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# Statics CE 221

MSU

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This 52 page Class Notes was uploaded by Jolie Shields on Saturday September 19, 2015. The Class Notes belongs to CE 221 at Michigan State University taught by Gilbert Baladi in Fall. Since its upload, it has received 95 views. For similar materials see /class/207370/ce-221-michigan-state-university in Civil Engineering at Michigan State University.

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Date Created: 09/19/15

1 I quot aquot l i ME 221 STATICS CE 221 Statics I ME 222 Mechanics of Deformable Solids I J CE 305 ME 423 CE 312 ME 425 CE 321 ME 426 CE 337 ME 477 ME 495 GENERAL REVIEW Trigonometry amp Geometry GEOMETRYTRIGONOMETRY TWO Parallel Lines 2007 by R C Hibbeler To be published by Pearson Prentice Hall Pearson Education Inc Upper Saddle Riven New Jersey All rights reserved GEOMETRYTRIGONOMETRY TRIGONOMETRY If A 90 and the lengths AB and AC are equal then gt Angle G Angle 3 450 gt Angle gt ngle l3 1 gt Angle G lt Angle 3 gt Cannot be determined TRIGONOMETRY h hypotenuse 0 Opposite 6 J 007 by R C Hibbeler To be published by Pearson Prentice Hall Pearson Education Inc Upper Saddle River New Jersey All rights reserved TRIGONOMETRY Right Triangle sides relative to angle b Opposite 39 YPOtenuse C C T Adjacent TRIGONOMETRY IfA 90 AB 4 and AC 3 then C BCV b BC 7 c BC 25 TRIGONOMETRY Right Triangle AC2 BC2 AB2 3 a b c 1800 B TRIGONOMETRY Sin b Oppositehypotenuse ACAB Cos b Adj acentHypotenuse CBAB Opposite ypotenuse Adjacent Tan b OppositeAdj acent ACCB SinbCosb Cot b Adj acentOpposite CBAC 1Tanb TRIGONOMETRY Ifoc 900 AB 3 and AC 4 then c b Angle 9 4860 c Angle 9 53130 d Angle 5 36870 4 TRIGONOMETRY Circle with a radius of unit length TRIGONOMETRY Circle with a unit diameter Dlameter Sin 3 TRIGONOMETRY gt Sinza Cosza 1 gt Sin a b sin acos b cos asin b gt Cos a b cos acos b sin asin b gt SinZa 2sinac0sa gt C0s2a Cos2a Sin2a In the gure below assume that the side OB is a unit length hence OB 10 S D C n c Eggs112501 TRIGONOMETRYCosine Law 3C2 3A2 AC2 2BAACc0sa TRIGONOMETRYCosine Law BC2 BD2 DC2 BA AD2 DC2 BC2 BA2 AD2 DC2 2BAAD BC2 BA2 AC2 DC2 DC2 2BAAD BC2 BAZ AC2 2BAAD BC2 BA2 AC2 2BAACcosa2 BC2 BA2 AC2 2BAACCosa BC2 BA2 AC2 2BAACc0sa TRIGONOMETRYCosine Law BC2 BA2 AC2 2BAACcosa From the right triangle BDC A BC2 DC2 BD2 1 9 From the right triangle DAC DC2 AC2 DA2 2 Substituting 2 in 1 B C BC2 AC2 DA2 BD2 3 BC2 AC2 DA2 BA DA 2 Draw CD at 90 degrees to BA BC2 AC2 DA2 BA2 DA2 2BADA BCZ AC2 13D2 2BACAc0sa TRIGONOMETRYsm Law AB BC AC Sin cSin aSin b A b B TRIANGLES To proof it use the parallel lines theory ABC 1800 The sum of the interior angles of a triangle is 1800 GEOMETRYTRIGONOMETRY a b d c For any closed polygon the sum of the interior angles is equal to 90 2S 4 Where S is the number of sides GEOMETRYTRIGONOMETRY A B E D The sum of the interior angles of a polygon 6 sides abcdef900264 GEOMETRYTRIGONOMETRY EC The sum of the interior angles of a polygon 6 sides abcdef900264 TRIGONOMETRIC EQUATION S W b 14 TM 0 A B a B From the ght mangle C TRIGONOMETRIC EQUATION Am b Multiply the left hand side of equation 1 by CC l i Q A C C Substituting yields 6 B 423 Mg C B TRIGONOMETRIC EQUATION S W b 25 Tahmnwdzhaa iz 1 2 0 go39 255 348111123 563 1 2 A From the right triangle TRIGONOMETRIC EQUATION 2 W b Multiply the left hand side of equation 1 by CC 2818 0 i 255 2 8 Substituting yields Solutions Manual Engineering Mechanics Statics lst Edition Michael E Plesha University of Wisconsin Madison Gary L Gray The Pennsylvania State University Francesco Costanzo The Pennsylvania State University With the assistance of Chris Punshon Andrew J Miller Justin High Chris O Brien Chandan Kumar Joseph Wyne Jonathan Fleischmann Version August 24 2009 The McGraw Hill Companies Inc Copyright 2002 2010 Michael E Plesha Gary L Gray and Francesco Costanzo This solutions manual in any print or electronic form remains the property of McGraW Hill Inc It may be used andor possessed only by permission of McGraw Hill and must be surrendered upon request of McGraw Hill Any duplication or distribution either in print or electronic form without the permission of McGraw Hill is prohibited Slalics le 3 Important Information about this Solutions Manual We encourage you to occasionally Visit httpwwwmhhecompgc to obtain the most uptodate version of this solutions manual Contact the Authors If you nd any errors andor have questions concerning a solution please do not hesitate to contact the authors and editors Via email at st atsolns email esm psu edu We welcome your input This solunons manual in any punt OI eleeuonie form iemains the property of McGraerlll Inc It may be used andor possessed only by pennission of McGraerlll and must be surrendered upon request of McGraerlll Any duplication or distribution eilnei m punt or electronic form Without the pennission of McGraerlll is prohlblted August 24 2009 4 Salun39uns Manual Accuracy of Numbers in Calculations Throughout this solutions manual we will generally assume that the data given for problems is accurate to 3 signi cant digits When calculations are performed all intermediate numerical results are reported to 4 signi cant digits Final answers are usually reported with 3 signi cant digits If you verify the calculations in this solutions manual using the rounded intermediate numerical results that are reported you should obtain the nal answers that are reported to 3 signi cant digits This solutions manual In any print or electronic form remains the property of Mchaerill Inc It may be used andox possessed only by permissto of McGraerill and must be surrendered upon request of McGraerill Any duplication or distribution either In print or electronic form Without an permission of Mchaerill IS prohibited August 24 2009 n e Chapter 1 Solutions Problem 11 a Consider a situation in which the force F applied to a particle of mass m is zero Multiply the scalar form of Eq 12 on page 7 ie a dvdt by dt and integrate both sides to show that the velocity 1 also a scalar is constant Then use the scalar form of Eq 11 to show that the scalar position r is a linear function of time b Repeat part a when the force applied to the particle is a non zero constant to show that the velocity and position are linear and quadratic functions of time respectively Solution Part 3 Consider the scalar form of Eq 13 on page 7 for the case with F 0 Fma z 0ma gt a0 1 Next consider the scalar form of Eq 12 on page 7 11 361 gt dvadt gt fdvvadt 2 Substituting a 0 into Eq 2 and evaluating the integral provides demonstrating that the velocity v is constant when the acceleration is zero Next consider the scalar form of Eq 11 d r E For the case with constant velocity given by Eq 3 it follows that rv0dtv0dtvot61i where Cl is a constant of integration Thus the position r is a linear function of time when the acceleration is zero Note that in the special case that 120 0 then the position r does not change with time 1 gt drvdt gt fdrrvdt 4 5 Part b When the force F is constant then Newton s second law provides F constant ma gt a Fm constant 6 Tnis so1nnons manual in any punt oi e1eononio form iemains the property of McGtaWrHtll Inc It may be used andoi possessed only by peimission August 24 2009 of McGtaWrHtll and must be surrendered upon request of McGtaWrHtll Any duplication oi distribution either in print or electronic form Without the permission of McGraertll is prohibited 6 Solutions Manual Following the same procedure as used in Part a we nd that F F vadtidtitcz 7 m m where CZ is a constant of integration and v is shown to be a linear function of time Likewise recalling Eq 4 F F u vdt2ilczdtit2czt63 8 m 2m which is a general quadratic function of time To determine the constants of integration requires that initial conditions be speci ed That is at some instant of time usually I 0 we must specify the position and velocity of the particle Tms so1nnons manual In any pnm or e1eonomo form remains the property of MCGIaWrHlll Inc It may be used andor possessed only by permisslo August 24 2009 of McGraerill and must be surrendered upon request of McGraerill Any duplication or distribution either In print o eleonomo form Without m permission of Mchaerill IS prohibited n e Statics l e 7 Problem 12 Using the length and force conversion factors in Table 12 on p 10 verify that 1 slug 1459 kg Solution lbszft 4448N ft k ms2 1slug 1slug 7 7 7 7 1459kg 1 slug 1b 0 3048 In N Tms soluuons manual In any punt or elecnomc form remaan the property of McGraerxll Inc It may be used andor possessed only by permlssxon August 24 2009 of McGraerxll and mus be surrendered upon request of McGraerxll Any duphcauon or dxsmbuuon exther m punt or electromc form whom the permlsslon of MchaerllL xs prohxbned 8 Solutions Manual Problems 13 through 15 Convert the numbers given in US Customary units to the corresponding SI units indicated Problem 13 i a Length Convertl 235 in to In b Mass Convert m 0156 slug to kg 0 Farce weight Convert F 100 lb to N d Mament torque Convert M 329 ftlb to Nm Problem 14g a Length Convertl 0001 in to M111 b Mass Convertm 0305 lbs2in to kg 0 Farce weight Convert F 256 kip to kN Recall lkip 1000 lb d Mass mament 0f inertia Convert mass 230 in lbs2 to Nms2 Problem 15 l a Pressure Convert p 251bft2 to Nmz b Elastic madnlns Convert E 30 X 106 lbin2 to GNmz c Area mament afinertia Convert Iar63 632 in4 to nun4 1 Mass mament a inertia Convert mass 154 inlbs2 to k m2 g Solution to 13 Part a l 235m m 00597111 1 in 103 111111 Part b m 0156slugMuggkg 228kg 2 Part c F lOOlb439448N 445N 3 This solutions manual in any punt oi e1eononio form iennnins the property of MCGIaWrHlll Inc It may be used andor possessed only by peinnission August 24 2009 of MCGIaerllll and must be sniiendeied upon request of MCGIaerllll Any duplication oi distribution either in print or electronic form Wimoni the peinnission of MeGraerlll is pionibiied S39mlics le 9 Part d 03048 4448N M 329ftlb 7m 7 446Nm 4 ft lb Solution to 14 Part a 254 106 l 0001m 7mm m m 2544m 1 111 103 Hull In Part b 4448N k m 2 m 03051b s2m 7 g S 7 534kg 2 1b N 00254m Part c 4448kN F 256k1ps 7 114kN 3 hp Part d 00254 4448N 1m 230in1bs2 7m 7 260Nms2 4 1n 1b Solution to 15 Part a 4448N ft 2 251b f8 7 7 120 103N 2 1 p 1b 03048m X n 0 Part b 2 4448N G E 30X1061bj112 7 7 N 207GNm2 2 00254111 1b 109 N Part c 254 4 Iarea 6321114 263X106mm4 3 111 Part d 00254 4448N k 2 mass 154 1111bs2 7m 174kg m2 4 111 1b N Tms soluuons manual In any punt or elecnomc form remaan the property of Mchaerxll Inc It may be used andor possessed only by permlssxon Augus124 2009 of Monerm and must be surrendered upon request of Monerm Any duphcauon or dxsmbuuon exther m punt or electromc form whom the permlsslon of 04onerle 15 prohxbxted 10 Solutions Manual Problems 16 through 18 Convert the numbers given in SI units to the corresponding US Customary units indicated Problem 163gt a Length Convertl 153 In to in b Mass Convertm 65 kg to slug 0 Farce weight Convert F 892 N to lb d Mament torque Convert M 329 Nm to in lb Problem 17 g a Length Convertl 122 mn to in b Mass Convertm 321 kg to lbs2in c Farce weight Convert F 132 kN to lb d Mass mament 0f inertia Convert mass 932 kgm2 to slugin2 Problem 181 a Pressure Convert p 25 kNm2 to lbin2 b Elastic madnlns Convert E 200 GNm2 to 1bin2 c Area mament afinertia Convert Iar63 235 X 105 min4 to in4 1 Mass mament a inertia Convert mass 123 k m2 to in1bs2 g Solution to 16 Part 3 ft 12 in l153 7 7 602 1 m03048m ft m Part b slu 65k 7 4461 2 m g1459kg sug 0 Part c 1b F 892 N 7 3 4 448 N ans so1nnons manual in any print or e1eononro form remains the property of McGraerill Inc it may be used andor possessed only by perrnrssron August 24 2009 of McGraerill and rnnsr be surrendered upon request of McGraerill Any duplication or drsrrrbnnon erner m prrnr or e1eorronro form without the perrnrssron of McGraerllL rs pronrbrred S39mlics le 1 1 Part d 1b 111 M 329N111 7 7 291111113 4 4448N 00254111 Solution to 17 Part a 1059 1 12211111 7m 7 480x10 6 m 1 nm 00254m Part b 1 lb 211 11 m 321kg amp S 7 001831bs2111 2 1459kg slug 12111 Part c 103N 1b F 132kN 2970115 3 kN 4448N Part d slug in 2 1 932k 27 7 9901031 2 4 3 g m 1459kg00254m X S g m Solution to 18 Part a 103N 1b 00254 2 p 2511N1112 7m 3631111112 1 kN 4448N 111 Part b 109 N 1b 00254 2 E 200GN1112 7m 290x106 115112 2 GN 4448N 111 Part c m 4 W 235x1051111114m 5651114 3 Part d 1b 211 11 2 mass 123kg 1112 S 7 7 1091bs2 111 4 1459 kg 12111 00254111 Th1s soluuons 1111111111 111 any 1111111 01 elecnomc form remams the property of McGraer l Inc 11 may be used ando1 possessed only by perm1ss1on Augu5124 2009 of 111611er11 and must be surrendered upon request of 111611er11 Any 1111111011101 01 d15mbut1on e1111e1 11 p1111 o1 electromc form wnhout the perm1ss1on of 111611er111 15 proh1b1ted 12 Solutions Manual Problem 19 a Convert the kinetic energy T 0379 kgm2s2 t0 slugin2sz b Convert the kinetic energy T 101 sluginzs2 t0 kgm2sz Solution Part a 1 2 T 0379kgm2s2 amp i 403slugin2sz 1 1459 kg 00254 In Part b 2 14 59 k 00254 T 101slugin2szlt7ggt 00951kg1n2sz 2 slug In This solutions manual in any punt oi eleononio form iemains the property of MCGIBWVHIH Inc It may be used andoi possessed only by peimission August 24 2009 of MCGIBWVHIH and must be sniiendeied upon request of MCGIBWVHIH Any duplication oi distribution either in print or electronic form Without the peimission of McGraWVHIIL is prohibited Statics l e 13 Problem 110 H If the weight of a certain object on the surface of the Earth is 0254 lb determine its mass in kilograms Solution We rst determine the mass of the object in slugs using w mg gt m wg 02541b322 ftsz 7888x 103 1b s2ft 7888x10 3 slug 1 Next we convert the mass into SI units such that 14 59 kg m 7888gtlt10 3Slug 7 0115kg 2 slug Tnis so1nnons manual in any punt oi e1eononio form iennnins the property of MCGIaWrHlll Inc It may be used andor possessed only by peinnission August 24 2009 of MCGIaerllll and must be smendeied upon request of MCGIaerllll Any duplication oi distribution either in print or electronic form Without Lh peinnission of McGraWrHlll is prohibited 14 Solutions Manual Problem 111 H If the mass of a certain object is 691 kg determine its weight on the surface of the Earth in pounds Solution We rst determine the weight of the object in newtons using w mg 691kg 981ms2 6779kg ms2 6779N Next we convert the weight into pounds using w 6779N 1521b 1b 4448N 1 2 ans so1unons manual in any prrnr or e1eouonro form remarns the properry of McGraerill Inc 1r may be used andor possessed only by permrssron of McGraerill and musr be surrendered upon request of McGraerill Any duplroanon or drsrrrbunon errner m prrnr or e1eorromo form wrmour the permrssron of MchaerilL rs pronrbrred August 24 2009 Statics l e 15 Problem 112 H Use Eq 17 on p 15 to compute a theoretical value of acceleration due to gravity g and compare this value with the actual acceleration due to gravity at the Earth s poles which is about 03 higher than the value reported in Eq 18 Comment on the agreement Solution The values G 6674gtlt10 12 m3kg s2 mEa h 59736gtlt1024 kg and rEa h 6371 X 106 m are given in the text in the discussion of Eqs 16 and 17 Using these values the theoretical value of acceleration due to gravity Eq 17 is 59736 1024k gtheory 6 6674X10 12m3kg s2 X g 9822ms2 1 Earth 6371x106m2 The commonly accepted value for acceleration due to gravity namely g 981ms2 is most accurate at 45O latitude it accounts for the Earth not being perfectly spherical and the Earth s rotation As stated in the problem description the acceleration due to gravity at the poles is about 03 higher than at 45O latitude and therefore at the poles the acceleration due to gravity is approximately giggles 981ms2l 0003 9839ms2 2 Note that the poles are useful locations for comparing the theoretical value of acceleration due to gravity Eq 1 with the actual Value given approximately by Eq 2 because the effects of the Earth s rotation ie centripetal acceleration are absent at the poles While the agreement between Eqs 1 and 2 is quite good the differences between these two values is due to the Earth not being perfectly spherical Tms so1nnons manual In any print or e1eonomo form remains the property of McGtaerltll Inc It may be used andor possessed only by permission August 24 2009 of McGtaerltll and must be surrendered upon request of McGtaerltll Any duplication or distribution either m print or electronic form Without the permission of McGraertll IS prohibited 16 Solutions Manual Problem 113 H Two identical asteroids travel side by side while touching one another If the asteroids are composed of homogeneous pure iron and are spherical what diameter in feet must they have for their mutual gravitational attraction to be 1 lb Solution Begin by considering Eq 16 on p 15 G 1 where F 1 lb according to the problem statement The mass of each asteroid is given by m p471r3 3 where p is the density and r is the radius of the spherical asteroids Thus Eq 1 becomes p4JTr332 4712szr4 F G W f 2 Solving for r we obtain 14 14 9F 9 4448 N 3 960 3 r 7 m 4712sz 47126674gtlt10 12m3kgs27860kgm32 where F 1 lb 4448 N Since the diameter d of the asteroid is twice its radius r ie d 2r it follows that ft d 23960mltm 260 ft 4 Tms solnnons manual In any punt or eleonomo form re alns the property of Mchaerlll Inc It may be used andor possessed only by petmlsslon August 24 2009 of McGraerlll and must be surrendered upon request of McGraerlll Any duplleatlon or distribution elthet m punt or eleon me form W petmlsslon of McGraerlll IS prohibited Statics l e 17 Problem 114 H The mass of the Moon is approximately 735 X 1022 kg and its mean distance from the Earth is about 380 X 108 km Determine the force of mutual gravitational attraction in newtons between the Earth and Moon In View of your answer discuss why the Moon does not crash into the Earth Solution Use Eq 16 to solve for the force F mimz 59736x1024 kg735gtlt1022 kg r2 F G 2 380X108 km103 mkm 6674x10 12m3kg s2 202X1014kg ms2 1 Although this force is large the force due to centripetal acceleration equilibrates this so that the Moon maintains its orbit so that the force may be written as Tms so1nnons manual In any print or electronic form remains the property of MCGIaWrHlll Inc It may be used andor possessed only by permission August 24 2009 of McGraerill and must be surrendered upon request of McGraerill Any duplication or distribution either m print o eleonomo form without m permission of MchaerilL IS prohibited 18 Solutions Manual Problem 115 H If a person standing at the rst oor entrance to the Sears Tower in Chicago weighs exactly 150 lb determine the weight while he or she is standing on top of the building which is 1450 ft above the rst oor entrance How high would the top of the building need to be for the person s weight to be 99 of its value at the rst oor entrance Solution We will use Eq 16 from p 15 m 1m2 T 1 where the person s weight is given by F in Eq 1 Let the person s weight on the rst oor be given by W1 the weight on top of the building be given by W101 and the height between the rst oor and the top of the building be h Using Eq 1 we may write these two relations FG W1 G mEanhszerson VVmp G mEanhmperso 2 rE31111 rEanh h Dividing me by W1 and solving for me leads to W101 r1521th r1521th gt W W 3 W1 mean h2 1 1 mean h2 such that 6371 106 2 mp 1501b X m 2 1501b099986 149979 1b 4 6371X105m 1450 ft03048mft For the second part of the problem we need to nd the height h needed to make me 099 W1 ie r2 h 32095x104m 15 099 gt h 0005038 rEmh 1053x105 ft 5 rEanh h2 03048 mft Therefore the top of the building must be h above the rst oor where h is given by h 321km 105x105n 19941111 6 This so1nnons manual in any punt or eleenome form remams the property of McGraerill Inc It may be used andor possessed only by permlsslon August 24 2009 of Mchaerill and must be surrendered upon request of Mchaerill Any dupheauon or dmnbuuon either in punt or electronic form without the permlsslon of Mchaerill 1s proh1b1ted S39mlics le Problem 116 El a Zinc die casting alloy 3 0242 lbin3 b Oil shale 30 galton rock 3 133 lbft3 c Styrofoam medium density 3 201bft3 d Silica glass 3 0079 lbin3 The speci c weights of several materials are given in US Customary units Convert these to speci c weights in SI units kNm3 and also compute the densities of these materials in SI units kgm3 Solution The equation to be used to convert the speci c weight to density is ypg gt pVg 1 Parta 4448N y 02421bm3 L 6569x103Nm3 657kNm3 2 1b 00254m 6569 103N 3 6569 103k 2 2 X n X gm S 670x103kgm3 3 981ms2 981ms2 Partb 4 448N ft 3 3 3 3 3 y 1331bftlt 1b gtlt0t3048mgt 2089x10 Nm 209kNm 4 2089 103N 3 2089 103k 2 2 X m X gm S2l3XlO3kgm3 5 981ms2 981ms2 Partc 3 4448N ft 20115 ft3 7 7 3142N 3 0314kN 3 6 y lt 1b gtlt0304Smgt n m 0 3142Nm3 3142 kgmz s2 3 7 320k 7 7 981nns2 981ms2 gm Partd 4448N m 3 3 3 3 y 0079lbm lt lb gtlt0t0254mgt 2144x10 Nm 214kNm 8 2144x103Nm3 2144X103kgm2s2 219 103k 3 p 981nns2 981ms2 X Wu 9 This solnnons manual In any print or electronic form remains the property of Mchaerill Inc It may be used andor possessed only by permission of McGraerill and must be surrendered upon request of McGraerill Any duplication or distribution either In print or electronic form Without the permission of Mchaerlll IS prohibited August 24 2009 20 Solutions Manual Problem 117 El The densities of several materials are given in SI units Convert these to densities in US Customary units slugft3 and also compute the speci c weights of these materials in US Customary units lbft3 a Lead pure p 1134 gcm3 b Ceramic alumina A1203 p 390 Mgm3 c Polyethylene high density p 960 kgm3 d Balsa wood p 02 Mgm3 Solution The equation to be used to determine the speci c weight is V pg 1 Part a k 1 100 3 0 3048 3 g s ug cm m 3 1134 3 7 7 2201 ft 2 p gcm 103g lt1459kggtlt 111 gt lt ft gt S ug 1b 2f1 y 2201s1ugf13322f1s2lt7gt 7091bft3 3 s ug Part b 103k slu 0 3048111 3 p 390Mgm3 g 7 7 7569 slugft3 757 slugft3 4 Mg 1459kg ft lb 2f1 y 7569slugft3322fts2 2441bf13 5 s ug Part c 1 0 3048 3 s ug m 3 3 960k 3 7 218631 ft 21861 ft 6 p gm 143959kggt ft gt sug 8 ug 1b 2f1 y 1863slugft3322fts2 6001bft3 7 s ug Part d 103k slu 0 3048111 3 p 02Mgm3 g 7 7 03882 slugft3 0388 slugft3 8 Mg 1459kg ft lb 2f1 y 03882s1ugf13322f1s2lt7gt 1251bft3 9 s ug This so1nnons manual in any print 01 electronic form remains the property of McGraerill Inc It may be used andor possessed only by pe11n1ss1on August 24 2009 of Monerm and must be smende1ed upon 1eques1 of Monerm Any duplication o1 amnbunon either in print 01 electronic form whom the pe11n1ss1on of Monerle 1s prohibited Sralics l e 21 Problem 118 H A Super Ball is a toy ball made of hard synthetic rubber called Zectron This material has a high coef cient of restitution so that if it is dropped from a certain height onto a hard xed surface it rebounds to a substantial portion of its original height If the Super Ball has 5 cm diameter and the density of Zectron is about 15 Mgm3 determine the weight of the Super Ball on the surface of the Earth in US Customary units Solution Keeping in mind that the mass m is given by m pV and the volume V for a sphere is given by V 471r3 3 where r is the radius it follows that the weight w is 4 r3pg 71d3pg 3 7 wmgpVg 6 1 where d is the diameter To obtain the weight in pounds 115cm3 15Mgm3981ms2 lt m 3 lt103 kglt lb f 0217 lb 2 102 cm Mg gt 4448 kg ms2 Tnis solutions manual in any print or e1eouonio form remains the property of McGraerill Inc It may be used andor possessed only by permission August 24 2009 of MoGrawHiu and must be surrendered upon requesr of MoGrawHiu Any duplication or disrriburion either in prinr or e1eorronio form Wimour the permission of MoGrawHilL is prohibited 22 Solutions Manual Problem 119 H An ice hockey puck is a short circular cylinder or disk of vulcanized rubber with 300 in diameter and 100 in thickness with weight between 55 and 60 oz 16 oz 1 lb Compute the range of densities for the rubber in conventional SI units that will provide for a puck that meets these speci cations Solution Based on the problem statement the weight w of the hockey puck should be in the range lb lb 55 oz 5 w 5 600z 7 gt 0343751b 5 w 5 03751b 1 16oz 16oz The weight w mass m density p and volume V of the hockey puck are related by w w mg pl5 p rzhg gt p f 2 r hg where in the above expressions we have used the volume V of a cylinder as V rzh where r is the radius and h is the thickness Multiplying all three terms of Eq 1 by 1 r2hg leads to with r 15 in andg 322fts2 0343751b lt lt 0375113 3 7115 in2lin322fts2 p 15m12f nl21m xmansz which simpli es to obtain 26101bs2ft4 5 p 5 28471bs2ft4 4 Since 1 slug 1 lb szft it follows that 2610 slugft3 5 p 5 2847 slugft3 5 Converting the above results to SI units provides 3 3 l 1459k ft l 1459k ft 2610g g 7 5 p 5 2847g 7g 7 6 ft3 slug 03048m ft3 slug 03048m which yields 1340 kgm3 5 p 5 1470kgm3 7 Th1s solnnons manual 1n any punt or e1eonon1o form remams the property of McGraer l Inc 11 may be used andor possessed only by perm1ss1on August 24 2009 of Mchaer l and must be surrendered upon request of Mchaer l Any duphcauon or absmbnnon enher 1n punt or electromc form wnhout the perm1ss1on of Mchaer L 1s proh1b1ted 71 rad S39Ialics le 23 Problem 1203 Convert the angles given to the units indicated a Convert 0 3560 to rad b Convert 0 108 X 10 30 to mrad 0 Convert 0 465 rad to degrees d Convert 0 0254 mrad to degrees Solution Part 3 0 356 ad 0621 d 1 1800 ra Part b 0 108 x10 3quot 7 rad 103mm 0 0188 d 2 39 180 rad 39 mquot Part c 1800 O 0 465rad 266 3 1 rad Part d d 1800 0 0254mrad L 00146 4 103 mrad Tnis solutions manual in any print oi electronic roim remains the property of McGraerill Inc It may be used andor possessed only by peimission of McGraerill and must be surrendered upon request of McGraerill Any duplication or distribution either in print or electronic form Without the peimission of McGraWrHill is prohibited August 24 2009

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