Solution Manual For Engineering Mechanics: Statics, 5th Edition

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Problem 1.1The value ofis 3.14159265. . .IfCisthe circumference of a circle andris its radius, deter-mine the value ofr/Cto four significant digits.Solution:CD2r)rCD12D0.159154943.To four significant digits we haverCD0.1592Problem 1.2The base of natural logarithms iseD2.718281828. . .(a)Expressetofive significant digits.(b)Determine the value ofe2tofive significant digits.(c)Use the value ofeyou obtained in part (a) to deter-mine the value ofe2tofive significant digits.[Part (c) demonstrates the hazard of using rounded-offvalues in calculations.]Solution:The value of e is:eD2.718281828(a)Tofive significantfigureseD2.7183(b)e2tofive significantfigures ise2D7.3891(c)Using the value from part (a) wefinde2D7.3892which isnot correct in thefifth digit.Problem 1.3A machinist drills a circular hole in apanel with a nominal radiusrD5 mm. The actual radiusof the hole is in the rangerD5š0.01 mm. (a) To whatnumber of significant digits can you express the radius?(b) To what number of significant digits can you expressthe area of the hole?Solution:a)The radius is in the ranger1D4.99 mm tor2D5.01 mm. Thesenumbers are not equal at the level of three significant digits, butthey are equal if they are rounded off to two significant digits.Two:rD5.0 mmb)The area of the hole is in the range fromA1Dr12D78.226 m2toA2Dr22D78.854 m2. These numbers are equal only if roundedto one significant digit:One:AD80 mm2Problem 1.4The opening in the soccer goal is 24 ftwide and 8 ft high, so its area is 24 ftð8 ftD192 ft2.What is its area in m2to three significant digits?Solution:AD192 ft2(1 m3.281 ft)2D17.8 m2AD17.8 m2Problem 1.5The Burj Dubai, scheduled for comple-tion in 2008, will be the world’s tallest building with aheight of 705 m. The area of its ground footprint will be8000 m2. Convert its height and footprint area to U.S.customary units to three significant digits.Solution:hD705 m(3.281 ft1 m)D2.31ð103ftAD8000 m2(3.218 ft1 m)2D8.61ð104ft2hD2.31ð103ft,AD8.61ð104ft2

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Problem 1.6Suppose that you have just purchaseda Ferrari F355 coupe and you want to know whetheryou can use your set of SAE (U.S. Customary Units)wrenches to work on it. You have wrenches with widthswD1/4 in, 1/2 in, 3/4 in, and 1 in, and the car has nutswith dimensionsnD5 mm, 10 mm, 15 mm, 20 mm,and 25 mm. Defining a wrench tofit ifwis no morethan 2% larger thann, which of your wrenches can youuse?nSolution:Convert the metric sizento inches, and compute thepercentage difference between the metric sized nut and the SAEwrench. The results are:5 mm(1 inch25.4 mm)D0.19685..in,(0.196850.250.19685)100D 27.0%10 mm(1 inch25.4 mm)D0.3937..in,(0.39370.50.3937)100D 27.0%15 mm(1 inch25.4 mm)D0.5905..in,(0.59050.50.5905)100D C15.3%20 mm(1 inch25.4 mm)D0.7874..in,(0.78740.750.7874)100D C4.7%25 mm(1 inch25.4 mm)D0.9843..in,(0.98431.00.9843)100D 1.6%A negative percentage implies that the metric nut is smaller than theSAE wrench; a positive percentage means that the nut is larger thenthe wrench. Thus within the definition of the 2%fit, the 1 in wrenchwillfit the 25 mm nut.The other wrenches cannot be used.Problem 1.7Suppose that the height of Mt. Everest isknown to be between 29,032 ft and 29,034 ft. Based onthis information, to how many significant digits can youexpress the height (a) in feet? (b) in meters?.Solution:a)h1D29032 fth2D29034 ftThe two heights are equal if rounded off to four significant digits.Thefifth digit is not meaningful.Four:hD29,030 ftb)In meters we haveh1D29032 ft(1 m3.281 ft)D8848.52 mh2D29034 ft(1 m3.281 ft)D8849.13 mThese two heights are equal if rounded off to three significantdigits. The fourth digit is not meaningful.Three:hD8850 mProblem 1.8The maglev (magnetic levitation) trainfrom Shanghai to the airport at Pudong reaches a speedof 430 km/h. Determine its speed (a) in mi/h; (b) ft/s.Solution:a)vD430 kmh(0.6214 mi1 km)D267 mi/hvD267 mi/hb)vD430 kmh(1000 m1 km) (1 ft0.3048 m) (1 h3600 s)D392 ft/svD392 ft/sProblem 1.9In the 2006 Winter Olympics, the men’s15-km cross-country skiing race was won by AndrusVeerpalu of Estonia in a time of 38 minutes, 1.3 seconds.Determinehisaveragespeed(thedistancetraveleddivided by the time required) to three significant digits(a) in km/h; (b) in mi/h.Solution:a)vD15 km(38C1.360)min(60 min1 h)D23.7 km/hvD23.7 km/hb)vD23.7 km/h(1 mi1.609 km)D14.7 mi/hvD14.7 mi/h

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Problem 1.10The Porsche’s engine exerts 229 ft-lb(foot-pounds) of torque at 4600 rpm. Determine thevalue of the torque in N-m (Newton-meters).Solution:TD229 ft-lb(1 N0.2248 lb) (1 m3.281 ft)D310 N-mTD310 N-mProblem 1.11Thekinetic energyof the man in ActiveExample 1.1isdefinedby12mv2,wheremishismass andvis his velocity. The man’s mass is 68 kgand he is moving at 6 m/s, so his kinetic energy is12(68 kg)(6 m/s)2D1224 kg-m2/s2. What is his kineticenergy in U.S. Customary units?Solution:TD1224 kg-m2/s2(1 slug14.59 kg) (1 ft0.3048m)2D903 slug-ft2/sTD903 slug-ft2/sProblem 1.12The acceleration due to gravity at sealevel in SI units isgD9.81 m/s2. By converting units,use thisvalue todeterminethe acceleration due togravity at sea level in U.S. Customary units.Solution:Use Table 1.2. The result is:gD9.81(ms2) (1 ft0.3048 m)D32.185. . .(fts2)D32.2(fts2)Problem 1.13Afurlong per fortnightis a facetiousunit of velocity, perhaps made up by a student as asatirical comment on the bewildering variety of unitsengineers must deal with. A furlong is 660 ft (1/8 mile).A fortnight is 2 weeks (14 nights). If you walk to classat 2 m/s, what is your speed in furlongs per fortnight tothree significant digits?Solution:vD2 m/s(1 ft0.3048 m) (1 furlong660 ft) (3600 shr) (24 hr1 day) (14 day1 fortnight)vD12,000 furlongsfortnightProblem 1.14Determine the cross-sectional area ofthe beam (a) in m2; (b) in in2.120 mmxy40 mm40 mm40mm200 mmSolution:AD200 mm2280 mm120 mmD20800 mm2a)AD20800 mm2(1 m1000 mm)2D0.0208 m2AD0.0208 m2b)AD20800 mm2(1 in25.4 mm)2D32.2 in2AD32.2 in2Problem 1.15The cross-sectional area of the C12ð30American Standard Channel steel beam isAD8.81 in2.What is its cross-sectional area in mm2?xyASolution:AD8.81 in2(25.4 mm1 in)2D5680 mm2

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Problem 1.16A pressure transducer measures a valueof 300 lb/in2. Determine the value of the pressure inpascals. A pascal (Pa) is one newton per meter squared.Solution:Convert the units using Table 1.2 and the definition ofthe Pascal unit. The result:300(lbin2) (4.448 N1 lb) (12 in1 ft)2(1 ft0.3048 m)2D2.0683. . . 106(Nm2)D2.07106PaProblem 1.17A horsepower is 550 ft-lb/s. A watt is1 N-m/s. Determine how many watts are generated bythe engines of the passenger jet if they are producing7000 horsepower.Solution:PD7000 hp(550 ft-lb/s1 hp) (1 m3.281 ft) (1 N0.2248 lb)D5.22ð106WPD5.22ð106WProblem 1.18Chapter 7discussesdistributedloadsthat are expressed in units of force per unit length. Ifthe value of a distributed load is 400 N/m, what is itsvalue in lb/ft?.Solution:wD400 N/m(0.2248 lb1 N) (1 m3.281 ft)D27.4 lb/ftwD27.4 lb/ftProblem 1.19The moment of inertia of the rectan-gular area about thexaxis is given by the equationID13bh3.The dimensions of the area arebD200 mm andhD100 mm. Determine the value ofIto four significantdigits in terms of (a) mm4; (b) m4; (c) in4.hbxySolution:(a)ID13200 mm100 mm3D66.7ð106mm4(b)ID66.7ð106mm4(1 m1000 mm)4D66.7ð106m4(c)ID66.7ð106mm4(1 in25.4 mm)4D160 in4Problem 1.20In Example 1.3, instead of Einstein’sequation consider the equationLDmc, where the massmis in kilograms and the velocity of lightcis in metersper second. (a) What are the SI units ofL? (b) If thevalue ofLin SI units is 12, what is its value in U.S.Customary base units?Solution:a)LDmc)UnitsLDkg-m/sb)LD12 kg-m/s(0.0685 slug1 kg) (3.281 ft1 m)D2.70 slug-ft/sLD2.70 slug-ft/s

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Problem 1.21The equationDMyIis used inthemechanics ofmaterialstodeterminenormal stresses in beams.(a)When this equation is expressed in terms of SI baseunits,Mis in newton-meters (N-m), y is in meters(m), andIis in meters to the fourth power (m4).What are the SI units of?(b)IfMD2000 N-m,yD0.1 m,andID7ð105m4, what is the value ofin U.S. Customarybase units?Solution:(a)DMyID(N-m)mm4DNm2(b)DMyID2000 N-m0.1 m7ð105m4(1 lb4.448 N) (0.3048 mft)2D59,700 lbft2Problem 1.22The acceleration due to gravity on thesurface of the moon is 1.62 m/s2. (a) What would themass of the C-clamp in Active Example 1.4 be on the surfaceof the moon? (b) What would the weight of the C-clampin newtons be on the surface of the moon?Solution:a)The mass does not depend on location. The mass in kg is0.0272 slug(14.59 kg1 slug)D0.397 kgmassD0.397 kgb)The weight on the surface of the moon isWDmgD0.397 kg1.62 m/s2D0.643 NWD0.643NProblem 1.23The 1 ftð1 ftð1 ft cube of ironweighs 490 lb at sea level. Determine the weight innewtons of a 1 mð1 mð1 m cube of the samematerial at sea level.1 ft1 ft1 ftSolution:The weight density isD490 lb1 ft3The weight of the 1 m3cube is:WDVD(490 lb1 ft3)1 m3(1 ft0.3048 m)3(1 N0.2248 lb)D77.0 kNProblem 1.24TheareaofthePacificOceanis64,186,000 square miles and its average depth is 12,925 ft.Assume that the weight per unit volume of ocean wateris 64 lb/ft3. Determine the mass of the Pacific Ocean(a) in slugs; (b) in kilogramsSolution:The volume of the ocean isVD64,186,000 mi212,925 ft(5,280 ft1 mi)2D2.312ð1019ft3(a)mDVD(64 lb/ft332.2 ft/s2)2.312ð1019ft3D4.60ð1019slugs(b)mD4.60ð1019slugs(14.59 kg1 slug)D6.71ð1020kgProblem 1.25Theaccelerationduetogravityatsea levelisgD9.81 m/s2.Theradiusoftheearthis6370 km.TheuniversalgravitationalconstantisGD6.67ð1011N-m2/kg2. Use this information todetermine the mass of the earth.Solution:Use Eq (1.3)aDGmER2. Solve for the mass,mEDgR2GD9.81 m/s26370 km2(103mkm)26.671011(N-m2kg2)D5.9679. . . 1024kgD5.971024kg

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Problem 1.26A person weighs 180 lb at sea level. Theradius of the earth is 3960 mi. What force is exerted onthe person by the gravitational attraction of the earth ifhe is in a space station in orbit 200 mi above the surfaceof the earth?Solution:Use Eq (1.5).WDmg(REr)2D(WEg)g(RERECH)2DWE(39603960C200)2D1800.90616D163 lbProblem 1.27The acceleration due to gravity on thesurface of the moon is 1.62 m/s2. The moon’s radius isRMD1738 km.(a)What is the weight in newtons on the surface ofthe moon of an object that has a mass of 10 kg?(b)Using the approach described in Example 1.5, deter-mine the force exerted on the object by the gravityof the moon if the object is located 1738 km abovethe moon’s surface.Solution:a)WDmgMD10 kg1.26 m/s2D12.6 NWD12.6 Nb)Adapting equation 1.4 we haveaMDgM(RMr)2. The force isthenFDmaMD10 kg1.62 m/s2(1738 km1738 kmC1738 km)2D4.05 NFD4.05 NProblem 1.28If an object is near the surface of theearth, the variation of its weight with distance from thecenter of the earth can often be neglected. The acceler-ation due to gravity at sea level isgD9.81 m/s2. Theradius of the earth is 6370 km. The weight of an objectat sea level ismg, wheremis its mass. At what heightabove the earth does the weight of the object decreaseto 0.99mg?Solution:Use a variation of Eq (1.5).WDmg(RERECh)2D0.99 mgSolve for the radial height,hDRE(1p0.991)D63701.00503781.0D32.09. . .kmD32,100 mD32.1 kmProblem 1.29The planet Neptune has an equatorialdiameter of 49,532 km and its mass is 1.0247ð1026kg.If the planet is modeled as a homogeneous sphere, whatis the acceleration due to gravity at its surface? (TheuniversalgravitationalconstantisGD6.67ð1011N-m2/kg2.)Solution:We have:WDG mNmrN2D(G mNr2)m)gNDG mNrN2Note that the radius of Neptune isrND1249,532 kmD24,766 kmThusgND(6.67ð1011N-m2kg2) (1.0247ð1026kg24766 km2) (1 km1000 m)2D11.1 m/s2gND11.1 m/s2Problem 1.30At a point between the earth and themoon, the magnitude of the force exerted on an objectby the earth’s gravity equals the magnitude of the forceexerted on the object by the moon’s gravity. What isthe distance from the center of the earth to that pointto three significant digits? The distance from the centerof the earth to the center of the moon is 383,000 km,and the radius of the earth is 6370 km. The radius of themoon is 1738 km, and the acceleration due to gravity atits surface is 1.62 m/s2.Solution:LetrEpbe the distance from the Earth to the point wherethe gravitational accelerations are the same and letrMpbe the distancefrom the Moon to that point. Then,rEpCrMpDrEMD383,000 km.The fact that the gravitational attractions by the Earth and the Moonat this point are equal leads to the equationgE(RErEp)2DgM(RMrMp)2,whererEMD383,000 km. Substituting the correct numerical valuesleads to the equation9.81(ms2) (6370 kmrEp)2D1.62(ms2) (1738 kmrEMrEp)2,whererEpis the only unknown. Solving, we getrEpD344,770 kmD345,000 km.

Solution Manual For Engineering Mechanics: Statics, 5th Edition - Page 8 preview image

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Problem 2.1In Active Example 2.1, suppose that thevectorsUandVare reoriented as shown. The vectorVis vertical. The magnitudes arejUj D8 andjVj D3.GraphicallydeterminethemagnitudeofthevectorUC2V.VU45Solution:Draw the vectors accurately and measure the resultant.RD jUC2Vj D5.7RD5.7Problem 2.2Suppose that the pylon in Example 2.2 ismoved closer to the stadium so that the angle betweenthe forcesFABandFACis 50°. Draw a sketch of thenew situation. The magnitudes of the forces arejFABj D100 kN andjFACj D60 kN. Graphically determine themagnitude and direction of the sum of the forces exertedon the pylon by the cables.Solution:Accurately draw the vectors and measure the magnitudeand direction of the resultantjFABCFACj D146 kN˛D32°

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Problem 2.3ThemagnitudejFAj D80 lbandtheangle˛D65°.ThemagnitudejFACFBj D120 lb.Graphically determine the magnitude ofFB.FBFAaFCSolution:Accurately draw the vectors and measure the magnitudeofFB.jFBj D62 lbProblem 2.4ThemagnitudesjFAj D40 N,jFBj D50 N, andjFCj D40 N. The angle˛D50°andˇD80°.Graphically determine the magnitude ofFACFBCFC.FBFAaFCSolution:Accurately draw the vectors and measure the magnitudeofFACFBCFC.RD jFACFBCFCj D83 N

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Problem 2.5ThemagnitudesjFAj D jFBj D jFCj D100 lb, and the angles˛D30°. Graphically determinethevalueoftheangleˇforwhichthemagnitudejFACFBCFCjis a minimum and the minimum valueofjFACFBCFCj.FBFAaFCSolution:For a minimum, the vectorFCmust point back to theorigin.RD jFACFBCFCj D93.2 lbˇD165°Problem 2.6The angleD50°. Graphically determinethe magnitude of the vectorrAC.60 mm150 mmACBrABrBCrACSolution:Draw the vectors accurately and then measurejrACj.jrACj D181 mm

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Problem 2.7ThevectorsFAandFBrepresenttheforcesexertedonthepulleybythebelt.Theirmagnitudes arejFAj D80 NandjFBj D60 N.Graphically determine the magnitude of the total forcethe belt exerts on the pulley.45FAFB10Solution:Draw the vectors accurately and then measurejFACFBj.jFACFBj D134 NProblem 2.8ThesumoftheforcesFACFBCFCD0. The magnitudejFAj D100 N and the angle˛D60°. Graphically determine the magnitudesjFBjandjFCj.30FBFAFCaSolution:Draw the vectors so that they add to zero.jFBj D86.6 N,jFCj D50.0 NProblem 2.9ThesumoftheforcesFACFBCFCD0.ThemagnitudesjFAj D100 NandjFBj D80 N. Graphically determine the magnitudejFCjand theangle˛.30FBFAFCaSolution:Draw the vectors so that they add to zero.jFCj D50.4 N, ˛D52.5°

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Problem 2.10The forces acting on the sailplane arerepresented by three vectors. The liftLand dragDare perpendicular. The magnitude of the weightWis500 lb. The sum of the forcesWCLCDD0. Graph-ically determine the magnitudes of the lift and drag.WDL25Solution:Draw the vectors so that they add to zero. Then measurethe unknown magnitudes.jLj D453 lbjDj D211 lbProblem 2.11A spherical storage tank is suspendedfrom cables. The tank is subjected to three forces, theforcesFAandFBexerted by the cables and its weightW.The weight of the tank isjWj D600 lb. The vector sumof the forces acting on the tank equals zero. Graphicallydetermine the magnitudes ofFAandFB.40˚FAWFB20˚20˚Solution:Draw the vectors so that they add to zero. Then measurethe unknown magnitudes.jFAj D jFBj D319 lb

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Problem 2.12The ropeABCexerts forcesFBAandFBCofequalmagnitudeontheblockatB.Themagnitude of the total force exerted on the block bythe two forces is 200 lb. Graphically determinejFBAj.20FBCFBABCABSolution:Drawthevectorsaccuratelyandthenmeasuretheunknown magnitudes.jFBAj D174 lbProblem 2.13Two snowcats tow an emergency shelterto a new location near McMurdo Station, Antarctica.(The top view is shown. The cables are horizontal.)The total forceFACFBexerted on the shelter is inthe direction parallel to the lineLand its magnitudeis 400 lb. Graphically determine the magnitudes ofFAandFB.LTop ViewFAFB5030Solution:Drawthevectorsaccuratelyandthenmeasuretheunknown magnitudes.jFAj D203 lbjFBj D311 lbProblem 2.14A surveyor determines that the horizon-tal distance fromAtoBis 400 m and the horizontaldistance fromAtoCis 600 m. Graphically determinethe magnitude of the vectorrBCand the angle˛.EastNorth6020CBArBCaSolution:Drawthevectorsaccuratelyandthenmeasuretheunknown magnitude and angle.jrBCj D390 m˛D21.2°

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Problem 2.15The vectorrextends from pointAtothe midpoint between pointsBandC. Prove thatrD12rABCrAC.ACBrACrrABrABSolution:The proof is straightforward:rDrABCrBM,andrDrACCrCM.Add the two equations and note thatrBMCrCMD0, since the twovectors are equal and opposite in direction.Thus 2rDrACCrAB,orrD(12)rACCrABrACrABrABMCProblem 2.16Bydrawingsketches ofthevectors,explain whyUCVCWDUCVCW.Solution:Additive associativity for vectors is usually given as anaxiom in the theory of vector algebra, and of course axioms are notsubject to proof. However we can by sketches show that associativityfor vector addition is intuitively reasonable: Given the three vectors tobe added, (a) shows the additionfirst ofVCW, and then the additionofU. The result is the vectorUCVCW.(b) shows the addition ofUCV, and then the addition ofW, leadingto the resultUCVCW.Thefinal vector in the two sketches is the same vector, illustrating thatassociativity of vector addition is intuitively reasonable.(a)UWVUWVV+WU+VU+[V+W][U+V]+W(b)Problem 2.17A forceFD40i20jN. What isits magnitudejFj?Strategy:The magnitude of a vector in terms of itscomponents is given by Eq. (2.8).Solution:jFj D p402C202D44.7 NProblem 2.18An engineer estimating the componentsof a forceFDFxiCFyjacting on a bridge abutmenthas determined thatFxD130 MN,jFj D165 MN, andFyis negative. What isFy?Solution:jFj D√jFxj2C jFyj2jFyj D√jFj2 jFxj2D√165 MN2130 MN2D101.6 MNFyD 102 MN

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Problem 2.19A support is subjected to a forceFDFxiC80j(N). If the support will safely support a forceof 100 N, what is the allowable range of values of thecomponentFx?Solution:Use the definition of magnitude in Eq. (2.8) and reducealgebraically.100½√Fx2C802, from which1002802½Fx2.ThusjFxj p3600, or60Fx C60 (N)Problem 2.20IfFAD600i800j(kip)andFBD200i200j(kip), what is the magnitude of the forceFDFA2FB?Solution:Take the scalar multiple ofFB, add the components ofthe two forces as in Eq. (2.9), and use the definition of the magnitude.FD6002200iC8002200jD200i400jjFj D√2002C4002D447.2 kipProblem 2.21The forces acting on the sailplane are itsweightWD 500jlb, the dragDD 200iC100j(lb)and the liftL. The sum of the forcesWCLCDD0.Determine the components and the magnitude ofL.yxWDLSolution:LD WDD 500j200iC100jD200iC400jlbjLj D√200 lb2C400 lb2D447 lbLD200iC400jlb,jLj D447 lb

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Problem 2.22Two perpendicular vectorsUandVliein thex-yplane. The vectorUD6i8jandjVj D20.What are the components ofV? (Notice that this problemhas two answers.)Solution:The two possible values ofVare shown in the sketch.The strategy is to (a) determine the unit vector associated withU,(b) express this vector in terms of an angle, (c) addš90°to thisangle, (d) determine the two unit vectors perpendicular toU, and(e) calculate the components of the two possible values ofV. Theunit vector parallel toUiseUD6i√62C828j√62C82D0.6i0.8jExpressed in terms of an angle,eUDicos˛jsin˛Dicos53.1°jsin53.1°Addš90°tofind the two unit vectors that are perpendicular to thisunit vector:ep1Dicos143.1°jsin143.1°D 0.8i0.6jep2Dicos36.9°jsin36.9°D0.8iC0.6jTake the scalar multiple of these unit vectors tofind the two vectorsperpendicular toU.V1D jVj0.8i0.6jD 16i12j.The components areVxD 16, VyD 12V2D jVj0.8iC0.6jD16iC12j.The components areVxD16, VyD12yx68UV2V1Problem 2.23Afish exerts a 10-lb force on the linethat is represented by the vectorF. ExpressFin termsof components using the coordinate system shown.yxF711Solution:Wecanusesimilartrianglestodeterminethecomponents ofF.FD10 lb(7p72C112i11p72C112j)D5.37i8.44jlbFD5.37i8.44jlb

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