WILLIAM F. RILEY LEROY D. STURGES [Solutions Manual|Volume 1¢ 2 ‘hapters 1-6 to Accompany + cnaPmeRs 7-14 ENGINEERING MECHANICS: STATICS secon eomon William F. Riley Professor: Eo Lowa State University Leroy D. Sturges toot Ses OEE JOHN WILEY & SONS, INC. NewYork Chichester Brishane Toronto Singapore Copyegbt © 1996 by Hb Wikey & Ses, ne ‘Tha terial may be repaced Lo esting oF Sumi! purpuses by people using etext, ISBNO-TI-I61688 Prioiedthe Uned Soca of America 19876543208 INTRODUCTION Thig manual was written with the same care and precision with which the book was wrilten, Included is a complete solution for every problem in the Book Each sohtion appears with the original problem statement and, where appropriate, the probleat fig- ‘ure, This is done for the convenience of the instructor, who ne longer will have Lo refer to both the bank and the solutions marual in preparing for class. ‘Asa guide to the problem material, problems with answers that are included in the back of the book are marked with an asterisk {*], and problems intended to be solved with the aid of a programmable calculator or a computer are marked with a [Ch Solutions Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter? Chapter 8 Chapter 9 Chapter 10 Chapter 1 TABLE OF CONTENTS, General Principles Cancustent Foxce Systems Statics of Porticlew Rigid Bodies: Equivalent Horce/ Moment Systems Distributed Forces: Centrids and Center of Gravity Bnuilibrium of Rigid Badtes Trusses, Prames, and Machines Internal Forces in Structural Members Friction Second Moments of Ares and Moments of Inestia Method of Vietwal Work 958 1097 FNOIUBEKING MECHANICS - STATICS, gud. Ed. W. R. RILEY AND L. D. STURGES J-1* Calculate the maga a of $ body that welghe GOO 1b aL the surface of the earth. ‘OLUTION 16.65 siug ane. 1-24 Calenlate the walght W of a body at the anrface of the earth if it haa a nace @ of B75 kg. SOLUTION = 6T519.807) = 6.62(107) N= 6.62 ke ans. 1-3 Calculate the gravitational constaut , in ll.S. Customary units, for fb location on the surface of the acon. SOLUTION Sey | 35.4391 10") (5.097 107" 2 2 azn & B03! b.0a8 rU/e® a 5.38 ft/e* ane. Pa (5. 702(30°11 —_ T-4 Calewlate the gravitational constant u, in SI unite, for location ou the surface of the gun. SOLUTZON 8.673(40" **)41.980) 107° 984 0.274L(207) e/a a 274 w/e” Aus, [6-960¢ 10°) 1 T-B¥ The planet Veous haa a dianeter of 7700 ai und a nase of 9.94(10"°} slug. Determine the gravitational acceleration at the surface of Ue] Hse. SOLUTION hy | scegotso"®y19.991000°9) Say. g peje” ane, vo 17700(5280)/2]7 f] NGEMERMING MRCHAEICS — STATICS, 2nd. Fd, fi. F, RILEY AND 2. D, STURGES 1-68 the gravitational acceleration yy lke surface of tars is 9.73 w/s* and the mies of Mars ip 6.39(10"") kg. Detarmine the radius of Mare. SOLUTION From fa. {1-3} Fa *1¢6.981410' BSc] (5.67310 a Sra a.ag1(10%) a @ 3.98(00") @ ane. 1-1 Determine the gtavieational force, in Kipa, exerted by the sua an the exer he ‘SDLUTION got10"® 110 (4, s0ar10"*y fc o.7974:708) Te = 7.971107) kip ana. 1-8 Uelermine the gravitationat force, in kilonetons, exerted by the earth of the wnon SouTTON 26731100 '2)15. 9761 (10"*y17.350) (1077 Ts.aascn0" 1? = Leagarto®) y= 2yggci0") kw ans. The equatorial radiye of the earth ix 20926410) ft and the polar tdi 2 20868(10"| ft. nlecermine the gravitational acceleration fat she <0 Lorat ione. SOLUTION seaycau™ (4.095100). 9p. 16 eure aes. [2,092500711 40"): 3.38 1/3 7 sas. §2,0a5a(107 17 2 TUNGINEERING NECGHANICS - QTATICS, 2nd, ed, W. F. RILEY AND L, D, STURGES 1-10¢ Two spherical bodles have masses of 60 kg and 80 kg, respectively. Detoruine the gravitational force of attraction between the spheres if the distance fron center to conter 19 600 an. SOLUTTON 2 _ 07310”! 9160180) 9 goc10 ss Ans. (0. 600) T-N Two solid spherical bodies have 12 in. and,10 iu. diameters and are node of a waterial that weiuhs 0.2] Ib/in: Determine the Gravitational force of attraction between the tno upheres when they are touching each other, wustor - (Gee) = ossen 5 nm, = 9.03690(6)° = 7.998 elue am, = 0.036985)" = 4.823 slug eat — TH12 A satellite da placed in orbit 1.8(10°) = aboye the surface of the hoon, If the uaa of the satellite ie 3.010.) kx, determine the Gravitational force exerted on the satellite hy the oon. SOLUTION Aro") + 1.798(10%) = 3.938010") @ s.g7910"!* 20") 7.501.302) £3.988(10")1* 2 13.21(207) wa 19,21 ie 3 ERCINGERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L, Dy STURGES - 1-19* Determine the Kelght W of @ satellite when it Ie dy orbit 8800 wiles abore the surface of the earth if the satellite Keighs 7600 Ib at the austaves ‘S0UITTOR Frem eq. (2-915 wet hte = re = Gon Fron Table It: 209018071 te +h = 2,090(10") + 950088200) = 6.578110") et Ana. — I-MM# Determine the weight W of a eatelLite when it is in orbit 20.2010) | w ahove tha surface of the earth if the satellite weighs 6430 4 at the surface. SoWTsON From Ra. tTats w= Wyte yet = cae 8.371010°) a From Table 1-1: b= 8,971(10") + 20,2(10") = 26.573110") = ofa | S45016. 3710 6 sas g pw 4061 ane Ee ras.srica0 4 I-18 If s vonan keighe 195 1b shen standing on the surface of the earth, how mich Would she weigh chen standing on the zarfane of the moon? SOLUTION . = Seaanc0"* 5.027 oh rs (5. 702(10" 3) eng, = Hg = DB asem) = « ee ee a TENGTYRERING MECHANICS - STATICS, 2ud, E,W. F. RILEY AND J.. D. STORGES 1-16 Neteraine the Keight W of » body Uimt bas a mass of 1000 ke Ta) At the surface of the eurtb. (E] AE Lie tap at Mt, MrKialey (6193 @ shove sea level). (r) Tn a satellite at an altitude of 250 km. SOLUTION From Eq. (1-3): wre = Hee = Ga ‘From Table 1-1: 37108) tar vy ses + 0001.987) «© s0or we 9.0838 son. Un) eg = ryt ey = BeATTD) + 6199 = 6,977C10%) a, = 2008 = MOH SEKI oppy we 909 a8 ne. ream le) cy = ry FRE BAATIIO") + 250(197) = 6.821110") o gso7(s.ar1410") 2 (6.621410")1" 9060 8 @ 9.08 KN ane. T-17* TF a man welahe 210 2b at ea level, determine the weight W of the (al at che tay of Wt. Bverest 129,028 [t above sea Level). (bl Ina sacetlite at an altitude of 200 mi. SOLUTION cnt a ok From Ba, (1-3): = Wyre = Wek = Gaye From Table 1-1 2.090(107) ft 2.anr1u") + 2oze = 2.0929028(107) ft ax0l2.090¢70 11. 209.4 Ib 209 Ib ans, 2.092808 10797 2.090010") + 200(8280) = 2.19561107) ft 080110" = 199.29 Ly a 190.3 28 Aas [2 1956¢107)] s