In the second case the contact force acceleratesa more massive block than in the first, so it must be larger.57(a) Take the positive direction to be upward for both the monkey and the package. Suppose themonkey pulls downward on the rope with a force of magnitude F. According to Newtons thirdlaw, the rope pulls upward on the monkey with a force of the same magnitude, so Newtonssecond law for the monkey is F - mrng - mrnarn, where mrn is the mass of the monkey ande,rn is its acceleration. Since the rope is massless F is the tension in the rope. The rope pullsupward on the package with a force of magnitude F, so Newtons second law for the packageis F + Flr - mpg : mpap, where mp is the mass of the package, a,p is its acceleration, and Fxis the normal force of the ground on it.Now suppose F is the minimum force required to lift the package. Then Flr - 0 and e,,p - 0.According to the second law equation for the package, this means F - mpg. Substitute wpg forF in the second law equation for the monkey, then solve for am. You should obtain : F - rrlrng : (mo - m,n)g (15 kg - 10kg)(9.8 An - _ r kg ^lr - +Y mf s 2 aTn ,rr-, *-, lo 24 Chapter 5
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(b) Newtons second law equations are F -mpg: ffipap for the package and F -rrLrng : TrL,n,arnfor the monkey. If the acceleration of the package is downward, then the acceleration of themonkey is upward, so o??" - -ap. Solve the first equation for -F: p - mp(g + op) : ffip(g - arn).Substitute the result into the second equation and solve for arni anL- (TY - *)g / mp*Tftm. 15kg+10kg(c) The result is positive, indicating that the acceleration of the monkey is upward.(d) Solve the second law equation for the package to obtain p- Trl,p(g - a,nr): (15kgX9.8mf s2 -2.0*ls2) - 120N61The forces on the balloon arc the force of gravity mj, down , and the force of the atr Fo, up.Take the positive direction to be up. When the mass is M (before the ballast is thrown out)the acceleration is downward and Newtons second law is Fo M g - - M a. After the ballastis thrown out the mass is lVI wL, where m is the mass of the ballast, and the accelerationis upward. Newtons second law is Fo (M m)g - (M m)a. The first equation givesFo: M(g a) and the second gives M(g a) (M m)g - (M m)a. Solve for m:m:ZMal(g + a).73 rTake the axis to be horizontal and positive in the direction that the crate slides. ThenFcosd facceleration (the only nonvanishing component). In part (a) the acceleration is e,r : F cos? - f - (450N)cos38" - 125N n n -^^ 1^2 -ut+m/s A m 310kg In part (b) m: Wlg _- (3 10N) lQ.8^ls) - 31.6kg and Fcos0-f (450N)cos38o 125N ,1 ..r r2 nr m 36.1 kg79Let F be the maglitude of the force, a4 (: L2.0mls2) be the acceleration of object I, and a,2(:3.30 mlt; be the acceleration of obj ect2. According to Newtons second law the masses arerrLl: Flol and TrL2- Ff e,2.(a) The acceleration of an object of mass rTL2 - mr is (L- F - - +O m/ S TrL2 - Tft1 or u, I2.0 Chapter 5 25
30.
(b) The acceleration of an object of mass TtLl 1 m2 is A- F _ F eta,z (12.0 ^ls2x3.3o *lt) - 2.6mf s2 Tft2 * mr (F lor) + (F lo) a1 a aZ 12.0 mlr + 3.3 0 mf s29l(a) Both pieces arc station dry, so you know that the net force on each of them is zero. The forceson the bottom piece are the downward force of gravity, with magnitude Tftzg, and the upwardtension force of the bottom cord, with magnitude T6. Since the net force is zero, T6: TTL27 : (4.5kgX9 .8^lr2) - 44N.(b) The forces on the top piece are the downward force of gravity, with magnitude r(Ltg, thedownward tension force of the bottom cord, with magnitude Ta, and the upward force tension ofthe top cord, with magnitude ?7. Since the net force is zero, T1 : Tb t mrg :44 N + (3.5 kg)(9.8 mlst) - 78 N .(c) The forces on the bottom piece are the downward force of gravity, with magnitude rrLsg, andthe upward tension force of the middle cord, with magnitude Trn Since the net force is zeto, : : ^ls) - T,n TTLsg (5.5 kgXg .8 54 N.(d) The forces on the top piece are the downward force of gravity, with magnitude w4g, theupward tension force of the top cord, with magnitude T7 (: I99 N), and the downward tensionforce of the middle cord, with magnitude Trn Since the net force is zero) T,n: Tt - Tft3g : IggN - (4.8 kg)(9 .8^ls2) - 152N.9s(a) According to Newtons second law the magnitude of the net force on the rider is It - ma, :(60.0kg)(3.0 mls) - 1.80 x 102N.(9) Take the t force to__b. thl vector sum of the force of the motorcycle and the force of Earth: t Fnet: Fr, + Fn. Thus Frn: 4r.t - Fs. Now the net force is parallel to the ramp and thereforemakes the angle 0 (: 10) with the horrzontal, so Fnt_ (F cos 0)i+(lrsin 0)j, where the r axisis taken to be horrzonta| and,the A axLS is taken to be vefitcal. The force of Earth is F" - -mgj,so F,- - (F cos il?+ (F sin o + milj.Thus F,n* - It cos 0 - (1.80 x I02 N) cos 10oand Frna - (1.80 x 102) sin 10o + (60.0kg)(g.g mls2) - 6.19 x 102 N.The magnitude of the force of the motorcycle is -6.44 x TA -r rn t02 N)2 r02 N)2 102N.26 Chapter 5
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99The free-body diagrams for the two boxes are shown below. Flrr FxzHere T is the tension in the cord, FxTr is the norrnal force of the left incline on box 1, and FNzis the norrnal force of the right incline on box 2. Different coordinate system are used for thetwo boxes but the positive r direction are chosen so that the accelerations of the boxes have thesame sign. The r component of Newtons second law for box 1 gives T - TTLI7 stn01 : TTL1& andthe r component of the law for box 2 gives mzg stn 02 - T : TTL27. These equations are solvedsimultaneously for T. The result is mtmzg (3.0 kgx2.0 kgX9.8 m/s2) : T: (sin 91 * sin oz) (sin 30o + sin 60.) 16 N ?TL1 * mZ 3.0 kg + 2.0kg ^101Free-body diagrams for the two tins are shown onthe right. T is the tension in the cord and Fy is Tthe normal force of the incline on tin 1. The posi-tive n direction for tin I is chosen to be down theincline and the positive r direction for tin 2 is cho-sen to be downward. The sign of the accelerations ofthe two tins arc both then positive. Newtons secondlaw for tin 1 gives T + Tft1g stn {3 - ma, and for tin 2 rgives mzg T F - TrL2cL. The second equation issolvedforT,withtheresultT- TrL2@_ a)-F-(2.0kgX9.8^lrt-5.5m1s2)-6.0N-2.6N.The first Newtons law equation is solved for sin p, with the result sin C:mra-T mts (1.0kgX9.8 m/s)The angle is I7o Chapter 5 27
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Chapter 51(a) The free-body diagram for the bureau is shown on the right.F is the applied force, "f- i, the force of friction, fi" is the normalforce of the flooq and mj is the force of gravity. Take the r axisto be honzontal and the A axis to be vertical. Assume the bureaudoes not move and write the Newtons second law equations. Ther componentis F- f :0 andthe A componentis Fl/ -rng:0.The force of friction is then equal in magnitude to the appliedforce: fforce of gravity: ,Fl/ - mg. As F increases, f increases untilf : ltrrFx. Then the bureau starts to move. The minimum forcethat must be applied to start the bureau moving is It - FrFx : ltr,mg - (0 .45)(45 kgX9.8 mls2) - 2.0 x I02 N .(b) The equation for F is the same but the mass is now 45 kg - 1,7 kg - 28 kg. Thus F - F,mg - (0 .45)(28 kgX9.8 mlst) - 1.2 x T02 N .3(a) The free-body dragram for the crate is shown on the right. F isthe force of the person on the crate,,f-ir the force of friction, Fr itthe noffnal force of the floor, and mj ts the force of gravity. Themagnitude of the force of friction is given by f - l.rrcFw, whereltr* is the coefficient of kinetic friction. The vertical componentof Newtons second law is used to find the normal force. Sincethe vertical component of the acceleration is zero, Fl/ - mg - 0and Fl/ - mg. Thus f: F;Fw : ltkmg - (0.35X55 kgX9.8 mls2) - l.g x I02 N .(b) IJse the horizontal component of Newtons second law to find the acceleration. SinceF-f:ffia, (F - f) (220N - 18eN) a: - - 0.5 6mf s2 m 55kg28 Chapter 6
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13(a) The free-body diagram for the crate is shown on the right. f isthe tension force of the rope on the crate, Fn is the norrnal force ofthe floor on the crate, rnfi rs the force of gravity, and ,i ir the forceof friction. Take the r axis to be horizontal on the right and the Aaxis to be vertically upward. Assume the crate is motionless. The rcomponent of Newtons second law is then 7 cos e - f : 0 and the Acomponent is Tsin 0+F^/ *mg:0, where 0 (: 15") is the angle be-tween the rope and the horizontal. The first equation gives / - T cos Iand the second gives Fx - mg - T sin 9. If the crate is to remain atrest, / must be less than lr"Fx, or T cos? <the tension force is sufficient to just start the crate moving T cos 0 - lrr(mg 7 sin d). Solvefor T: T : Fmg cos 0 + p,, stn0 cos 15o + 0.50 sin 15o(b) The second law equations for the moving crate are 7 cos 0 f : ma and -Fl/ + T sin 0mg:0. Now ff - p,n(mg 7 sin 0). This expression is substituted for f in the first equation to obtain? cos 0 - pn(mg - T sin 0) : ffia, so the acceleration is T(cos 0 + p,7" sin 0) a: -ltt 9 *Its numerical value ls e, = - (0.35X9 .8^ls2) - 1.3 ^lr23The free-body diagrams for block B and for the knotjust above block ,4 arc shown on the right. T1 is themagnitude of the tension force of the rope pulling on blo ck B , Tz is the magnitude of the tension force of the other rope, f is the magnitude of the force of friction exerted by the horizontal surface on block B, FAr is the magnitude of the norrnal force exertedby the surface on block B, We is the weight ofblock A, and W n is the weight of block B . 0 (- 30") is the angle between the second rope and the horizontal.For each object take the r axis to be horizontal and the A axis to be vertical. The fr componentof Newtons second law for block B is then T1 f :0 and the A component is Fy Wn:0.The r aomponent of Newtons second law for the knot is T2 eos 0 - Tr - 0 and the A componentis Tzsin? Wt:0. Eliminate the tension forces and find expressions for f and F11r in terms Chapter 6 29
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of Wt and Wn, then select Wt so f : F"Fx. The second Newtons law equation givesFyr - Ws immediately. The third gives Tz - Tr I cos 9. Substitute this expression into thefourth equation to obtain T1to obtain f - Waf tan?. For the blocks to remain stationary f must be less than F"Fx orWaltan? <Solve for We: We: FrWntanT - (0.25)(711N) tan3Oo - 1.0 x I02 N.27(a) The free-body dragrams for the two blocks areshown on the right. T is the magnitude of the ten-sion force of the string , Fx t is the magnitude of thenormal force on block A, Fw s is the magnitude ofthe nonnal force on blo ck B , f a is the magnitude ofthe friction force on block A, f e is the magnitudeof the friction force on block B, TTLy is the mass ofblock A, and TTLs is the mass of block B . 0 is theangle of the incline (30). We have assumed that theincline goes down from right to left and that blockA is leading. It is the 3.6-N block.For each block take the r axis to be down the plane and the A axis to be in the direction of thenormal force. For block A the r component of Newtons second law is mtgsin0- fa-T:TrLAa,gand the A component is Fnt-Tftagcos0:0.Here a,s is the acceleration of the block. The magnitude of the frictional force is ft : FkAFxt : FkAmtgcos 0,where Fxa- mtg cos9, from the second equation, is substituted. Fnt is the coefficient ofkinetic friction for block A. When the expression for f a is substituted into the first equation theresult is Tftsg sin 0 - FntThsg cos 0 - T - ?rLga,n .The same analysis applied to block B leads to Trlpg sin 0 - FnsTrlyg cos 0 + T - rTL Ba, p .We must first find out if the rope is taut or slack. Assume the blocks are not joined by a rope andcalcul ate the acceleration of each. If the acceleration of A is greater than the acceleration of B,then the rope is taut when it is attached. If the acceleration of B is greater than the accelerationof A, then even when the rope is attached B gains speed at a greater rate than A and the ropeis slack.30 Chapter 6
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Set T:0 in the equation you derived above and solve for as and a,p. The results arc a,A- g(sin0 - ltp7cos 0): (9.8 mlrt;1ritt30o - 0.10cos30") : 4.05^ltand a,s - g(sin0 - ltn6 coS 0): (9.8 mlr;1ritr30o - 0 .20cos30o): 3.20mlrt.We have learned that when the blocks are joined, the rope is taut, the tension force is not zero,and the two blocks have the same acceleration.Now go back to mAg sin 0 - Fntmeg cos 0 -T - TrL4a, and ThBgsrn) - FnsmBgcos 0 +T -TTL6a, where a has been substituted for both a,s and a,s. Solve the first expression for T,substitute the result into the second, and solve for a. The result is a: gsin g llnaml* H"nnm6 a - rYL4 g cos ^ + mB N)l (9.g N) + (020X72 - (9 .g^lrt; 30o ,i,r [(010X36 L 3.6N+7.2N l-/ mlrr; ,os 30o - 3.5 ^lrt .Strictly speakirg, values of the masses rather than weights should be substituted, but the factorg cancels from the numerator and denominator.(b) IJse ffLsg sin 0 - Fneme7 cos 0 - T - rTLAa to find the tension force of the rope: T - mAg sin 0 - FntTTLsgcos e - TtLAa, - (3.6 N) sin 30o - (0. 10X3.6 N) cos 30o - (3.6 N/9.8 mls2x3 .49 mlst) - 0.21 N .35Let the magnitude of the frictional force be au, where e, : 70 N . s/m. Take the direction of theboats motion to be positive. Newtons second law is then -aa : m du ldt Thus f du: Ju, u mJo /where us is the velocity at time zero and u is the velocity at time t. The integrals can beevaluated, with the result -uat tn*- -A.Take 1) : ,ol2 and solve for f : loookg t -m rn2 - 70N.s/m hz - g.9s. a49(a) At the highest point the seat pushes up on the student with a force of magnitude F^/ (: 556 N).Earth pulls down with a force of magnitude W (: 667 N). The seat is pushing up with a forcethat is smaller than the students weight in magnitude. The student feels light at the highestpoint. Chapter 6 31
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(b) When the student is at the highest point, the net force toward the center of the circular orbitis W - Fy and, according to Newtons second law, this must be mulR, where u is the speed ofthe student and R is theradius of the orbit. Thus mu2f R-W - f^i:667N- 556N- 111].{.The force of the seat when the student is at the lowest point is upward, so the net force towardthe center of the circle is F^i W and lr^/ W - mu2 I R. Solve for Fn: ,.,/,/- 667N- 77gN. rytW:ltlN+(c) At the highest point W F^/ : mu2 f R, so f^imu2 lR increases by a factor of 4, to 444N. Then F^/ :667 N - 444N - 223NI. :(d) At the lowest point W + Fl/ muI R, so F^/ -W-mu2fR,so-F1/ :mu2 I R is still 444 N, F^r 667 N + 444N 1.11 - x 103 N.53The free-body diagram for the plane is shown on the right" F is themagnitude of the lift on the wings and m LS the mass of the plane. Since thewings are tilted by 40" to the honzontal and the lift force is perpendicularto the wings, the angle 0 is 50o. The center of the circular orbit is to theright of the plane, the dashed line along r being a portion of the radius.Take the r axis to be to the right and the A axis to be upward. Thenthe r component of Newtons second law is F cos g - mu2 I R and the Acomponent is F sin I mg - 0, where R is the radius of the orbit. Thefirst equation gives It - mu2 I Rcos 0 and when this is substituted into thesecond, (*rlR)tan? - mg results. Solve for R: tr n12 R- tanl. IThe speed of the plane is tr :480 kmfh - 133 mls, so _ 033 mls)2 : 2.2 x 103 m R tansoo . 9.8 mls59(a) The free-body dragram for the ball is shown on the right. f,is the tension force of the upper string , ip is the tension forceof the lower string, and m is the mass of the ball. Note thatthe tension force of the upper string is greater than the tensionforce of the lower string. It must balance the downward pullof gravity and the force of the lower string.32 Chapter 6
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Take the r axis to be to the left, toward the center of the circular orbit, and the A axis to beupward. Since the magnitude of the acceleration is a -- ,2 I R, the r component of Newtonssecond law is Tucosg+ Tacos0-ry,where u is the speed of the ball and R is the radius of its orbit. The A component is 2,, sin 0 - Tzsrn? - mg - 0.The second equation gives the tension force of the lower string: Tt : Tu - mg f stn 0 . Since thetriangle is equilateral 0 - 30o. Thus (13a kgX9"9 m/s2) Tr: 35 N - - 8.74N . srn 30o(b) The net force is radially inward and has magnitude Fnet, str - (7, + TD cos 08.74N) cos 30" - 37.9 N.(c) [Jse Fn"t,str: nr,uz f R. The radius of the orbit is [(1.70m) l2)Jtan30" - 1.47 m. Thus RFn"t, str (1.a7 mX37.eN) 1.3 4kg - 6.45 mls .65The first sentence of the problem statement tells us that the maximum force of static frictionbetween the two block is f ,,^u*:12N.When the force F is applied the only horizontal force on the upper block is the frictional force ofthe lower block, which has magnitude f and is in the forward direction. According to Newtonsthird law the upper block exerts a force of magnitude f on the lower block and this force is inthe rearward direction. The net force on the lower block is F - f .Since the blocks move together their accelerations are the same. Newtons second law for theupper block gives f - Tftt& and the second law for the lower block gives F - f : TTL6a, where cris the common acceleration. The first equation gives a - f l*t Use this to substitute for a, inthe second equation and obtain F - f : (mal^t)f .Thus Ii-(r*e) f mt/ If f has its maximum value then F has its maximum value, So the maximum force that can beapplied with the block moving together is F- (t.ffi) lzN):27NThe acceleration is then LL: f l2N ._ ,) TrLl 4.0 N Chapter 6 33
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77(a) The force of friction is the only horizontal force on the bicycle and provides the centripetalforce need for the bicycle to round the circle. The magnitude of this force is f : mu2 f r, wherem is the mass of the bicycle and rider together, u is the speed of the bicycle, and r is the radiusof the circle. Thus : (85.0 kgXg.oo ml s)2 f - ,n 25.0m(b) In addition to the frictional force the road also pushes up with a nonnal force that is equalin magnitude to the weight of the bicycle and rider together. The magnitude of this force is.F1r/ - mg: (85.0kg)(9.8 The frictional and normal forces are pe{pendicular to ^lt)-833N.each other, so the magnitude of the net force of the road on the bicycle is 4r"t - , E vL * ** - Va (275 N)2 + (833 N)2 - 877 N.81The free-body diagrams are shown on theright. T is the tension in the cord, F* isthe norrnal force of the incline on block A,^F* u is the normal force of the platfonn onblock B, I is the angle that the incline makeswith the horizontal (which is also the anglebetween the normal force and the vertic aI),and ,f-ir the frictional force of the platforrnon block B . The r axis for each block is alsoshown.The r component of Newtons second law for block A gives mgsinO T- Trlye,, the frcomponent of the second law for block B is T fFxn - TTLsg - 0.Note that the blocks have the same acceleration.The magnitude of the frictional force LS ptrFxs: ltkTftsg,where ?rLpg was substituted for Fxn,and the r component for B becomes T - Fnmng - TTLsa,. The equations mgsin 0 - T - TLnaand T - Fnmng: TTLsa are solved simultaneously for T and a. The results are + (4.0 kgX2.0 kg)(sin 30o + 0.50) t _ TftlTrLB(sin 0 Fil rF r rJl 1 *^+r", 10roand (4.0kg) sin3Oo - (0.50x2.0kg) ., , r 2 e,- Masin 0 - Fnrns g: :1.6m/s ms+ mB r 4.0k9 + 2.0k98s(a) If u is the speed of the car, m ts its mass, and r is the radius of the curve, then the magnitudeof the frictional force on the tires of the car must be f : muz f r or else the cat does not negotiatethe curve. Since m - W I g, where W is the weight of the car, r. wu2 (10.7 x 103 N(13.4m1s)2 . ,_,1 1 a gr JL-J./rLzlWl. (9.g mls2x61.0 m)34 Chapter 6
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(b) The norrnal force of the road on the car is l7^/ - W and the maximum possible force of staticfriction is .f", -u* - FrFxforce that is required is less than the maximum possible, the car successfully rounds the curve.9tLet F be the magnitude of the applied force and f be the magnitude of the frictional force.Assume the cabinet does not move. Then its acceleration is zero and, according to Newtonssecond law, F - f . The norrnal force is F1,maximum force of static friction is Fr,maxif F is less than 378 N the cabinet does not move and the frictional force is fgreater than 378 N, then the cabinet does move and the frictional force is f(0.56X556 N) - 311 N.(a) The cabinet does not move and f : 222I.(b) The cabinet does not move and f : 334 N.(c) The cabinet does move and f : 311 N.(d) The cabinet does move and f : 311 N.(e) The cabinet moves in attempts (c) and (d).99(a) The free-body diagram for the block is shown on the right. Themagnitude of the frictional force is denoted by f , the magnitude -. Fl/of the normal force is denoted by lrl/ , and the angle between theincline and the horizontal is denoted by 0. Since the block is sliditgdown the incline the frictional force is up the incline. The positive /0r direction is taken to be down the incline. For the block when frit is sliding with constant velocity the n component of Newtonssecond law gives mg sin 0 f - 0 and the A component givesmg cos 0 -Fl/ - 0. The second equation gives F1/ - mg cos9, so the magnitude of the frictionalforce is fobtain mg sin 0 - Tftuamg cos 0 : 0. Thus the coefficient of kinetic friction is Fp - tan 0 .When the block is sliditrg up the incline the frictional force has the same magnitude but is directeddown the plane. The r component of the second law equation becomes mg sin 0 + Fnmg cos e -rrLa, where a, is the acceleration of the block. Thus a - (sin 0 + [L7"cos0)g - 29 sind, where tan?was substituted for Fn and tan 0 - sin 0 f cos 0 was used.If d is the displacement of an object with constant acceleration a, us is its initial speed and uis its final speed, then u2 ulr: Zad. Set u equal to zero and a, equal to 29 stn? and obtaind - -ufil2n -- -ufrlag sin9. The negative sign indicates that the displacement is up the plane.(b) Since the coefficient of static friction is greater than the coefficient of kinetic friction themaximum possible static frictional force is greater than the actual frictional force and the blockremains at rest once it stops. Chapter 6 3s
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105The box is subjected to two horizontal forces: the applied force of the worker, with magnitudeF, and the frictional force, with magnitude f .Newtons second law gives F - f : ma) where mis the mass of the box and a is the magnitude of its acceleration. The magnitude of the frictionalforce is fthe normal force of the floor. In this case F1y - mg and fbecomes F - Ltnmg - me so ltt - (lr - ma) l*g.Let u be the final speed of the box and d, be the distance it moves. Then u2 - Zad, anda,- u2 lza- (1.0 mlilz l2(1 .4m):0.3 57mf s2. The coefficient of kinetic friction is F - ma (85 N- (a0kgX0.357 m/s2) ltt - m9 - (40 kgX9.8 m/s2) 0.1836 Chapter 6
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