The properties of force systems must be thoroughly understood by ...

c02.qxd 10/29/07 1:38 PM Page 2626 Chapter 2 Force SystemsF– FVector ComponentsIn addition to combining forces to obtain their resultant, we oftenneed to replace a force by its vector components in directions which areconvenient for a given application. The vector sum of the componentsmust equal the original vector. Thus, the force R in Fig. 2/3a may be replacedby, or resolved into, two vector components F 1 and F 2 with thespecified directions by completing the parallelogram as shown to obtainthe magnitudes of F 1 and F 2 .The relationship between a force and its vector components alonggiven axes must not be confused with the relationship between a forceand its perpendicular* projections onto the same axes. Figure 2/3eshows the perpendicular projections F a and F b of the given force R ontoaxes a and b, which are parallel to the vector components F 1 and F 2 ofFig. 2/3a. Figure 2/3e shows that the components of a vector are not necessarilyequal to the projections of the vector onto the same axes. Furthermore,the vector sum of the projections F a and F b is not the vectorR, because the parallelogram law of vector addition must be used toform the sum. The components and projections of R are equal onlywhen the axes a and b are perpendicular.F 2F 1 R1R 2R 1R 2RFigure 2/4A Special Case of Vector AdditionTo obtain the resultant when the two forces F 1 and F 2 are parallelas in Fig. 2/4, we use a special case of addition. The two vectors are combinedby first adding two equal, opposite, and collinear forces F and Fof convenient magnitude, which taken together produce no external effecton the body. Adding F 1 and F to produce R 1 , and combining withthe sum R 2 of F 2 and F yield the resultant R, which is correct in magnitude,direction, and line of action. This procedure is also useful forgraphically combining two forces which have a remote and inconvenientpoint of concurrency because they are almost parallel.It is usually helpful to master the analysis of force systems in twodimensions before undertaking three-dimensional analysis. Thus the remainderof Chapter 2 is subdivided into these two categories.SECTION ATWO-DIMENSIONAL FORCE SYSTEMSjF yyθF xFix2/3 RECTANGULAR C OMPONENTSThe most common two-dimensional resolution of a force vector isinto rectangular components. It follows from the parallelogram rulethat the vector F of Fig. 2/5 may be written asF F x F y(2/1)where F x and F y are vector components of F in the x- and y-directions.Each of the two vector components may be written as a scalar times theFigure 2/5*Perpendicular projections are also called orthogonal projections.

c02.qxd 10/29/07 1:38 PM Page 27Article 2/3 Rectangular Components 27appropriate unit vector. In terms of the unit vectors i and j of Fig. 2/5,F x F x i and F y F y j, and thus we may writeF F x i F y j(2/2)where the scalars F x and F y are the x and y scalar components of the vectorF.The scalar components can be positive or negative, depending onthe quadrant into which F points. For the force vector of Fig. 2/5, the xand y scalar components are both positive and are related to the magnitudeand direction of F by2 2F x F cos F F x F yFyβxF y F sin tan 1 F yF x(2/3)F x = F sin βF y = F cos βConventions for Describing Vector ComponentsWe express the magnitude of a vector with lightface italic type inprint; that is, F is indicated by F, a quantity which is always nonnegative.However, the scalar components, also denoted by lightface italictype, will include sign information. See Sample Problems 2/1 and 2/3 fornumerical examples which involve both positive and negative scalarcomponents.When both a force and its vector components appear in a diagram, itis desirable to show the vector components of the force with dashedlines, as in Fig. 2/5, and show the force with a solid line, or vice versa.With either of these conventions it will always be clear that a force andits components are being represented, and not three separate forces, aswould be implied by three solid-line vectors.Actual problems do not come with reference axes, so their assignmentis a matter of arbitrary convenience, and the choice is frequentlyup to the student. The logical choice is usually indicated by the way inwhich the geometry of the problem is specified. When the principal dimensionsof a body are given in the horizontal and vertical directions, forexample, you would typically assign reference axes in these directions.Determining the Components of a ForceDimensions are not always given in horizontal and vertical directions,angles need not be measured counterclockwise from the x-axis,and the origin of coordinates need not be on the line of action of a force.Therefore, it is essential that we be able to determine the correct componentsof a force no matter how the axes are oriented or how the anglesare measured. Figure 2/6 suggests a few typical examples of vectorresolution in two dimensions.Memorization of Eqs. 2/3 is not a substitute for understanding theparallelogram law and for correctly projecting a vector onto a referenceaxis. A neatly drawn sketch always helps to clarify the geometry andavoid error.F xβyF x = – F cos βF y = – F sin βyFβxF x = F sin ( π – β )F y = – F cos ( π – β )yFαβxF x = F cos( β – α )F y = F sin( β – α )Figure 2/6

c02.qxd 10/29/07 1:38 PM Page 2828 Chapter 2 Force SystemsRectangular components are convenient for finding the sum or resultantR of two forces which are concurrent. Consider two forces F 1and F 2 which are originally concurrent at a point O. Figure 2/7 showsthe line of action of F 2 shifted from O to the tip of F 1 according to thetriangle rule of Fig. 2/3. In adding the force vectors F 1 and F 2 , we maywriteorfrom which we conclude thatR F 1 F 2 (F 1xi F 1yj) (F 2xi F 2yj)R x i R y j (F 1x F 2x)i (F 1y F 2y) jR x F 1x F 2x ΣF xR y F 1y F 2y ΣF y(2/4)The term ΣF x means “the algebraic sum of the x scalar components”.For the example shown in Fig. 2/7, note that the scalar component F 2ywould be negative.© Vince Streano/Photographer’s Choice/Getty ImagesF 1yF 2yyjR yFigure 2/7OF 1xF 1F 2xF 2RixThe structural elements in the foregroundtransmit concentrated forcesto the brackets at both ends.R x

c02.qxd 10/29/07 1:38 PM Page 29Article 2/3 Rectangular Components 29Sample Problem 2/1The forces F 1 , F 2 , and F 3 , all of which act on point A of the bracket, arespecified in three different ways. Determine the x and y scalar components ofeach of the three forces.Solution.The scalar components of F 1 , from Fig. a, areF 1x 600 cos 35 491 NF 1y 600 sin 35 344 NThe scalar components of F 2 , from Fig. b, areF 2x 500 45 400 NF 2y 500 35 300 NAns.Ans.Ans.Ans.Note that the angle which orients F 2 to the x-axis is never calculated. The cosineand sine of the angle are available by inspection of the 3-4-5 triangle. Also notethat the x scalar component of F 2 is negative by inspection.The scalar components of F 3 can be obtained by first computing the angle of Fig. c.yF 2 = 500 N F 1 = 600 N3 A35°40.1 mx0.2 m0.3 mF 3 = 800 N B0.4 mFF 1 = 600 N1yFA 3x35°AF 1x(a)αF 3 = 800 N0.4 m tan 1 0.20.4 26.6Then F 3x F 3 sin 800 sin 26.6 358 NF 3y F 3 cos 800 cos 26.6 716 NAns.Ans.F 2 = 500 N34F 2x(b)AF 2yF 3y0.2 m(c)BAlternatively, the scalar components of F 3 can be obtained by writing F 3 asa magnitude times a unit vector n AB in the direction of the line segment AB.Thus,F 3 F 3 n AB F 3 ABlAB 800 0.2i 0.4j (0.2) 2 (0.4) 2The required scalar components are thenwhich agree with our previous results.F 3x 358 N 800 [0.447i 0.894j] 358i 716j NF 3y 716 NAns.Ans.Helpful Hints You should carefully examine thegeometry of each component determinationproblem and not rely onthe blind use of such formulas asF x F cos and F y F sin . A unit vector can be formed by dividingany vector, such as the geometricposition vector AB l, by itslength or magnitude. Here we usethe overarrow to denote the vectorwhich runs from A to B and theoverbar to determine the distancebetween A and B.

c02.qxd 10/29/07 1:38 PM Page 3030 Chapter 2 Force SystemsSample Problem 2/2Combine the two forces P and T, which act on the fixed structure at B, intoa single equivalent force R.P = 800 NT = 600 NBy6 mxGraphical solution. The parallelogram for the vector addition of forces T and P is constructed as shown in Fig. a. The scale used here is 1 cm 400 N; a scaleof 1 cm 100 N would be more suitable for regular-size paper and would givegreater accuracy. Note that the angle a must be determined prior to constructionof the parallelogram. From the given figureAα C 60°3 mDtan BDAD 6 sin 60 0.866 40.93 6 cos 60Measurement of the length R and direction of the resultant force R yields theapproximate resultsR 525 N 49Ans.T600 NαB800 NθR(a)PGeometric solution. The triangle for the vector addition of T and P is shownin Fig. b. The angle is calculated as above. The law of cosines givesR 524 NFrom the law of sines, we may determine the angle which orients R. Thus,Algebraic solution.we may writeR 2 (600) 2 (800) 2 2(600)(800) cos 40.9 274,300600sin 524sin 40.9sin 0.750 48.6Ans.Ans.By using the x-y coordinate system on the given figure,R x ΣF x 800 600 cos 40.9 346 NR y ΣF y 600 sin 40.9 393 NThe magnitude and dipection of the resultant force R as shown in Fig. c are thenHelpful Hints Note the repositioning of P to permitparallelogram addition at B.BRθ800 NαT(b)600 N Note the repositioning of F so as topreserve the correct line of action ofthe resultant R.yαPR R2x R2y (346) 2 (393) 2 524 N tan R 1 y 393 tan1R x 346 48.6Ans.Ans.R y = – 393 N BR x = 346 NθxThe resultant R may also be written in vector notation asR R x i R y j 346i 393j NAns.(c)R

c02.qxd 10/29/07 1:38 PM Page 31Article 2/3 Rectangular Components 31Sample Problem 2/3yThe 500-N force F is applied to the vertical pole as shown. (1) Write F interms of the unit vectors i and j and identify both its vector and scalar components.(2) Determine the scalar components of the force vector F along thex- and y-axes. (3) Determine the scalar components of F along the x- and y-axes.jAj′ y′30°xiSolution. Part (1). From Fig. a we may write F asF (F cos )i (F sin )j (500 cos 60)i (500 sin 60)j30°i′F = 500 Nx′ (250i 433j) NAns.yy′The scalar components are F x 250 N and F y 433 N. The vector componentsare F x 250i N and F y 433j N.Part (2). From Fig. b we may write F as F 500i N, so that the requiredscalar components areAF yθ = 60°FxAj′FF x 500 N F y 0Ans.i′Part (3). The components of F in the x- and y-directions are nonrectangularand are obtained by completing the parallelogram as shown in Fig. c. Themagnitudes of the components may be calculated by the law of sines. Thus,F x sin 90 500sin 30F y sin 60 500sin 30The required scalar components are thenF x 1000 NF x 1000 NF y 866 NF y 866 NAns.(a)F xy′ F x60° 30°F90° y′90°30° 60°F = 500 NHelpful Hint(c)(b) Obtain F x and F y graphically andcompare your results with the calculatedvalues.x′xSample Problem 2/4F 1 = 100 NaForces F 1 and F 2 act on the bracket as shown. Determine the projection F bof their resultant R onto the b-axis.C30°20°F 2 = 80 NbSolution. The parallelogram addition of F 1 and F 2 is shown in the figure.Using the law of cosines gives usR 2 (80) 2 (100) 2 2(80)(100) cos 130The figure also shows the orthogonal projection F b of R onto the b-axis. Itslength isF b 80 100 cos 50 144.3 NR 163.4 NAns.C100 NF 150°F 280 NR50°aNote that the components of a vector are in general not equal to the projectionsof the vector onto the same axes. If the a-axis had been perpendicular tothe b-axis, then the projections and components of R would have been equal.F bb

c02.qxd 10/29/07 1:38 PM Page 3232 Chapter 2 Force SystemsPROBLEMSyIntroductory ProblemsF = 4.8 kN2/1 The force F has a magnitude of 800 N. Express F as avector in terms of the unit vectors i and j. Identify thex and y scalar components of F.Ans. F 459i 655j NF x 459 N, F y 655 NyO34xF = 800 N35°Problem 2/3Ox2/4 The line of action of the 9.6-kN force F runs throughthe points A and B as shown in the figure. Determinethe x and y scalar components of F.y, mmProblem 2/1B (300, 100)2/2 The magnitude of the force F is 600 N. Express F as avector in terms of the unit vectors i and j. Identifyboth the scalar and vector components of F.OF = 9.6 kNx, mmy, mA (–150, –200)A (–3, 1)Problem 2/430°OF = 600 Nx, m2/5 A cable stretched between the fixed supports A and Bis under a tension T of 900 N. Express the tension asa vector using the unit vectors i and j, first, as a forceT A acting on A and second, as a force T B acting on B.Ans. T A 749i 499j N, T B 749i 499j NProblem 2/2A3 m2/3 The slope of the 4.8-kN force F is specified as shownin the figure. Express F as a vector in terms of theunit vectors i and j.Ans. F 2.88i 3.84j kNy2 mxBProblem 2/5

c02.qxd 10/29/07 1:38 PM Page 33Article 2/3 Problems 332/6 The 1800-N force F is applied to the end of the I-beam. Express F as a vector using the unit vectors iand j.F 1 = 800 NyF 2 = 425 NyF = 1800 N70°θA3O4xzxzProblem 2/62/7 The two structural members, one of which is in tensionand the other in compression, exert the indicatedforces on joint O. Determine the magnitude of the resultantR of the two forces and the angle which Rmakes with the positive x-axis.Ans. R 3.61 kN, 206Representative ProblemsProblem 2/82 kN3 kN2/9 In the design of a control mechanism, it is determinedthat rod AB transmits a 260-N force P to the crankBC. Determine the x and y scalar components of P.Ans. P x 240 NP y 100 NP = 260 N30° 60°OxtBA125Problem 2/72/8 Two forces are applied to the construction bracket asshown. Determine the angle which makes the resultantof the two forces vertical. Determine the magnitudeR of the resultant.nCy30°xProblem 2/92/10 For the mechanism of Prob. 2/9, determine thescalar components P t and P n of P which are tangentand normal, respectively, to crank BC.

c02.qxd 10/29/07 1:38 PM Page 3434 Chapter 2 Force Systems2/11 The t-component of the force F is known to be 75N. Determine the n-component and the magnitudeof F.Ans. F n 62.9 N, F 97.9 NntF 1 = 3 kN30°AF10°45°F 2 = 7 kN30°yProblem 2/112/12 A force F of magnitude 800 N is applied to point C ofthe bar AB as shown. Determine both the x-y andthe n-t components of F.AynxProblem 2/132/14 To satisfy design limitations it is necessary to determinethe effect of the 2-kN tension in the cable onthe shear, tension, and bending of the fixed I-beam.For this purpose replace this force by its equivalentof two forces at A, F t parallel and F n perpendicularto the beam. Determine F t and F n .xtAC20°40°F = 800 N30°2 kN60°BProblem 2/122/13 The two forces shown act at point A of the bent bar.Determine the resultant R of the two forces.Ans. R 2.35i 3.45j kNProblem 2/14

c02.qxd 10/29/07 1:38 PM Page 35Article 2/3 Problems 352/15 Determine the magnitude F s of the tensile springforce in order that the resultant of F s and F is a verticalforce. Determine the magnitude R of this verticalresultant force.Ans. F s 250 N, R 433 NbF = 2 kNF = 500 N60°45°Problem 2/17aA60°2/18 Determine the scalar components R a and R b of theforce R along the nonrectangular axes a and b. Alsodetermine the orthogonal projection P a of R ontoaxis a.Problem 2/15a2/16 The ratio of the lift force L to the drag force D forthe simple airfoil is L/D 10. If the lift force on ashort section of the airfoil is 200 N, compute themagnitude of the resultant force R and the angle which it makes with the horizontal.O30°R = 800 NL110°bAir flowCDProblem 2/16Problem 2/182/19 Determine the resultant R of the two forces shownby (a) applying the parallelogram rule for vector additionand (b) summing scalar components.Ans. R 520i 700j N2/17 Determine the components of the 2-kN force alongthe oblique axes a and b. Determine the projectionsof F onto the a- and b-axes.Ans. F a 0.598 kN, F b 1.633 kNP a 1.414 kN, P b 1.932 kNyx60°600 N400 NProblem 2/19

c02.qxd 10/29/07 1:38 PM Page 3636 Chapter 2 Force Systems2/20 It is desired to remove the spike from the timber byapplying force along its horizontal axis. An obstructionA prevents direct access, so that two forces, one1.6 kN and the other P, are applied by cables asshown. Compute the magnitude of P necessary toensure a resultant T directed along the spike. Alsofind T.yxO60 mmA40 mm200 mmP100 mm80 mmθnktA150 mmPProblem 2/22Problem 2/202/21 At what angle must the 800-N force be applied inorder that the resultant R of the two forces has amagnitude of 2000 N? For this condition, determinethe angle between R and the vertical.Ans. 51.3, 18.191400 Nθ1.6 kN800 N2/23 Refer to the statement and figure of Prob. 2/22.When pin P is in the position 20, determine then- and t-components of the force F which the springof modulus k 1.2 kN/m exerts on the pin.Ans. F n 19.18 N, F t 13.84 N2/24 The cable AB prevents bar OA from rotating clockwiseabout the pivot O. If the cable tension is 750 N,determine the n- and t-components of this force actingon point A of the bar.nAtBO60°1.5 mProblem 2/212/22 The unstretched length of the spring of modulus k 1.2 kN/m is 100 mm. When pin P is in the position 30, determine the x- and y-components of theforce which the spring exerts on the pin. (The forcein a spring is given by F kx, where x is the deflectionfrom the unstretched length.)1.2 mProblem 2/24

c02.qxd 10/29/07 1:38 PM Page 37Article 2/3 Problems 372/25 At what angle must the 400-N force be applied inorder that the resultant R of the two forces have amagnitude of 1000 N? For this condition what willbe the angle between R and the horizontal?Ans. 51.3, 18.19400 N2/27 The guy cables AB and AC are attached to the top ofthe transmission tower. The tension in cable AC is 8kN. Determine the required tension T in cable ABsuch that the net effect of the two cable tensions is adownward force at point A. Determine the magnitudeR of this downward force.Ans. T 5.68 kN, R 10.21 kNA40 m700 NθBO20 mC50 m40 mProblem 2/252/26 In the design of the robot to insert the small cylindricalpart into a close-fitting circular hole, the robotarm must exert a 90-N force P on the part parallel tothe axis of the hole as shown. Determine the componentsof the force which the part exerts on the robotalong axes (a) parallel and perpendicular to the armAB, and (b) parallel and perpendicular to the arm BC.Problem 2/272/28 The gusset plate is subjected to the two forcesshown. Replace them by two equivalent forces, F x inthe x-direction and F a in the a-direction. Determinethe magnitudes of F x and F a . Solve geometrically orgraphically.B15°yA60°45°CDP = 90 NA800 N10°x900 N25° 45°aProblem 2/28Problem 2/26

c02.qxd 10/29/07 1:38 PM Page 3838 Chapter 2 Force SystemsdF2/4 MOMENTIn addition to the tendency to move a body in the direction of its application,a force can also tend to rotate a body about an axis. The axismay be any line which neither intersects nor is parallel to the line of actionof the force. This rotational tendency is known as the moment M ofthe force. Moment is also referred to as torque.As a familiar example of the concept of moment, consider the pipewrench of Fig. 2/8a. One effect of the force applied perpendicular tothe handle of the wrench is the tendency to rotate the pipe about itsvertical axis. The magnitude of this tendency depends on both themagnitude F of the force and the effective length d of the wrenchhandle. Common experience shows that a pull which is not perpendicularto the wrench handle is less effective than the right-angle pullshown.Moment about a PointAOMdr(a)FαFigure 2/8b shows a two-dimensional body acted on by a force F inits plane. The magnitude of the moment or tendency of the force to rotatethe body about the axis O-O perpendicular to the plane of thebody is proportional both to the magnitude of the force and to the momentarm d, which is the perpendicular distance from the axis to theline of action of the force. Therefore, the magnitude of the moment isdefined asM Fd(2/5)+yO (b)M(c)xM = FdAd(d)Figure 2/8FThe moment is a vector M perpendicular to the plane of the body. Thesense of M depends on the direction in which F tends to rotate thebody. The right-hand rule, Fig. 2/8c, is used to identify this sense. Werepresent the moment of F about O-O as a vector pointing in the directionof the thumb, with the fingers curled in the direction of the rotationaltendency.The moment M obeys all the rules of vector combination and maybe considered a sliding vector with a line of action coinciding with themoment axis. The basic units of moment in SI units are newton-meters(N m), and in the U.S. customary system are pound-feet (lb-ft).When dealing with forces which all act in a given plane, we customarilyspeak of the moment about a point. By this we mean the momentwith respect to an axis normal to the plane and passing through thepoint. Thus, the moment of force F about point A in Fig. 2/8d has themagnitude M Fd and is counterclockwise.Moment directions may be accounted for by using a stated sign convention,such as a plus sign () for counterclockwise moments and aminus sign () for clockwise moments, or vice versa. Sign consistencywithin a given problem is essential. For the sign convention of Fig. 2/8d,the moment of F about point A (or about the z-axis passing throughpoint A) is positive. The curved arrow of the figure is a convenient wayto represent moments in two-dimensional analysis.

c02.qxd 10/29/07 1:38 PM Page 39Article 2/4 Moment 39The Cross ProductIn some two-dimensional and many of the three-dimensional problemsto follow, it is convenient to use a vector approach for moment calculations.The moment of F about point A of Fig. 2/8b may berepresented by the cross-product expressionM r F(2/6)where r is a position vector which runs from the moment referencepoint A to any point on the line of action of F. The magnitude of this expressionis given by*M Fr sin Fd(2/7)which agrees with the moment magnitude as given by Eq. 2/5. Note thatthe moment arm d r sin does not depend on the particular point onthe line of action of F to which the vector r is directed. We establish thedirection and sense of M by applying the right-hand rule to the sequencer F. If the filgers of the right hand are curled in the directionof rotation from the positive sense of r to the positive sense of F, thenthe thumb points in the positive sense of M.We must maintain the sequence r F, because the sequence F rwould produce a vector with a sense opposite to that of the correctmoment. As was the case with the scalar approach, the moment Mmay be thought of as the moment about point A or as the momentabout the line O-O which passes through point A and is perpendicularto the plane containing the vectors r and F. When we evaluate themoment of a force about a given point, the choice between using thevector cross product or the scalar expression depends on how thegeometry of the problem is specified. If we know or can easily determinethe perpendicular distance between the line of action of theforce and the moment center, then the scalar approach is generallysimpler. If, however, F and r are not perpendicular and are easily expressiblein vector notation, then the cross-product expression isoften preferable.In Section B of this chapter, we will see how the vector formulationof the moment of a force is especially useful for determining the momentof a force about a point in three-dimensional situations.Varignon’s TheoremOne of the most useful principles of mechanics is Varignon’s theorem,which states that the moment of a force about any point is equal tothe sum of the moments of the components of the force about the samepoint.*See item 7 in Art. C/7 of Appendix C for additional information concerning the crossproduct.

c02.qxd 10/29/07 1:38 PM Page 4040 Chapter 2 Force SystemsTo prove this theorem, consider the force R acting in the plane ofthe body shown in Fig. 2/9a. The forces P and Q represent any two nonrectangularcomponents of R. The moment of R about point O isBecause R P Q, we may writeM O r Rr R r (P Q)Using the distributive law for cross products, we haveM O r R r P r Q(2/8)which says that the moment of R about O equals the sum of the momentsabout O of its components P and Q. This proves the theorem.Varignon’s theorem need not be restricted to the case of two components,but it applies equally well to three or more. Thus we could haveused any number of concurrent components of R in the foregoingproof.*Figure 2/9b illustrates the usefulness of Varignon’s theorem. Themoment of R about point O is Rd. However, if d is more difficult to determinethan p and q, we can resolve R into the components P and Q,and compute the moment asM O Rd pP qQwhere we take the clockwise moment sense to be positive.Sample Problem 2/5 shows how Varignon’s theorem can help us tocalculate moments.PRPRBQqBQOrdOp(a)(b)Figure 2/9*As originally stated, Varignon’s theorem was limited to the case of two concurrent componentsof a given force. See The Science of Mechanics, by Ernst Mach, originally publishedin 1883.

c02.qxd 10/29/07 1:38 PM Page 4242 Chapter 2 Force SystemsSample Problem 2/6TBThe trap door OA is raised by the cable AB, which passes over the small frictionlessguide pulleys at B. The tension everywhere in the cable is T, and this tensionapplied at A causes a moment M O about the hinge at O. Plot the quantity M O /Tas a function of the door elevation angle over the range 0 90 and note minimumand maximum values. What is the physical significance of this ratio?0.4 mO0.5 mθ0.3 mASolution. We begin by constructing a figure which shows the tension force Tacting directly on the door, which is shown in an arbitrary angular position . Itshould be clear that the direction of T will vary as varies. In order to deal withthis variation, we write a unit vector n AB which “aims” T:Using the x-y coordinates of our figure, we can writeSoandThe desired unit vector isOur tension vector can now be written asThe moment of T about point O, as a vector, is M O r OB T, where r OB 0.4j m, orThe magnitude of M O isand the requested ratio isn AB r ABr AB r OB r OAr ABr OB 0.4j m and r OA 0.5(cos i sin j) mr AB r OB r OA 0.4j (0.5)(cos i sin j) 0.5 cos i (0.4 0.5 sin )j mr AB (0.5 cos ) 2 (0.4 0.5 sin ) 2 0.41 0.4 sin mn AB r AB 0.5 cos i (0.4 0.5 sin )jrAB 0.41 0.4 sin T Tn AB T 0.5 cos i (0.4 0.5 sin )j0.41 0.4 sin M O 0.4j T 0.5 cos i (0.4 0.5 sin )j0.41 0.4 sin 0.2T cos 0.41 0.4 sin k0.2T cos M O 0.41 0.4 sin M OT 0.2 cos 0.41 0.4 sin Ans.which is plotted in the accompanying graph. The expression M O /T is the momentarm d (in meters) which runs from O to the line of action of T. It has a maximumvalue of 0.4 m at 53.1 (at which point T is horizontal) and a minimum value of0 at 90 (at which point T is vertical). The expression is valid even if T varies.This sample problem treats moments in two-dimensional force systems, andit also points out the advantages of carrying out a solution for an arbitrary position,so that behavior over a range of positions can be examined.Br OBOyd0.50.40.3M O——–, mT 0.20.1r ABr OATθAHelpful Hints Recall that any unit vector can bewritten as a vector divided by itsmagnitude. In this case the vector inthe numerator is a position vector.x00 10 20 30 40 50 60 70 80 90θ, deg Recall that any vector may be writtenas a magnitude times an “aiming”unit vector. In the expression M r F, the positionvector r runs from the momentcenter to any point on the lineof action of F. Here, r OB is more convenientthan r OA .

c02.qxd 10/29/07 1:38 PM Page 43Article 2/4 Problems 43PROBLEMSIntroductory Problems2/29 The 10-kN force is applied at point A. Determine themoment of F about point O. Determine the points onthe x- and y-axes about which the moment of F iszero.Ans. M O 16 kN m CW(x, y) (2.67, 0) m and (0, 2) m2/31 Determine the moment of the 50-N force (a) aboutpoint O by Varignon’s theorem and (b) about point Cby a vector approach.Ans. M O 519 N mm CCW, M C 1616k N mmy, mmC (0, 25)B (40, 10)A (–4, 5)y, mOF = 50 Nx, mm34F = 10 kNA (–15, –20)OProblem 2/29x, mProblem 2/312/32 The force of magnitude F acts along the edge of thetriangular plate. Determine the moment of F aboutpoint O.2/30 Determine the moment of the 800-N force aboutpoint A and about point O.hFAy875 mmBObProblem 2/32625 mmOF = 800 N 30°x2/33 In steadily turning the water pump, a person exertsthe 120-N force on the handle as shown. Determinethe moment of this force about point O.Ans. M O 14.74 N m CW15°F = 120 NAProblem 2/3020°150 mmOProblem 2/33

c02.qxd 10/29/07 1:38 PM Page 4444 Chapter 2 Force Systems2/34 The throttle-control sector pivots freely at O. If aninternal torsional spring exerts a return moment M 1.8 N m on the sector when in the positionshown, for design purposes determine the necessarythrottle-cable tension T so that the net momentabout O is zero. Note that when T is zero, the sectorrests against the idle-control adjustment screw at R.30 mm200 mm15°250 NQTPProblem 2/36R50 mm2/37 A mechanic pulls on the 13-mm combination wrenchwith the 140-N force shown. Determine the momentof this force about the bolt center O.Ans. M O 13.10 N m CCWMOF = 140 N25°Problem 2/34O15°2/35 A force F of magnitude 60 N is applied to the gear.Determine the moment of F about point O.Ans. M O 5.64 N m CW95 mmAOr = 100 mm20°F = 60 NProblem 2/372/38 As a trailer is towed in the forward direction, theforce F 500 N is applied as shown to the ball of thetrailer hitch. Determine the moment of this forceabout point O.F = 500 NProblem 2/3530°2/36 Calculate the moment of the 250-N force on the handleof the monkey wrench about the center of thebolt.32 mmAO38 mm275 mmProblem 2/38

c02.qxd 10/29/07 1:38 PM Page 45Article 2/4 Problems 45Representative Problems2/39 A portion of a mechanical coin sorter works as follows:Pennies and dimes roll down the 20 incline,the last triangular portion of which pivots freelyabout a horizontal axis through O. Dimes are lightenough (2.28 grams each) so that the triangular portionremains stationary, and the dimes roll into theright collection column. Pennies, on the other hand,are heavy enough (3.06 grams each) so that the triangularportion pivots clockwise, and the penniesroll into the left collection column. Determine themoment about O of the weight of the penny in termsof the slant distance s in millimeters.Ans. M O 0.1335 0.0282s N mm (s in mm)2/40 Elements of the lower arm are shown in the figure.The mass of the forearm is 2.3 kg with mass centerat G. Determine the combined moment about theelbow pivot O of the weights of the forearm and the3.6-kg homogeneous sphere. What must the bicepstension force be so that the overall moment about Ois zero?TsO50 mm3.5 mmLIBERTYTRUSTWE GOD IN INO 20°19899.5 mm55°150 mmGA2.3(9.81) N325 mm3.6(9.81) NProblem 2/40penniesProblem 2/39dimes2/41 A 150-N pull T is applied to a cord, which is woundsecurely around the inner hub of the drum. Determinethe moment of T about the drum center C. Atwhat angle should T be applied so that the momentabout the contact point P is zero?Ans. M C 18.75 N m CW, 51.3T = 150 Nθ125 mmC200 mmPProblem 2/41

c02.qxd 10/29/07 1:38 PM Page 4646 Chapter 2 Force Systems2/42 A force of 200 N is applied to the end of the wrenchto tighten a flange bolt which holds the wheel to theaxle. Determine the moment M produced by thisforce about the center O of the wheel for the positionof the wrench shown.450 mm200 N2/44 The uniform work platform, which has a mass perunit length of 28 kg/m, is simply supported by crossrods A and B. The 90-kg construction worker startsfrom point B and walks to the right. At what locations will the combined moment of the weights ofthe man and platform about point B be zero?1 m 4 m 3 ms90 kg650 mmO20°28 kg/m125 mmABProblem 2/422/43 In order to raise the flagpole OC, a light frame OABis attached to the pole and a tension of 3.2 kN is developedin the hoisting cable by the power winch D.Calculate the moment M O of this tension about thehinge point O.Ans. M O 6.17 kN m CCWAProblem 2/442/45 In raising the pole from the position shown, the tensionT in the cable must supply a moment about O of72 kN m. Determine T.Ans. T 8.65 kN10 mD20°3 m 3 m3 m BOCT30 mProblem 2/4312 mO60°Problem 2/45

c02.qxd 10/29/07 1:38 PM Page 47Article 2/4 Problems 472/46 The force exerted by the plunger of cylinder AB onthe door is 40 N directed along the line AB, and thisforce tends to keep the door closed. Compute the momentof this force about the hinge O. What force F Cnormal to the plane of the door must the door stopat C exert on the door so that the combined momentabout O of the two forces is zero?2/48 Calculate the moment M A of the 200-N force aboutpoint A by using three scalar methods and one vectormethod.200 N15°BOA B75100 C400 40025280mmAyx400 mmProblem 2/48Dimensions in millimetersProblem 2/462/47 The 10-N force is applied to the handle of the hydrauliccontrol valve as shown. Calculate the momentof this force about point O.Ans. M O 2.81 N m CWA2/49 An exerciser begins with his arm in the relaxed verticalposition OA, at which the elastic band is unstretched.He then rotates his arm to the horizontalposition OB. The elastic modulus of the band is k 60 N/m—that is, 60 N of force is required to stretchthe band each additional meter of elongation. Determinethe moment about O of the force which theband exerts on the hand B.Ans. M O 26.8 N m CCW20°F = 10 N250 mmB635 mmO60°37.5 mmBOFProblem 2/47A740 mmCProblem 2/49

c02.qxd 10/29/07 1:38 PM Page 4848 Chapter 2 Force Systems2/50 (a) Calculate the moment of the 90-N force aboutpoint O for the condition 15. Also, determinethe value of for which the moment about O is (b)zero and (c) a maximum.2/52 Design criteria require that the robot exert the 90-Nforce on the part as shown while inserting a cylindricalpart into the circular hole. Determine the momentabout points A, B, and C of the force which thepart exerts on the robot.O150 mmB15°800 mmF = 90 N600 mmθAA550 mm450 mm60°45°CDP = 90 NProblem 2/502/51 The small crane is mounted along the side of apickup bed and facilitates the handling of heavyloads. When the boom elevation angle is 40, theforce in the hydraulic cylinder BC is 4.5 kN, and thisforce applied at point C is in the direction from B toC (the cylinder is in compression). Determine themoment of this 4.5-kN force about the boom pivotpoint O.Ans. M O 0.902 kN m CWProblem 2/522/53 The masthead fitting supports the two forces shown.Determine the magnitude of T which will cause nobending of the mast (zero moment) at point O.Ans. T 4.04 kNA90mm120mm785mm110mmCθ340mmO360mmC60 mm 2O30°5B5 kNTProblem 2/51Problem 2/53

c02.qxd 10/29/07 1:38 PM Page 49Article 2/4 Problems 492/54 The piston, connecting rod, and crankshaft of adiesel engine are shown in the figure. The crankthrow OA is half the stroke of 200 mm, and thelength AB of the rod is 350 mm. For the position indicated,the rod is under a compression along AB of16 kN. Determine the moment M of this force aboutthe crankshaft axis O.B2/56 If the combined moment of the two forces aboutpoint C is zero, determine(a) the magnitude of the force P(b) the magnitude R of the resultant of the twoforces(c) the coordinates x and y of the point A on the rimof the wheel about which the combined momentof the two forces is a maximum(d) the combined moment M A of the two forcesabout A.Ans. (a) P 61.6 N(b) R 141.3 N(c) x 49.0 mm, y 63.2 mm(d) M A 15.77 N m CWyOA30°100 N60°OC40 mmP40 mmx40 mmProblem 2/54Problem 2/562/55 The 120-N force is applied as shown to one end ofthe curved wrench. If 30, calculate the momentof F about the center O of the bolt. Determine thevalue of which would maximize the moment aboutO; state the value of this maximum moment.Ans. M O 41.5 N m CW 33.2, (M O ) max 41.6 N m CWF = 120 Nα25 mmA150 mm70mm70mm25 mmOProblem 2/55

c02.qxd 10/29/07 1:38 PM Page 5050 Chapter 2 Force SystemsOOOad(a)r Ar B(b)(c)–FFB –FrFAM2/5 COUPLEThe moment produced by two equal, opposite, and noncollinearforces is called a couple. Couples have certain unique properties andhave important applications in mechanics.Consider the action of two equal and opposite forces F and F a distanced apart, as shown in Fig. 2/10a. These two forces cannot be combinedinto a single force because their sum in every direction is zero. Their onlyeffect is to produce a tendency of rotation. The combined moment of thetwo forces about an axis normal to their plane and passing through anypoint such as O in their plane is the couple M. This couple has a magnitudeorM F(a d) FaM FdIts direction is counterclockwise when viewed from above for the case illustrated.Note especially that the magnitude of the couple is independentof the distance a which locates the forces with respect to themoment center O. It follows that the moment of a couple has the samevalue for all moment centers.Vector Algebra MethodWe may also express the moment of a couple by using vector algebra.With the cross-product notation of Eq. 2/6, the combined momentabout point O of the forces forming the couple of Fig. 2/10b isMMM r A F r B (F) (r A r B ) Fwhere r A and r B are position vectors which run from point O to arbitrarypoints A and B on the lines of action of F and F, respectively. Becauser A r B r, we can express M asM r FMCounterclockwisecouple(d)Figure 2/10MClockwisecoupleHere again, the moment expression contains no reference to the momentcenter O and, therefore, is the same for all moment centers. Thus,we may represent M by a free vector, as show in Fig. 2/10c, where thedirection of M is normal to the plane of the couple and the sense of M isestablished by the right-hand rule.Because the couple vector M is always perpendicular to the plane ofthe forces which constitute the couple, in two-dimensional analysis wecan represent the sense of a couple vector as clockwise or counterclockwiseby one of the conventions shown in Fig. 2/10d. Later, when we dealwith couple vectors in three-dimensional problems, we will make fulluse of vector notation to represent them, and the mathematics will automaticallyaccount for their sense.Equivalent CouplesChanging the values of F and d does not change a given couple aslong as the product Fd remains the same. Likewise, a couple is not affectedif the forces act in a different but parallel plane. Figure 2/11

c02.qxd 10/29/07 1:38 PM Page 5252 Chapter 2 Force SystemsSample Problem 2/7The rigid structural member is subjected to a couple consisting of the two100-N forces. Replace this couple by an equivalent couple consisting of the twoforces P and P, each of which has a magnitude of 400 N. Determine the properangle .40M–P θ θPSolution. The original couple is counterclockwise when the plane of the forcesis viewed from above, and its magnitude is[M Fd]M 100(0.1) 10 N m100100The forces P and P produce a counterclockwise coupleM 400(0.040) cos 100 N10060Equating the two expressions gives100 NHelpful Hint10 (400)(0.040) cos cos1 1016 51.3Ans. Since the two equal couples are parallel free vectors, the only dimensionswhich are relevant are those which give the perpendicular distances betweenthe forces of the couples.Dimensions in millimetersP = 400 Nθ40 mmdθθP = 400 NSample Problem 2/8Replace the horizontal 400-N force acting on the lever by an equivalent systemconsisting of a force at O and a couple.200 mm400 NSolution. We apply two equal and opposite 400-N forces at O and identify thecounterclockwise coupleO60°[M Fd]M 400(0.200 sin 60) 69.3 N mThus, the original force is equivalent to the 400-N force at O and thecouple as shown in the third of the three equivalent figures.Ans.69.3-N mHelpful Hint The reverse of this problem is often encountered, namely, the replacementof a force and a couple by a single force. Proceeding in reverse is the same asreplacing the couple by two forces, one of which is equal and opposite to the400-N force at O. The moment arm to the second force would be M/F 69.3/400 0.1732 m, which is 0.2 sin 60, thus determining the line of actionof the single resultant force of 400 N.400 N 400 N≡ ≡O O O400 N 400 N 400 N69.3 N.m

c02.qxd 10/29/07 1:38 PM Page 53Article 2/5 Problems 53PROBLEMSIntroductory Problems2/57 The caster unit is subjected to the pair of 400-Nforces shown. Determine the moment associatedwith these forces.Ans. M 14 N m CW35mm2/59 The top view of a revolving entrance door is shown.Two persons simultaneously approach the door andexert force of equal magnitudes as shown. If the resultingmoment about the door pivot axis at O is 25N m, determine the force magnitude F.Ans. F 16.18 NF15°0.8 m400 NO0.8 m15°– FProblem 2/59400 NProblem 2/572/58 A force F 60 N acts along the line AB. Determinethe equivalent force–couple system at point C.2/60 The indicated force–couple system is applied to asmall shaft at the center of the plate. Replace thissystem by a single force and specify the coordinate ofthe point on the x-axis through which the line of actionof this resultalt force passes.M = 400 N·my10 mmBFyxOF = 6 kNxCAProblem 2/60Problem 2/58

c02.qxd 10/29/07 1:38 PM Page 5454 Chapter 2 Force Systems2/61 The bracket is spot welded to the end of the shaft atpoint O. To show the effect of the 900-N force on theweld, replace the force by its equivalent of a forceand couple M at O. Express M in vector notation.Ans. M 90i N m2/63 Replace the 10-kN force acting on the steel columnby an equivalent force–couple system at point O.This replacement is frequently done in the design ofstructures.Ans. R 10 kN, M O 0.75 kN m CCWy10 kN75 mmxO100 mmz1 m900 NOProblem 2/612/62 As part of a test, the two aircraft engines are revvedup and the propeller pitches are adjusted so as to resultin the fore and aft thrusts shown. What force Fmust be exerted by the ground on each of the mainbraked wheels at A and B to counteract the turningeffect of the two propeller thrusts? Neglect any effectsof the nose wheel C, which is turned 90 andunbraked.Problem 2/632/64 Each propeller of the twin-screw ship develops a fullspeedthrust of 300 kN. In maneuvering the ship,one propeller is turning full speed ahead and theother full speed in reverse. What thrust P must eachtug exert on the ship to counteract the effect of theship’s propellers?50 m2 kNFF12 mAC2 kNB3 m 5 m120 mProblem 2/64Problem 2/62

c02.qxd 10/29/07 1:38 PM Page 55Article 2/5 Problems 55Representative Problems2/65 A lug wrench is used to tighten a square-head bolt. If250-N forces are applied to the wrench as shown, determinethe magnitude F of the equal forces exertedon the four contact points on the 25-mm bolt head sothat their external effect on the bolt is equivalent tothat of the two 250-N forces. Assume that the forcesare perpendicular to the flats of the bolt head.Ans. F 3500 N25 mm2/67 The 180-N force is applied to the end of body OAB. If 50, determine the equivalent force–couple systemat the shaft axis O.Ans. F 169.1i 61.6j N, M O 41.9 N m CCWF = 180 NθBy30°120 mmAx150 mm250 NA350 mmC350 mmProblem 2/652/66 During a steady right turn, a person exerts the forcesshown on the steering wheel. Note that each forceconsists of a tangential component and a radiallyinwardcomponent. Determine the moment exertedabout the steering column at O.30°250 NB8 NView C Detail(clearances exaggerated)Problem 2/672/68 A force F of magnitude 50 N is exerted on the automobileparking-brake lever at the position x 250mm. Replace the force by an equivalent force–couplesystem at the pivot point O.O100 mm15°xFO20°10°15°AOProblem 2/68B15°30°8 N375 mmProblem 2/66

c02.qxd 10/29/07 1:38 PM Page 5656 Chapter 2 Force Systems2/69 The tie-rod AB exerts the 250-N force on the steeringknuckle AO as shown. Replace this force by anequivalent force–couple system at O.Ans. F 43.4i 246j N, M O 60.0 N m CW10° F = 250 N235 mmy2/71 The system consisting of the bar OA, two identicalpulleys, and a section of thin tape is subjected to thetwo 180-N tensile forces shown in the figure. Determinethe equivalent force–couple system at point O.Ans. M 21.7 N m CCW180 NArBxr = 25 mmA50 mmr100 mmOO45°180 N50 mmProblem 2/692/70 The combined drive wheels of a front-wheel-driveautomobile are acted on by a 7000-N normal reactionforce and a friction force F, both of which areexerted by the road surface. If it is known that theresultant of these two forces makes a 15 angle withthe vertical, determine the equivalent force–couplesystem at the car mass center G. Treat this as a twodimensionalproblem.500 mmGProblem 2/712/72 Calculate the moment M B of the 900-N force aboutthe bolt at B. Simplify your work by first replacingthe force by its equivalent force–couple system at A.200 mmCAB30°600 mm900 NA F1000mm7000 NProblem 2/70B40°Problem 2/72

c02.qxd 10/29/07 1:38 PM Page 57Article 2/5 Problems 572/73 The bracket is fastened to the girder by means of thetwo rivets A and B and supports the 2-kN force. Replacethis force by a force acting along the centerlinebetween the rivets and a couple. Then redistributethis force and couple by replacing it by two forces,one at A and the other at B, and ascertain the forcessupported by the rivets.Ans. F A 0.8 kNF B 2.8 kN2/75 The weld at O can support a maximum of 2500 N offorce along each of the n- and t-directions and a maximumof 1400 N m of moment. Determine the allowablerange for the direction of the 2700-N forceapplied at A. The angle is restricted to 0 90.Ans. 22.2 47.8O0.7 mAθtA250 mmnProblem 2/752700 NxB300 mmProblem 2/73100 mm2 kN2/74 The angle plate is subjected to the two 250-N forcesshown. It is desired to replace these forces by anequivalent set consisting of the 200-N force appliedat A and a second force applied at B. Determine they-coordinate of B.250 N200mmAxA200 N30°2/76 The device shown is a part of an automobile seatback-releasemechanism. The part is subjected to the4-N force exerted at A and a 300-N mm restoringmoment exerted by a hidden torsional spring. Determinethe y-intercept of the line of action of the singleequivalent force.10 mm40 mm300 N·mmyOAx15°F = 4 N240 mmProblem 2/76250 NB160 mmyyProblem 2/74

c02.qxd 10/29/07 1:38 PM Page 5858 Chapter 2 Force Systems2/6 RESULTANTSF 1F 2F 1RF 2R 1R 1F 3The properties of force, moment, and couple were developed in theprevious four articles. Now we are ready to describe the resultant action ofa group or system of forces. Most problems in mechanics deal with a systemof forces, and it is usually necessary to reduce the system to its simplestform to describe its action. The resultant of a system of forces is the simplestforce combination which can replace the original forces without alteringthe external effect on the rigid body to which the forces are applied.Equilibrium of a body is the condition in which the resultant of allforces acting on the body is zero. This condition is studied in statics. Whenthe resultant of all forces on a body is not zero, the acceleration of the bodyis obtained by equating the force resultant to the product of the mass andacceleration of the body. This condition is studied in dynamics. Thus, thedetermination of resultants is basic to both statics and dynamics.The most common type of force system occurs when the forces allact in a single plane, say, the x-y plane, as illustrated by the system ofthree forces F 1 , F 2 , and F 3 in Fig. 2/13a. We obtain the magnitude anddirection of the resultant force R by forming the force polygon shown inpart b of the figure, where the forces are added head-to-tail in any sequence.Thus, for any system of coplanar forces we may write(a)yF FF 2y 2 F 33yF 1R F y 1yRθF 1xF 2x F 3xR x(b)Figure 2/13F 3xR F 1 F 2 F 3 ΣFR x ΣF x R y ΣF y R (ΣF x ) 2 (ΣF y ) 2 tan 1 R yR x tan 1 ΣF yΣF x(2/9)Graphically, the correct line of action of R may be obtained by preservingthe correct lines of action of the forces and adding them by theparallelogram law. We see this in part a of the figure for the case ofthree forces where the sum R 1 of F 2 and F 3 is added to F 1 to obtain R.The principle of transmissibility has been used in this process.Algebraic MethodWe can use algebra to obtain the resultant force and its line of actionas follows:1. Choose a convenient reference point and move all forces to thatpoint. This process is depicted for a three-force system in Figs.2/14a and b, where M 1 , M 2 , and M 3 are the couples resulting fromthe transfer of forces F 1 , F 2 , and F 3 from their respective originallines of action to lines of action through point O.2. Add all forces at O to form the resultant force R, and add all couplesto form the resultant couple M O . We now have the single force–couple system, as shown in Fig. 2/14c.3. In Fig. 2/14d, find the line of action of R by requiring R to have amoment of M O about point O. Note that the force systems of Figs.2/14a and 2/14d are equivalent, and that Σ(Fd) in Fig. 2/14a is equalto Rd in Fig. 2/14d.

c02.qxd 10/29/07 1:38 PM Page 59Article 2/6 Resultants 59F 1F 2M 1 = F 1 d 1F 1d1 d 3O d 2F 2OF 3M 2 = F 2 d 2M 3 = F 3 d 3(a)F 3(b)M O = Σ(Fd)OR = ΣFOdRd = –— M OR(c)(d)Figure 2/14Principle of MomentsThis process is summarized in equation form byR ΣFM O ΣM Σ(Fd)(2/10)Rd M OThe first two of Eqs. 2/10 reduce a given system of forces to a force–couple system at an arbitrarily chosen but convenient point O. The lastequation specifies the distance d from point O to the line of action of R,and states that the moment of the resultant force about any point Oequals the sum of the moments of the original forces of the system aboutthe same point. This extends Varignon’s theorem to the case of nonconcurrentforce systems; we call this extension the principle of moments.For a concurrent system of forces where the lines of action of allforces pass through a common point O, the moment sum ΣM O aboutthat point is zero. Thus, the line of action of the resultant R ΣF, determinedby the first of Eqs. 2/10, passes through point O. For a parallelforce system, select a coordinate axis in the direction of the forces.If the resultant force R for a given force system is zero, the resultantof the system need not be zepo because the resultant may be a couple.The three forces in Fig. 2/15, for instance, have a zero resultant forcebut have a resultant clockwise couple M F 3 d.F 1F 1 + F 2 = –F 3F 1F 2F 2dF 3Figure 2/15

c02.qxd 10/29/07 1:38 PM Page 6060 Chapter 2 Force SystemsSample Problem 2/9Determine the resultant of the four forces and one couple which act on theplate shown.60 N2 m50 N5 mySolution. Point O is selected as a convenient reference point for the force–couplesystem which is to represent the given system.[R x ΣF x ][R y ΣF y ][R R x2 R y 2 ] tan1 R yR x[M O Σ(Fd)]R x 40 80 cos 30 60 cos 45 66.9 NR y 50 80 sin 30 60 cos 45 132.4 NR (66.9) 2 (132.4) 2 148.3 N tan1 132.466.9 63.2M O 140 50(5) 60 cos 45(4) 60 sin 45(7) 237 N mAns.Ans.The force–couple system consisting of R and M O is shown in Fig. a.We now determine the final line of action of R such that R alone representsthe original system.[Rd M O ]148.3d 237 d 1.600 mAns.Hence, the resultant R may be applied at any point on the line which makes a63.2 angle with the x-axis and is tangent at point A to a circle of 1.600-m radiuswith center O, as shown in part b of the figure. We apply the equation Rd M O inan absolute-value sense (ignoring any sign of M O ) and let the physics of the situation,as depicted in Fig. a, dictate the final placement of R. Had M O been counterclockwise,the correct line of action of R would have been the tangent at point B.The resultant R may also be located by determining its intercept distance bto point C on the x-axis, Fig. c. With R x and R y acting through point C, only R yexerts a moment about O so thatR y b M O and b 237132.4 1.792 mAlternatively, the y-intercept could have been obtained by noting that the momentabout O would be due to R x only.A more formal approach in determining the final line of action of R is to usethe vector expressionr R M Owhere r xi yj is a position vector running from point O to any point on theline of action of R. Substituting the vector expressions for r, R, and M O and carryingout the cross product result in(xi yj) (66.9i 132.4j) 237k(132.4x 66.9y)k 237kThus, the desired line of action, Fig. c, is given by132.4x 66.9y 237By setting y 0, we obtain x 1.792 m, which agrees with our earlier calculationof the distance b.45°140 N·m2 m80 N2 m40 N O 30° x1 m(a)(b)(c)|M O | =237 N·mOR = 148.3 N1.600 mA132.4x – 66.9y =–237RCOObyyR = 148.3 Nθ = 63.2°x63.2°Helpful Hints We note that the choice of point O asa moment center eliminates any momentsdue to the two forces whichpass through O. Had the clockwisesign convention been adopted, M Owould have been 237 N m, withthe plus sign indicating a sensewhich agrees with the sign convention.Either sign convention, ofcourse, leads to the conclusion of aclockwise moment M O . Note that the vector approachyields sign information automatically,whereas the scalar approachis more physically oriented. Youshould master both methods.Bxx

c02.qxd 10/29/07 1:38 PM Page 61Article 2/6 Problems 61PROBLEMSIntroductory Problems2/77 Calculate the magnitude of the tension T and theangle for which the eye bolt will be under a resultantdownward force of 15 kN.Ans. T 12.85 kN, 38.92/79 Determine the equivalent force–couple system at theorigin O for each of the three cases of forces beingapplied to the edge of a circular disk. If the resultantcan be so expressed, replace this force–couple systemwith a stand-alone force.Ans. (a) R 2Fj along x r(b) R Fi along y 3r(c) R Fi along y rFyFyFFFyxxr Or OrOxxTθ30°6 kNF(a)FFF(b)Problem 2/79(c)FProblem 2/772/78 Determine the resultant R of the four forces actingon the gusset plate. Also find the magnitude of Rand the angle x which the resultant makes with thex-axis.30 kN20°50 kN20°y8 kN60 kN2/80 Determine the height h above the base B at whichthe resultant of the three forces acts.600 mm650 N600 mm600 mmProblem 2/80300 N250 N2/81 Where does the resultant of the two forces act?Ans. 10.70 m to the left of AB40°680 Nx300mm500mmA40 kNProblem 2/78660 NProblem 2/81

c02.qxd 10/29/07 1:38 PM Page 6262 Chapter 2 Force Systems2/82 Under nonuniform and slippery road conditions, thetwo forces shown are exerted on the two rear-drivewheels of the pickup truck, which has a limited-sliprear differential. Determine the y-intercept of the resultantof this force system.y2/84 Determine the magnitude of the force F applied tothe handle which will make the resultant of thethree forces pass through O.F250 N750 mmGx300 mm750 mm500 NOProblem 2/822/83 If the resultant of the two forces and couple Mpasses through point O, determine M.Ans. M 148.0 N m CCWMO1675 mm150mm1125 mm150mm30°60°320 NProblem 2/842/85 Determine and locate the resultant R of the twoforces and one couple acting on the I-beam.Ans. R 4 kN down at x 5 mx150 100mm120 Nmm240 N52 kN·m3 m 3 m 3 m10 kN160mm400 NProblem 2/836 kNProblem 2/85

c02.qxd 10/29/07 1:38 PM Page 63Article 2/6 Problems 632/86 A commercial airliner with four jet engines, eachproducing 90 kN of forward thrust, is in a steady,level cruise when engine number 3 suddenly fails.Determine and locate the resultant of the three remainingengine thrust vectors. Treat this as a twodimensionalproblem.90 kN349 m12 mRepresentative Problems2/88 The directions of the two thrust vectors of an experimentalaircraft can be independently changed fromthe conventional forward direction within limits. Forthe thrust configuration shown, determine theequivalent force–couple system at point O. Then replacethis force–couple system by a single force andspecify the point on the x-axis through which theline of action of this resultant passes. These resultsare vital to assessing design performance.10 my90 kN90 kN2112 m9 mT3 m3 m15°TABOxProblem 2/862/87 Replace the three forces acting on the bent pipe by asingle equivalent force R. Specify the distance xfrom point O to the point on the x-axis throughwhich the line of action of R passes.Ans. R 200i 80j N, x 1.625 m (off pipe)yO160 N200 N250 mmxProblem 2/882/89 Determine the resultant R of the three forces actingon the simple truss. Specify the points on the x- andy-axes through which R must pass.Ans. R 15i 47.3j kNx 7.42 m, y 23.4 my25 kN5 m20 kN30°250 mm250 mm240 N125 mmO3 m 6 m 3 mxProblem 2/8730 kNProblem 2/89

c02.qxd 10/29/07 1:38 PM Page 6464 Chapter 2 Force Systems2/90 The gear and attached V-belt pulley are turningcounterclockwise and are subjected to the tooth loadof 1600 N and the 800-N and 450-N tensions in theV-belt. Represent the action of these three forces bya resultant force R at O and a couple of magnitudeM. Is the unit slowing down or speeding up?Problem 2/902/91 The design specifications for the attachment at A forthis beam depend on the magnitude and location ofthe applied loads. Represent the resultant of thethree forces and couple by a single force R at A and acouple M. Specify the magnitude of R.Ans. R 1.879i 1.684j kN, R 2.52 kNM 14.85 kN m CWy450 N800 N1.8 mx15°15°4 kN280mm 150mm1.2 mOy20°20°30°1600 NProblem 2/922/93 Two integral pulleys are subjected to the belt tensionsshown. If the resultant R of these forces passesthrough the center O, determine T and the magnitudeof R and the counterclockwise angle it makeswith the x-axis.Ans. T 60 N, R 193.7 N, 34.6T80 N120 N30°200 mm160 N200 mm30°100mmOyOθP250 mm20°200 N200 mmx150 N1.5 mA0.8 m3 kNxProblem 2/918 kN·m2 kN2/92 In the equilibrium position shown, the resultant ofthe three forces acting on the bell crank passesthrough the bearing O. Determine the vertical forceP. Does the result depend on ?Problem 2/932/94 While sliding a desk toward the doorway, three studentsexert the forces shown in the overhead view.Determine the equivalent force–couple system atpoint A. Then determine the equation of the line ofaction of the resultant force.100 Ny1.5 m60 N80 NA0.75 mxProblem 2/94

c02.qxd 10/29/07 1:38 PM Page 65Article 2/6 Problems 652/95 Under nonuniform and slippery road conditions, thefour forces shown are exerted on the four drivewheels of the all-wheel-drive vehicle. Determine theresultant of this system and the x- and y-interceptsof its line of action. Note that the front and reartracks are equal (i.e., AB CD).Ans. R 903i 175j N, (0, 0.253) m, (1.308, 0) m300 N300 NProblem 2/952/96 The rolling rear wheel of a front-wheel-drive automobilewhich is accelerating to the right is subjectedto the five forces and one moment shown. The forcesA x 240 N and A y 2000 N are forces transmittedfrom the axle to the wheel, F 160 N is the frictionforce exerted by the road surface on the tire, N 2400 N is the normal reaction force exerted by theroad surface, and W 400 N is the weight of thewheel/tire unit. The couple M 3 N m is the bearingfriction moment. Determine and locate the resultantof the system.yBAxMA200 N1520 mm 1120 mmyA yNCDA x30°750 mm760 mm760 mm2/97 A rear-wheel-drive car is stuck in the snow betweenother parked cars as shown. In an attempt to freethe car, three students exert forces on the car atpoints A, B, and C while the driver’s actions result ina forward thrust of 200 N acting parallel to the planeof rotation of each rear wheel. Treating the problemas two-dimensional, determine the equivalentforce–couple system at the car center of mass G andlocate the position x of the point on the car centerlinethrough which the resultant passes. Neglect allforces not shown.Ans. R 925i 567j NM G 690 N m CCWx 1.218 m200 N825mm400 N 825mm200 NProblem 2/972/98 An exhaust system for a pickup truck is shown inthe figure. The weights W h , W m , and W t of the headpipe,muffler, and tailpipe are 10, 100, and 50 N, respectively,and act at the indicated points. If theexhaust-pipe hanger at point A is adjusted so that itstension F A is 50 N, determine the required forces inthe hangers at points B, C, and D so that theforce–couple system at point O is zero. Why is a zeroforce–couple system at O desirable?F AAG250 N 30°F ByBF Cx30°C1650mm350 N1800mm30°F DWNFOABW h W mW tCDr = 0.350.5 0.2 0.65 0.650.5 0.4Problem 2/96Dimensions in metersProblem 2/98

c02.qxd 10/29/07 1:38 PM Page 66⎭66 Chapter 2 Force SystemsSECTION BTHREE-DIMENSIONAL FORCE SYSTEMSz2/7 RECTANGULAR C OMPONENTSMany problems in mechanics require analysis in three dimensions,and for such problems it is often necessary to resolve a force into itsthree mutually perpendicular components. The force F acting at point Oin Fig. 2/16 has the rectangular components F x , F y , F z , whereF z kFyF x F cos x F F2x F2y F2zF y F cos yF F x i F y j F z k(2/11)θ zθ yF z F cos z F F(i cos x j cos y k cos z )kiOjF x iθ xF y jxThe unit vectors i, j, and k are in the x-, y-, and z-directions, respectively.Using the direction cosines of F, which are l cos x , m cos y ,and n cos z , where l 2 m 2 n 2 1, we may write the force asFigure 2/16F F(li mj nk)(2/12)We may regard the right-side expression of Eq. 2/12 as the forcemagnitude F times a unit vector n F which characterizes the direction ofF, orF Fn F(2/12a)zFA (x 1 , y 1 , z 1 )⎫⎪⎪⎪⎬⎪⎪⎪⎭B (x 2 , y 2 , z 2 )(z 2 – z 1 )It is clear from Eqs. 2/12 and 2/12a that n F li mj nk, whichshows that the scalar components of the unit vector n F are the directioncosines of the line of action of F.In solving three-dimensional problems, one must usually find the x,y, and z scalar components of a force. In most cases, the direction of aforce is described (a) by two points on the line of action of the force or(b) by two angles which orient the line of action.(a) Specification by two points on the line of action of the force.If the coordinates of points A and B of Fig. 2/17 are known, the force Fmay be written asx⎫⎪(y 2 – y 1 )⎪⎪⎬⎪⎪⎪Figure 2/17⎫ ⎬⎭(x 2 – x 1 )yF Fn F F ABlAB F (x 2 x 1 )i (y 2 y 1 )j (z 2 z 1 )k(x 2 x 1 ) 2 (y 2 y 1 ) 2 (z 2 z 1 ) 2Thus the x, y, and z scalar components of F are the scalar coefficients ofthe unit vectors i, j, and k, respectively.

c02.qxd 10/29/07 1:38 PM Page 67Article 2/7 Rectangular Components 67(b) Specification by two angles which orient the line of action ofthe force. Consider the geometry of Fig. 2/18. We assume that the angles and are known. First resolve F into horizontal and vertical components.zF zθF xy F cos F z F sin Then resolve the horizontal component F xy into x- and y-components.F x F xy cos F cos cos F y F xy sin F cos sin The quantities F x , F y , and F z are the desired scalar components of F.The choice of orientation of the coordinate system is arbitrary, withconvenience being the primary consideration. However, we must use aright-handed set of axes in our three-dimensional work to be consistentwith the right-hand-rule definition of the cross product. When we rotatefrom the x- to the y-axis through the 90 angle, the positive direction forthe z-axis in a right-handed system is that of the advancement of aright-handed screw rotated in the same sense. This is equivalent to theright-hand rule.xφFF yF xF xyyFigure 2/18Dot ProductWe can express the rectangular components of a force F (or anyother vector) with the aid of the vector operation known as the dot orscalar product (see item 6 in Art. C/7 of Appendix C). The dot product oftwo vectors P and Q, Fig. 2/19a, is defined as the product of their magnitudestimes the cosine of the angle between them. It is written asP Q PQ cos We can view this product either as the orthogonal projection P cos ofP in the direction of Q multiplied by Q, or as the orthogonal projectionQ cos of Q in the direction of P multiplied by P. In either case the dotproduct of the two vectors is a scalar quantity. Thus, for instance, wecan express the scalar component F x F cos x of the force F in Fig.2/16 as F x F i, where i is the unit vector in the x-direction.PFαQF n = F·nnF n = F·nn (unit vector)(a)(b)Figure 2/19

c02.qxd 10/29/07 1:38 PM Page 6868 Chapter 2 Force SystemsIn more general terms, if n is a unit vector in a specified direction,the projection of F in the n-direction, Fig. 2/19b, has the magnitudeF n F n. If we want to express the projection in the n-direction as avector quantity, then we multiply its scalar component, expressed byF n, by the unit vector n to give F n (F n)n. We may write this asF n F nn without ambiguity because the term nn is not defined, andso the complete expression cannot be misinterpreted as F (nn).If the direction cosines of n are , , and , then we may write n invector component form like any other vector asn i j kwhere in this case its magnitude is unity. If the direction cosines of Fwith respect to reference axes x-y-z are l, m, and n, then the projectionof F in the n-direction becomesF n F n F(li mj nk) (i j k) F(l m n)becauseandi i j j k k 1i j j i i k k i j k k j 0The latter two sets of equations are true because i, j, and k have unitlength and are mutually perpendicular.Angle between Two VectorsIf the angle between the force F and the direction specified by theunit vector n is , then from the dot-product definition we have F n Fn cos F cos , where n n 1. Thus, the angle between F and nis given by cos 1 F nF(2/13)PIn general, the angle between any two vectors P and Q is cos 1 P QPQ(2/13a)P′αQFigure 2/20If a force F is perpendicular to a line whose direction is specified by theunit vector n, then cos 0, and F n 0. Note that this relationshipdoes not mean that either F or n is zero, as would be the case withscalar multiplication where (A)(B) 0 requires that either A or B (orboth) be zero.The dot-product relationship applies to nonintersecting vectors aswell as to intersecting vectors. Thus, the dot product of the nonintersectingvectors P and Q in Fig. 2/20 is Q times the projection of P on Q,or PQ cos PQ cos because P and P are the same when treated asfree vectors.

c02.qxd 10/29/07 1:38 PM Page 69Article 2/7 Rectangular Components 69Sample Problem 2/10A force F with a magnitude of 100 N is applied at the origin O of the axesx-y-z as shown. The line of action of F passes through a point A whose coordinatesare 3 m, 4 m, and 5 m. Determine (a) the x, y, and z scalar components ofF, (b) the projection F xy of F on the x-y plane, and (c) the projection F OB of Falong the line OB.Solution. Part (a). We begin by writing the force vector F as its magnitudeF times a unit vector n OA .F Fn OA F OAlOA 100 3i 4j 5k 3 2 4 2 5 2 100[0.424i 0.566j 0.707k] 42.4i 56.6j 70.7k Nz5 mOz4 m3 mF = 100 NAy6 m6 mxFB2 mThe desired scalar components are thusF zF x 42.4 N F y 56.6 N F z 70.7 NAns.yPart (b).The cosine of the angle xy between F and the x-y plane isF y3 2 4 2cos xy 3 2 4 2 5 0.7072so that F xy F cos xy 100(0.707) 70.7 NAns.OF xθxyF xy = 70.7 NPart (c).The unit vector n OB along OB isxn OB OBlOB6i 6j 2k 0.688i 0.688j 0.229k6 2 6 2 2 2FThe scalar projection of F on OB isF OB F n OB (42.4i 56.6j 70.7k) (0.688i 0.688j 0.229k) (42.4)(0.688) (56.6)(0.688) (70.7)(0.229)z 84.4NAns.yn OBIf we wish to express the projection as a vector, we writeF OB = 84.4 NF OB F n OB n OB 84.4(0.688i 0.688j 0.229k) 58.1i 58.1j 19.35k NOHelpful Hintsx In this example all scalar componentsare positive. Be prepared for the casewhere a direction cosine, and hencethe scalar component, are negative. The dot product automatically findsthe projection or scalar componentof F along line OB as shown.

c02.qxd 10/29/07 1:38 PM Page 7070 Chapter 2 Force SystemsPROBLEMSz, mmIntroductory ProblemsB (–25, 50, 40)2/99 Express F as a vector in terms of the unit vectorsi, j, and k. Determine the angle between F and they-axis.Ans. F 18.86i 23.6j 51.9k N y 113.1zx, mmOA (15, –20, –25)F = 750 Ny, mm110F = 60 N40O50 yxDimensions in millimetersProblem 2/99Problem 2/1012/102 Express F as a vector in terms of the unit vectors i,j, and k. Determine the projection, both as a scalarand as a vector, of F onto line OA, which lies in thex-y plane.z40°F = 5 kN2/100 The 70-m microwave transmission tower is steadiedby three guy cables as shown. Cable AB carriesa tension of 12 kN. Express the corresponding forceon point B as a vector.zxO30° 35°AyAProblem 2/10235 mB25 mx70 m10 mO20 m30 mCD20 m40 my2/103 The force F has a magnitude of 900 N and acts alongthe diagonal of the parallelepiped as shown. Express Fin terms of its magnitude times the appropriate unitvector and determine its x-, y-, and z-components.Ans. F 900 [ 1 3 i 2 3 j 2 k] 3 NF x = 300 N, F y = 600 N, F z = 600 NzProblem 2/1002/101 Express the force F as a vector in terms of theunit vectors i, j, and k. Determine the angles x , y , and z which F makes with the positive x-, y-,and z-axes.Ans. F 290i 507j 471k N x 112.7, y 47.5, z 51.14 m2 mF = 900 N4 myxProblem 2/103

c02.qxd 10/29/07 1:38 PM Page 71Article 2/7 Problems 712/104 The turnbuckle is tightened until the tension inthe cable AB equals 2.4 kN. Determine the vectorexpression for the tension T as a force acting onmember AD. Also find the magnitude of the projectionof T along the line AC.z2/106 Use the result cited for Prob. 2/105 and determinethe magnitude T GF of the projection of Tonto line GF.Representative ProblemsDA2/107 The tension in the supporting cable AB is 10 kN.Write the force which the cable exerts on the boomBC as a vector T. Determine the angles x , y , and z which the line of action of T forms with the positivex-, y-, and z-axes.Ans. T 10[0.406i 0.761j 0.507k] kN x 66.1, y 139.5, z 59.55 mzxProblem 2/1042/105 The rigid pole and cross-arm assembly is supportedby the three cables shown. A turnbuckle at D istightened until it induces a tension T in CD of 1.2kN. Express T as a vector. Does it make any differencein the result which coordinate system is used?Ans. T 0.321i 0.641j 0.962k kN, NoFxx′1 mC2 m3 mG1 mBAzOOB1.5 m1.5 m3 m3 mz′1.5 m2 mCE1 my′T = 1.2 kNDyyProblem 2/1072/108 The cable BC carries a tension of 750 N. Write thistension as a force T acting on point B in terms ofthe unit vectors i, j, and k. The elbow at A forms aright angle.zxAC0.7 m5 m4 m1.2 mCOyT = 10 kN5 m 2.5 mProblem 2/1081.6 mB30°AB0.8 myxProblem 2/105

c02.qxd 10/29/07 1:38 PM Page 7272 Chapter 2 Force Systems2/109 Derive the expression for the projection F DC of theforce F onto the line directed from D to C.(b 2 a 2 )FAns. F DC a 2 b 2 a 2 b 2 c 2D2/112 The rectangular plate is supported by hinges alongits side BC and by the cable AE. If the cable tensionis 300 N, determine the projection onto line BC ofthe force exerted on the plate by the cable. Notethat E is the midpoint of the horizontal upper edgeof the structural support.FBCAbCProblem 2/109acE25°T = 300 N400 mm1200 mmD2/110 If M is the center of the face ABCD of the rectangularsolid, express F as a vector in terms of the unitvectors i, j, and k. Determine the scalar projectionof F along line AE. The quantities F, a, b, and c areknown.EzBBAProblem 2/1122/113 The access door is held in the 30 open position bythe chain AB. If the tension in the chain is 100 N,determine the projection of the tension force ontothe diagonal axis CD of the door.Ans. T CD 46.0 NcOAMFBxbProblem 2/1102/111 Determine the angle between the 200-N force andline OC.Ans. 54.9DaCyC1200 mm30°A900mm900 mmDzOF = 200 NBProblem 2/11380 mmxA240 mmC120 mmyProblem 2/111

c02.qxd 10/29/07 1:38 PM Page 73Article 2/7 Problems 732/114 The spring of constant k 2.6 kN/m is attached tothe disk at point A and to the end fitting at point Bas shown. The spring is unstretched when A and B are both zero. If the disk is rotated 15 clockwiseand the end fitting is rotated 30 counterclockwise,determine a vector expression for the force F whichthe spring exerts at point A. The magnitude of thespring force is the constant k multiplied by the deflection(lengthening or shortening) of the spring.Ans. F 3.12i 41.1j 7.52k N2/116 A force F is applied to the surface of the sphere asshown. The angles and locate point P, and pointM is the midpoint of ON. Express F in vector form,using the given x-, y-, and z-coordinates.Ans.F F (2 sin 1)(i cos j sin ) k(2 cos )5 4 sin zFzxOA250 mmOφRPθB = 30°A = 15°θMyy900 mm k = 2.6 kN/m200 mmNθBxProblem 2/114Problem 2/1162/115 Determine the x-, y-, and z-components of force Fwhich acts on the tetrahedron as shown. The quantitiesa, b, c, and F are known, and M is the midpointof edge AB.2acFAns. F x a 2 b 2 a 2 b 2 4c 2z2bcFF y a 2 b 2 a 2 b 2 4c 2F z F a 2 b 2a 2 b 2 4c 2CcObBFyaMAxProblem 2/115

c02.qxd 10/29/07 1:38 PM Page 7474 Chapter 2 Force Systems2/8 MOMENT AND C OUPLEIn two-dimensional analyses it is often convenient to determine amoment magnitude by scalar multiplication using the moment-armrule. In three dimensions, however, the determination of the perpendiculardistance between a point or line and the line of action of the forcecan be a tedious computation. A vector approach with cross-productmultiplication then becomes advantageous.Moments in Three DimensionsConsider a force F with a given line of action acting on a body, Fig.2/21a, and any point O not on this line. Point O and the line of F establisha plane A. The moment M O of F about an axis through O normal tothe plane has the magnitude M O Fd, where d is the perpendicular distancefrom O to the line of F. This moment is also referred to as the momentof F about the point O.The vector M O is normal to the plane and is directed along the axisthrough O. We can describe both the magnitude and the direction of M Oby the vector cross-product relation introduced in Art. 2/4. (Refer toitem 7 in Art. C/7 of Appendix C.) The vector r runs from O to any pointon the line of action of F. As described in Art. 2/4, the cross product of rand F is written r F and has the magnitude (r sin )F, which is thesame as Fd, the magnitude of M O .The correct direction and sense of the moment are established bythe right-hand rule, described previously in Arts. 2/4 and 2/5. Thus,with r and F treated as free vectors emanating from O, Fig. 2/21b, thethumb points in the direction of M O if the fingers of the right hand curlin the direction of rotation from r to F through the angle . Therefore,we may write the moment of F about the axis through O asAM OOF(a)M OdrFαAO(b)αrM O r F(2/14)The order r F of the vectors must be maintained because F rwould produce a vector with a sense opposite to that of M O ; that is,F r M O .Figure 2/21Evaluating the Cross ProductThe cross-product expression for M O may be written in the determinantformM O i r xF xjr yF ykzr zF(2/15)(Refer to item 7 in Art. C/7 of Appendix C if you are not already familiarwith the determinant representation of the cross product.) Note thesymmetry and order of the terms, and note that a right-handed coordinatesystem must be used. Expansion of the determinant givesM O (r y F z r z F y )i (r z F x r x F z )j (r x F y r y F x )k

c02.qxd 10/29/07 1:38 PM Page 75Article 2/8 Moment and Couple 75To gain more confidence in the cross-product relationship, examinethe three components of the moment of a force about a point asFF zobtained from Fig. 2/22. This figure shows the three components of aforce F acting at a point A located relative to O by the vector r. Thezscalar magnitudes of the moments of these forces about the positiveAF yx-, y-, and z-axes through O can be obtained from the moment-armM zrule, and arer y zF xrM x r y F z r z F y M y r z F x r x F z M z r x F y r y F xwhich agree with the respective terms in the determinant expansion forthe cross product r F.OMr yxr yM xxMoment about an Arbitrary AxisFigure 2/22We can now obtain an expression for the moment M of F about anyaxis through O, as shown in Fig. 2/23. If n is a unit vector in the-direction, then we can use the dot-product expression for the componentof a vector as described in Art. 2/7 to obtain M O n, the componentof M O in the direction of . This scalar is the magnitude of the momentM of F about .To obtain the vector expression for the moment M of F about ,multiply the magnitude by the directional unit vector n to obtainM OλnM λrFM (r F n)n(2/16)Owhere r F replaces M O . The expression r F n is known as a triplescalar product (see item 8 in Art. C/7, Appendix C). It need not be written(r F) n because a cross product cannot be formed by a vector anda scalar. Thus, the association r (F n) would have no meaning.The triple scalar product may be represented by the determinantFigure 2/23r xM M F xr yF yr z F z(2/17)where , , are the direction cosines of the unit vector n.Varignon’s Theorem in Three DimensionsIn Art. 2/4 we introduced Varignon’s theorem in two dimensions.The theorem is easily extended to three dimensions. Figure 2/24 shows asystem of concurrent forces F 1 , F 2 , F 3 ,.... The sum of the momentsabout O of these forces isOrF 3AF 2F 1r F 1 r F 2 r F 3 r (F 1 F 2 F 3 ) r ΣFFigure 2/24

c02.qxd 10/29/07 1:38 PM Page 7676 Chapter 2 Force Systemswhere we have used the distributive law for cross products. Using thesymbol M O to represent the sum of the moments on the left side of theabove equation, we haveM O Σ(r F) r R(2/18)This equation states that the sum of the moments of a system of concurrentforces about a given point equals the moment of their sum aboutthe same point. As mentioned in Art. 2/4, this principle has many applicationsin mechanics.Couples in Three DimensionsThe concept of the couple was introduced in Art. 2/5 and is easilyextended to three dimensions. Figure 2/25 shows two equal and oppositeforces F and F acting on a body. The vector r runs from any point Bon the line of action of F to any point A on the line of action of F.Points A and B are located by position vectors r A and r B from any pointO. The combined moment of the two forces about O isB–FrMAdFr Br AM r A F r B (F) (r A r B ) FOFigure 2/25However, r A r B r, so that all reference to the moment center O disappears,and the moment of the couple becomesM r F(2/19)Thus, the moment of a couple is the same about all points. The magnitudeof M is M Fd, where d is the perpendicular distance between thelines of action of the two forces, as described in Art. 2/5.The moment of a couple is a free vector, whereas the moment of aforce about a point (which is also the moment about a defined axisthrough the point) is a sliding vector whose direction is along the axisthrough the point. As in the case of two dimensions, a couple tends toproduce a pure rotation of the body about an axis normal to the plane ofthe forces which constitute the couple.Couple vectors obey all of the rules which govern vector quantities.Thus, in Fig. 2/26 the couple vector M 1 due to F 1 and F 1 may be addedM 1MM 1M 2F 2–F 1–FF 1≡M 2F–F 2Figure 2/26

c02.qxd 10/29/07 1:38 PM Page 77Article 2/8 Moment and Couple 77as shown to the couple vector M 2 due to F 2 and F 2 to produce the coupleM, which, in turn, can be produced by F and F.In Art. 2/5 we learned how to replace a force by its equivalentforce–couple system. You should also be able to carry out this replacementin three dimensions. The procedure is represented in Fig. 2/27,where the force F acting on a rigid body at point A is replaced by anequal force at point B and the couple M r F. By adding the equaland opposite forces F and F at B, we obtain the couple composed ofF and the original F. Thus, we see that the couple vector is simply themoment of the original force about the point to which the force is beingmoved. We emphasize that r is a vector which runs from B to any pointon the line of action of the original force passing through A.M = r × FFFBA≡BFrA≡BF–FFigure 2/27

c02.qxd 10/29/07 1:38 PM Page 7878 Chapter 2 Force SystemsSample Problem 2/11zDetermine the moment of force F about point O (a) by inspection and (b) bythe formal cross-product definition M O r F.abFSolution. (a) Because F is parallel to the y-axis, F has no moment about thataxis. It should be clear that the moment arm from the x-axis to the line of actionof F is c and that the moment of F about the x-axis is negative. Similarly, themoment arm from the z-axis to the line of action of F is a and the moment of Fabout the z-axis is positive. So we havexcOyM O cFi aFk F(ci ak)Ans.z(b) Formally,abM O r F (ai ck) Fj aFk cFiFHelpful Hint F(ci ak)Ans.crOy Again we stress that r runs from the moment center to the line of action of F.Another permissible, but less convenient, position vector is r ai bj ck.xSample Problem 2/121.6 mzThe turnbuckle is tightened until the tension in cable AB is 2.4 kN. Determinethe moment about point O of the cable force acting on point A and themagnitude of this moment.A2 mSolution.We begin by writing the described force as a vector.T Tn AB 2.4 0.8i 1.5j 2k0.8 2 1.5 2 2 2 0.731i 1.371j 1.829k kNx1.5 mBO0.8 mzyThe moment of this force about point O is1.6 mM O r OA T (1.6i 2k) (0.731i 1.371j 1.829k)A 2.74i 4.39j 2.19k kN mAns.r OAThis vector has a magnitude2 mTM O 2.74 2 4.39 2 2.19 2 5.62 kN mAns.Helpful Hint The student should verify by inspection the signs of the moment components.x1.5 mBO0.8 my

c02.qxd 10/29/07 1:38 PM Page 79Article 2/8 Moment and Couple 79Sample Problem 2/13yA tension T of magnitude 10 kN is applied to the cable attached to the top Aof the rigid mast and secured to the ground at B. Determine the moment M z of Tabout the z-axis passing through the base O.ASolution (a). The required moment may be obtained by finding the componentalong the z-axis of the moment M O of T about point O. The vector M O isnormal to the plane defined by T and point O, as shown in the accompanying figure.In the use of Eq. 2/14 to find M O , the vector r is any vector from point O tothe line of action of T. The simplest choice is the vector from O to A, which iswritten as r 15j m. The vector expression for T isFrom Eq. 2/14,[M O r F]T Tn AB 10 12i 15j 9k(12) 2 (15) 2 (9) 2 10(0.566i 0.707j 0.424k) kNM O 15j 10(0.566i 0.707j 0.424k) 150(0.566k 0.424i) kN mThe value M z of the desired moment is the scalar component of M O in thez-direction or M z M O k. Therefore,M z 150(0.566k 0.424i) k 84.9 kN mAns.The minus sign indicates that the vector M z is in the negative z-direction. Expressedas a vector, the moment is M z 84.9k kN m.z15 mO12 mHelpful HintsT = 10 kNB9 m We could also use the vector from Oto B for r and obtain the same result,but using vector OA is simpler. It is always helpful to accompany yourvector operations with a sketch of thevectors so as to retain a clear pictureof the geometry of the problem. Sketch the x-y view of the problemand show d.yxSolution (b). The force of magnitude T is resolved into components T z and T xyin the x-y plane. Since T z is parallel to the z-axis, it can exert no moment about this axis. The moment M z is, then, due only to T xy and is M z T xy d, where d isthe perpendicular distance from T xy to O. The cosine of the angle between T andT xy is 15 2 12 2 / 15 2 12 2 9 2 0.906, and therefore,T xy 10(0.906) 9.06 kNThe moment arm d equals OA multiplied by the sine of the angle between T xyand OA, orzArOTM zBxM od 151212 2 15 2 9.37 mHence, the moment of T about the z-axis has the magnitudeyAT xM z 9.06(9.37) 84.9 kN mAns.and is clockwise when viewed in the x-y plane.T15 mSolution (c). The component T xy is further resolved into its components T x and T y .It is clear that T y exerts no moment about the z-axis since it passes through it, sothat the required moment is due to T x alone. The direction cosine of T with respectto the x-axis is 12/9 2 12 2 15 2 0.566 so that T x 10(0.566) 5.66 kN. Thus,M z 5.66(15) 84.9 kN mAns.zT zT yT xyxO12 mB9 m

c02.qxd 10/29/07 1:38 PM Page 8080 Chapter 2 Force SystemsSample Problem 2/14Determine the magnitude and direction of the couple M which will replacethe two given couples and still produce the same external effect on the block.Specify the two forces F and F, applied in the two faces of the block parallel tothe y-z plane, which may replace the four given forces. The 30-N forces act parallelto the y-z plane.x 100 mm50 mm30 N60°60 mm30 N60°40 mmy25 NSolution. The couple due to the 30-N forces has the magnitude M 1 30(0.06) 1.80 N m. The direction of M 1 is normal to the plane defined by the two forces,and the sense, shown in the figure, is established by the right-hand convention.The couple due to the 25-N forces has the magnitude M 2 25(0.10) 2.50 N mwith the direction and sense shown in the same figure. The two couple vectorscombine to give the components25 NM 2 = 2.5 N·mzThus,withAns.Ans.The forces F and F lie in a plane normal to the couple M, and their momentarm as seen from the right-hand figure is 100 mm. Thus, each force has themagnitude[M = Fd] F 2.230.10 22.3 NAns.and the direction 44.3.M y 1.80 sin 60 1.559 N mM z 2.50 1.80 cos 60 1.600 N mM (1.559) 2 (1.600) 2 2.23 N m tan1 1.5591.600 tan1 0.974 44.3zθ60°MyxM 1 = 1.8 N·mHelpful Hintθ–F Bear in mind that the couple vectorsare free vectors and therefore haveno unique lines of action.zθFySample Problem 2/15zA force of 400 N is applied at A to the handle of the control lever which is attachedto the fixed shaft OB. In determining the effect of the force on the shaftat a cross section such as that at O, we may replace the force by an equivalentforce at O and a couple. Describe this couple as a vector M.Solution. The couple may be expressed in vector notation as M r F,where r OA l 0.2j 0.125k m and F 400i N. Thus,M (0.2j 0.125k) (400i) 50j 80k N m75 mmOxBz200 mmyA400 N50 mmAlternatively we see that moving the 400-N force through a distance d 0.125 2 0.2 2 0.236 m to a parallel position through O requires the additionof a couple M whose magnitude isM Fd 400(0.236) 94.3 N mAns.The couple vector is perpendicular to the plane in which the force is shifted, andits sense is that of the moment of the given force about O. The direction of M inthe y-z plane is given byxMθO400 Ndθ200 mm(400 N)A125 mmy1 125 tan200 32.0Ans.

c02.qxd 10/29/07 1:38 PM Page 81Article 2/8 Problems 81PROBLEMSIntroductory Problems2/117 Determine the moment of force F about point O.Ans. M O F(cj bk)z2/119 Determine the moment about O of the force ofmagnitude F for the case (a) when the force F isapplied at A and for the case (b) when F is appliedat B.Ans. (a) M O FLi(b) M O F(Li Dk)ycaObFF(b)BF(a)ALOxyDzxProblem 2/1172/118 Determine the moment of force F about point A.xProblem 2/1192/120 In opening a door which is equipped with a heavydutyreturn mechanism, a person exerts a force Pof magnitude 32 N as shown. Force P and the normaln to the face of the door lie in a vertical plane.Compute the moment of P about the z-axis.zAaFbz900 mmPy30°Problem 2/118ny20°xProblem 2/120

c02.qxd 10/29/07 1:38 PM Page 8282 Chapter 2 Force Systems2/121 The two forces acting on the handles of the pipewrenches constitute a couple M. Express the coupleas a vector.Ans. M 75i 22.5j N m250 mm150 mm250 mm150 Ny2/123 A helicopter is shown here with certain threedimensionalgeometry given. During a ground test,a 400-N aerodynamic force is applied to the tailrotor at P as shown. Determine the moment of thisforce about point O of the airframe.Ans. M O 480i 2400k N mz6 mx0.8 mxO150 NProblem 2/121y1.2 m400 NP2/122 The bent bar has a mass per unit of length. Determinethe moment of the weight of the bar aboutpoint O.zProblem 2/1232/124 The 24-N force is applied at point A of the crank assembly.Determine the moment of this force aboutpoint O.dzF = 24 NA30°hO18 mmyOyx36 mm65°Problem 2/124xProblem 2/122

c02.qxd 10/29/07 1:38 PM Page 83Article 2/8 Problems 832/125 The right-angle pipe OAB of Prob. 2/108 is shownagain here. Replace the 750-N tensile force whichthe cable exerts on point B by a force–couple systemat point O.Ans. R 598i 411j 189.5k NM O 361i 718j 419k N m2/127 If the magnitude of the moment of F about line CDis 50 N m, determine the magnitude of F.Ans. F 228 NByFzC0.7 m1.2 mO1.6 m30°A0.8 mx0.2 mC0.2 mA0.2 m 0.4 mProblem 2/127Representative ProblemsDProblem 2/1252/126 Determine the moment associated with the pair of400-N forces applied to the T-shaped structure.B2/128 A mechanic applies the horizontal 100-N force perpendicularto the wrench as indicated in the figure.Determine the moment of this force about the boltcenter O. The wrench centerline lies in the x-y plane.yyOx15°B400 NzO100 NA15°x0.45 mA0.25 m0.25 mzOA = 185 mmProblem 2/128400 N15°Problem 2/126

c02.qxd 10/29/07 1:38 PM Page 8484 Chapter 2 Force Systems2/129 Two 4-N thrusters on the nonrotating satellite aresimultaneously fired as shown. Compute the momentassociated with this couple and state aboutwhich satellite axes rotations will begin to occur.Ans. M 5i 4k N m625 mmx4 N500mmzG500mm4 N625 mmy2/131 A space shuttle orbiter is subjected to thrusts fromfive of the engines of its reaction control system.Four of the thrusts are shown in the figure; thefifth is an 850-N upward thrust at the right rear,symmetric to the 850-N thrust shown on the leftrear. Compute the moment of these forces aboutpoint G and show that the forces have the samemoment about all points.Ans. M 3400i 51 000j 51 000k N m1700 N1700 NyG3.2 m3.2 m1700 N850 N12 m2 mProblem 2/1292/130 Compute the moment M O of the 1.2-kN force aboutthe axis O-O.O300 mm60°40°1.2 kNxProblem 2/1312/132 In picking up a load from position B, a cable tensionT of magnitude 24 kN is developed. Calculatethe moment which T produces about the base O ofthe construction crane.zz18 m18 m105 mm200 mm OAProblem 2/130T30 mxOBy5 m 6 mProblem 2/132

c02.qxd 10/29/07 1:38 PM Page 85Article 2/8 Problems 852/133 The specialty wrench shown is used for difficult-toaccessbolts which have a low torque specification.Determine the moment about O of the 40-N forceapplied at point A of the wrench handle. Discussany shortcomings of this wrench.Ans. M O 12i 9k N mAzy125 mmO100mm125mmz40 NAx300 mmx100 mm50 mmB50 mm60°400 NyProblem 2/1332/134 In order to decrease the undesirable moment aboutthe x-axis, a mechanic uses his left hand to supportthe wrench handle at point B. What upward force Fmust he exert in order that there be no momentabout the x-axis? What then is the net momentabout O?yProblem 2/1352/136 The specialty wrench shown in the figure is designedfor access to the hold-down bolt on certainautomobile distributors. For the configurationshown where the wrench lies in a vertical planeand a horizontal 200-N force is applied at A perpendicularto the handle, calculate the moment M O appliedto the bolt at O. For what value of thedistance d would the z-component of M O be zero?OA100mm125mmzB40 N FAx150 mm300 mm200 Ν20°d = 125 mmProblem 2/134200 mm2/135 Determine the moment of the 400-N force aboutpoint A by (a) using the vector cross-product relationand (b) resolving the force into its componentsand finding their respective moments.Ans. M A 9.64i 17.32j 10k N mxzO70 mmyProblem 2/136

c02.qxd 10/29/07 1:38 PM Page 8686 Chapter 2 Force Systems2/137 The moment M applied to the pulley wheel causesa tension T 80 N in the cable which is secured tothe wheel at D and to the ground at E. Determinethe moment about O of this 80-N force as appliedat C.Ans. M O 15.49i 4k N mAz200 mmDM2/139 Using the principles to be developed in Chapter 3on equilibrium, one can determine that the tensionin cable AB is 143.4 N. Determine the momentabout the x-axis of this tension force acting onpoint A. Compare your result with the moment ofthe weight W of the 15-kg uniform plate about thex-axis. What is the moment of the tension force actingat A about line OB?Ans. M x 31.1 N m, (M x ) W 31.1 N mM OB 0zCB100 mmB0.35 mO400 mm0.4 m20°yOG15 kgxEyx0.45 m0.7 mProblem 2/137AW2/138 An 320-N force is applied to the end of the wrenchas the sprocket wheel is attached to an enginecrankshaft. The force is perpendicular to the planecontaining points O, A, and B. Determine the momentof this force about point O.Problem 2/139F = 320 Ny20°x10°B400 mmO100mmAzProblem 2/138

c02.qxd 10/29/07 1:38 PM Page 87Article 2/8 Problems 872/140 The rigid pole and cross-arm assembly of Prob.2/105 is shown again here. Determine the vectorexpression for the moment of the 1.2-kN tension(a) about point O and (b) about the pole z-axis. Findeach moment in two different ways.Fxx′1 m2 mGB1 mAzO1.5 m1.5 m3 mz′1.5 m3 mCEy′T = 1.2 kNDy2/142 Determine the vector expression for the momentM O of the 600-N force about point O. The designspecification for the bolt at O would require thisresult.50 mmz150 mmy60° A130 mm600 NO45°Problem 2/142140 mmxProblem 2/1402/141 A 5-N vertical force is applied to the knob of thewindow-opener mechanism when the crank BC ishorizontal. Determine the moment of the forceabout point A and about line AB.Ans. M A 375i 325j N mmM AB 281i 162.4k N mm25mmz50 mm75 mmyC5 NDAB30°xProblem 2/141

c02.qxd 10/29/07 1:38 PM Page 8888 Chapter 2 Force Systems2/9 RESULTANTS© Raymond Forbes/Age Fotostock America, Inc.In Art. 2/6 we defined the resultant as the simplest force combinationwhich can replace a given system of forces without altering the externaleffect on the rigid body on which the forces act. We found themagnitude and direction of the resultant force for the two-dimensionalforce system by a vector summation of forces, Eq. 2/9, and we located theline of action of the resultant force by applying the principle of moments,Eq. 2/10. These same principles can be extended to three dimensions.In the previous article we showed that a force could be moved to aparallel position by adding a corresponding couple. Thus, for the systemof forces F 1 , F 2 , F 3 . . . acting on a rigid body in Fig. 2/28a, we may moveeach of them in turn to the arbitrary point O, provided we also introducea couple for each force transferred. Thus, for example, we maymove force F 1 to O, provided we introduce the couple M 1 r 1 F 1 ,where r 1 is a vector from O to any point on the line of action of F 1 .When all forces are shifted to O in this manner, we have a system ofconcurrent forces at O and a system of couple vectors, as represented inpart b of the figure. The concurrent forces may then be added vectoriallyto produce a resultant force R, and the couples may also be added toproduce a resultant couple M, Fig. 2/28c. The general force system,then, is reduced toThe cables of a suspension bridgeexert a three-dimensional system ofconcentrated forces on this bridgetower.R F 1 F 2 F 3 ΣFM M 1 M 2 M 3 Σ(r F)(2/20)The couple vectors are shown through point O, but because they arefree vectors, they may be represented in any parallel positions. Themagnitudes of the resultants and their components areR x ΣF x R y ΣF y R z ΣF zR (ΣF x ) 2 (ΣF y ) 2 (ΣF z ) 2M x Σ(r F) x M y Σ(r F) y M z Σ(r F) zM M x2 M y2 M z2(2/21)M2M 3F 1F 2F 1M 1F 2MRO≡O≡OF 3F 3(a)(b)Figure 2/28(c)

c02.qxd 10/29/07 1:38 PM Page 89Article 2/9 Resultants 89The point O selected as the point of concurrency for the forces is arbitrary,and the magnitude and direction of M depend on the particularpoint O selected. The magnitude and direction of R, however, are thesame no matter which point is selected.In general, any system of forces may be replaced by its resultantforce R and the resultant couple M. In dynamics we usually select themass center as the reference point. The change in the linear motion ofthe body is determined by the resultant force, and the change in the angularmotion of the body is determined by the resultant couple. In statics,the body is in complete equilibrium when the resultant force R iszero and the resultant couple M is also zero. Thus, the determination ofresultants is essential in both statics and dynamics.We now examine the resultants for several special force systems.Concurrent Forces. When forces are concurrent at a point, only thefirst of Eqs. 2/20 needs to be used because there are no moments aboutthe point of concurrency.Parallel Forces. For a system of parallel forces not all in the sameplane, the magnitude of the parallel resultant force R is simply the magnitudeof the algebraic sum of the given forces. The position of its line ofaction is obtained from the principle of moments by requiring thatr R M O . Here r is a position vector extending from the force–couplereference point O to the final line of action of R, and M O is the sum ofthe moments of the individual forces about O. See Sample Problem 2/17for an example of parallel-force systems.Coplanar Forces. Article 2/6 was devoted to this force system.Wrench Resultant. When the resultant couple vector M is parallelto the resultant force R, as shown in Fig. 2/29, the resultant is called awrench. By definition a wrench is positive if the couple and force vectorspoint in the same direction and negative if they point in opposite directions.A common example of a positive wrench is found with the applicationof a screwdriver, to drive a right-handed screw. Any general forcesystem may be represented by a wrench applied along a unique line ofaction. This reduction is illustrated in Fig. 2/30, where part a of the figurerepresents, for the general force system, the resultant force R actingat some point O and the corresponding resultant couple M. Although Mis a free vector, for convenience we represent it as acting through O.In part b of the figure, M is resolved into components M 1 along the directionof R and M 2 normal to R. In part c of the figure, the couple M 2 isreplaced by its equivalent of two forces R and R separated by a distanceMRMRPositive wrenchNegative wrenchFigure 2/29

c02.qxd 10/29/07 1:38 PM Page 9090 Chapter 2 Force SystemsMMM 2M 1 RROO(a)(b)OdM 1RROR–RM 1(c)(d)Figure 2/30d M 2 /R with R applied at O to cancel the original R. This step leavesthe resultant R, which acts along a new and unique line of action, and theparallel couple M 1 , which is a free vector, as shown in part d of the figure.Thus, the resultants of the original general force system have been transformedinto a wrench (positive in this illustration) with its unique axis definedby the new position of R.We see from Fig. 2/30 that the axis of the wrench resultant lies in aplane through O normal to the plane defined by R and M. The wrench isthe simplest form in which the resultant of a general force system maybe expressed. This form of the resultant, however, has limited application,because it is usually more convenient to use as the reference pointsome point O such as the mass center of the body or another convenientorigin of coordinates not on the wrench axis.

c02.qxd 10/29/07 1:38 PM Page 91Article 2/9 Resultants 91Sample Problem 2/16zDetermine the resultant of the force and couple system which acts on therectangular solid.50 N70 N . m100 N . mSolution. We choose point O as a convenient reference point for the initialstep of reducing the given forces to a force–couple system. The resultant force isR ΣF (80 80)i (100 100)j (50 50)k 0 NThe sum of the moments about O is80 N80 NO50 N96N . m1.2 mM O [50(1.6) 70]i [80(1.2) 96]j [100(1) 100]k 10i N mx1.6 m1 m 100 NHence, the resultant consists of a couple, which of course may be applied at anypoint on the body or the body extended.100 NyHelpful Hints Since the force summation is zero, we conclude that the resultant, if it exists,must be a couple. The moments associated with the force pairs are easily obtained by using theM Fd rule and assigning the unit-vector direction by inspection. In manythree-dimensional problems, this may be simpler than the M rFapproach.Sample Problem 2/1750 NDetermine the resultant of the system of parallel forces which act on theplate. Solve with a vector approach.x0.5 mz0.5 mSolution.Transfer of all forces to point O results in the force–couple systemO0.35 mR ΣF (200 500 300 50)j 350j N500 N0.35 mM O [50(0.35) 300(0.35)]i [50(0.50) 200(0.50)]k 87.5i 125k N mThe placement of R so that it alone represents the above force–couple system isdetermined by the principle of moments in vector formr R M O(xi yj zk) 350j 87.5i 125k350xk 350zi 87.5i 125kFrom the one vector equation we may obtain the two scalar equations350x 125 and 350z 87.5y300 NxRRy200 Nzxzr OM OHence, x 0.357 m and z 0.250 m are the coordinates through which theline of action of R must pass. The value of y may, of course, be any value, aspermitted by the principle of transmissibility. Thus, as expected, the variable ydrops out of the above vector analysis.Helpful Hint You should also carry out a scalarsolution to this problem.

c02.qxd 10/29/07 1:38 PM Page 9292 Chapter 2 Force SystemsSample Problem 2/18Replace the two forces and the negative wrench by a single force R appliedat A and the corresponding couple M.Solution.The resultant force has the components[R x ΣF x ] R x 500 sin 40 700 sin 60 928 N[R y ΣF y ] R y 600 500 cos 40 cos 45 871 N[R z ΣF z ] R z 700 cos 60 500 cos 40 sin 45 621 N100mmx30mmA50 mm60 mm120 mm700 N60°500 N25 N·m40°z B80 mm45°40 mm 600 NyThus,andR 928i 871j 621k NR (928) 2 (871) 2 (621) 2 1416 NAns.MRThe couple to be added as a result of moving the 500-N force is[M r F] M 500 (0.08i 0.12j 0.05k) 500(i sin 40 j cos 40 cos 45 k cos 40 sin 45)where r is the vector from A to B.AThe term-by-term, or determinant, expansion givesM 500 18.95i 5.59j 16.90k N mThe moment of the 600-N force about A is written by inspection of its x- and z-components, which givesM 600 (600)(0.060)i (600)(0.040)k 36.0i 24.0k N mThe moment of the 700-N force about A is easily obtained from the moments ofthe x- and z-components of the force. The result becomesM 700 (700 cos 60)(0.030)i [(700 sin 60)(0.060) (700 cos 60)(0.100)]j (700 sin 60)(0.030)k 10.5i 71.4j 18.19k N mAlso, the couple of the given wrench may be writtenM 25.0(i sin 40 j cos 40 cos 45 k cos 40 sin 45) 16.07i 13.54j 13.54k N mTherefore, the resultant couple on adding together the i-, j-, and k-terms of thefour M’s isM 49.4i 90.5j 24.6k N mHelpful Hints Suggestion: Check the cross-productresults by evaluating the momentsabout A of the components of the500-N force directly from the sketch. For the 600-N and 700-N forces it iseasier to obtain the components oftheir moments about the coordinatedirections through A by inspectionof the figure than it is to set up thecross-product relations. The 25-N m couple vector of thewrench points in the direction oppositeto that of the 500-N force, andwe must resolve it into its x-, y-, andz-components to be added to theother couple-vector components. Although the resultant couple vectorM in the sketch of the resultantsis shown through A, we recognizethat a couple vector is a free vectorand therefore has no specified lineof action.andM (49.4) 2 (90.5) 2 (24.6) 2 106.0 N mAns.

c02.qxd 10/29/07 1:38 PM Page 93Article 2/9 Resultants 93Sample Problem 2/19Determine the wrench resultant of the three forces acting on the bracket.Calculate the coordinates of the point P in the x-y plane through which the resultantforce of the wrench acts. Also find the magnitude of the couple M of thewrench.Solution. The direction cosines of the couple M of the wrench must be the same as those of the resultant force R, assuming that the wrench is positive. Theresultant force is40 N20 Nz80 mm60 mmy100 mmxPy40 NxR 20i 40j 40k Nand its direction cosines areR (20) 2 (40) 2 (40) 2 60 Nz80 mmcos x 20/60 1/3 cos y 40/60 2/3 cos z 40/60 2/360 mmyRThe moment of the wrench couple must equal the sum of the moments ofthe given forces about point P through which R passes. The moments about P ofthe three forces are(M) Rx 20yk N mm(M) Ry 40(60)i 40xk N mmx = 60 mmP100 mmMxy = 40 mm(M) Rz 40(80 y)i 40(100 x)j N mmand the total moment isM (800 40y)i (4000 40x)j (40x 20y)k N mmThe direction cosines of M areHelpful Hint We assume initially that the wrenchis positive. If M turns out to be negative,then the direction of the couplevector is opposite to that of theresultant force.cos x (800 40y)/Mcos y (4000 40x)/Mcos z (40x 20y)/Mwhere M is the magnitude of M. Equating the direction cosines of R and M gives800 40y M 34000 40x 2M 340x 20y 2M 3Solution of the three equations givesM 2400 N mm x 60 mm y 40 mmAns.We see that M turned out to be negative, which means that the couple vector ispointing in the direction opposite to R, which makes the wrench negative.

c02.qxd 10/29/07 1:38 PM Page 9494 Chapter 2 Force SystemsPROBLEMSIntroductory Problems2/143 Three forces act at point O. If it is known that they-component of the resultant R is 5 kN and thatthe z-component is 6 kN, determine F 3 , , and R.Ans. F 3 10.82 kN, 33.7, R 10.49 kN2/145 The thin rectangular plate is subjected to the fourforces shown. Determine the equivalent force–couple system at O. What is the resultant of thesystem?Ans.R 0, M O Fb 1 32 izF 3zFθOF 2 = 7 kNF 1 = 4 kN—2bF—2b Ob FzxyProblem 2/1432/144 Three equal forces are exerted on the equilateralplate as shown. Reduce the force system to anequivalent force–couple system at point O. Showthat R is perpendicular to M O .Fx30°F2bF30°Problem 2/1452/146 The spacecraft of Prob. 2/129 is repeated here. Theplan is to fire four 4-N thrusters as shown in orderto spin up the spacecraft about its z-axis, but thethruster at A fails. Determine the equivalentforce–couple system at G for the remaining threethrusters.zObFyxby4 N4 NProblem 2/144625 mm500mmG500mm625 mmyxA4 NProblem 2/146

c02.qxd 10/29/07 1:38 PM Page 95Article 2/9 Problems 952/147 An oil tanker moves away from its docked positionunder the action of reverse thrust from screw A,forward thrust from screw B, and side thrust fromthe bow thruster C. Determine the equivalentforce–couple system at the mass center G.Ans. R 8i kN, M G 48j 820k kN mz2/149 Determine the force–couple system at O which isequivalent to the two forces applied to the shaftAOB. Is R perpendicular to M O ?Ans. R 266j 1085k NM O 48.9j 114.5k N my40 my600 NA6 mCG8 kN7 m50 kN xA B50 kN5 m 5 mProblem 2/147z30°80 mmO160 mm45°Bx2/148 The pulley and gear are subjected to the loadsshown. For these forces, determine the equivalentforce–couple system at point O.800 NzyO220mm75mmx330mm10°1200 N100mm800 N200 NRepresentative ProblemsProblem 2/1492/150 The commercial airliner of Prob. 2/86 is redrawnhere with three-dimensional information supplied.If engine 3 suddenly fails, determine the resultantof the three remaining engine thrust vectors, eachof which has a magnitude of 90 kN. Specify the y-and z-coordinates of the point through which theline of action of the resultant passes. This informationwould be critical to the design criteria of performancewith engine failure.Problem 2/148y49 mx2 m 33 m90 kN3 m22 m12 mz90 kN12 m90 kN9 m1Problem 2/150

c02.qxd 10/29/07 1:38 PM Page 9696 Chapter 2 Force Systems2/151 Represent the resultant of the force system actingon the pipe assembly by a single force R at A and acouple M.Ans. R 120i 180j 100k NM A 100j 50k N m300 mmz120 N2/153 A 25-kg computer and a 10-kg laser printer rest ona horizontal tabletop. Determine the resultant ofthe system of two corresponding weights and specifythe coordinates of the point P in the x-y planethrough which the resultant passes. The twoweights act at the points G 1 and G 2 .Ans. R 343k N, M O 343i 176.6j N m(x, y) (514, 1000) mmz100 N50 N·m250 mm300 mm O10 kg25 kgx200 mm160 N100mmA160 N180 Nyx500mm1200mmBG 2600 mm yG 1A25°150mmProblem 2/1512/152 Two upward loads are exerted on the small threedimensionaltruss. Reduce these two loads to a singleforce–couple system at point O. Show that R isperpendicular to M O . Then determine the point inthe x-z plane through which the resultant passes.y0.9 mProblem 2/1532/154 The motor mounted on the bracket is acted on byits 160-N weight, and its shaft resists the 120-Nthrust and 25-N m couple applied to it. Determinethe resultant of the force system shown in terms ofa force R at A and a couple M.z200 mm0.9 m160 N75 mmzO1.2 mAy1.2 mProblem 2/1520.9 m1600 N800 Nxx75mm25 N·m120 N100mmProblem 2/154

c02.qxd 10/29/07 1:38 PM Page 97Article 2/9 Problems 972/155 In tightening a bolt whose center is at point O, aperson exerts a 180-N force on the ratchet handlewith his right hand. In addition, with his left handhe exerts a 90-N force as shown in order to securethe socket onto the bolt head. Determine the equivalentforce–couple system at O. Then find the pointin the x-y plane through which the line of action ofthe resultant force of the wrench passes.Ans. R 90j 180k N, M O 6.3i 36j N mx 160 mm, y 35 mmz200 mm35 mm180 N2/157 Replace the two forces acting on the frame by awrench. Write the moment associated with thewrench as a vector and specify the coordinates ofthe point P in the y-z plane through which the lineof action of the wrench passes. Note that the forceof magnitude F is parallel to the x-axis.Ans. R F(i 3k), M 3aF (i 3k)10y a , z 2a10z3FaFxO90 NyProblem 2/1552a2/156 Replace the two forces acting on the pole by awrench. Write the moment M associated with thewrench as a vector and specify the coordinates ofthe point P in the y-z plane through which the lineof action of the wrench passes.zxOyTProblem 2/157Ta3a2/158 For the system of two forces in Prob. 2/149, determinethe coordinates of the point in the x-z planethrough which the line of action of the resultant ofthe system passes.OyxProblem 2/156

c02.qxd 10/29/07 1:38 PM Page 9898 Chapter 2 Force Systems2/159 The resultant of the two forces and couple may berepresented by a wrench. Determine the vector expressionfor the moment M of the wrench and findthe coordinates of the point P in the x-z planethrough which the resultant force of the wrenchpasses.Ans. M 10i 10j N mx z 0.1 m100 Ny300 mmz20 N·m400 mm400mmx100 N2/160 For the position shown, the crankshaft of a smalltwo-cylinder compressor is subjected to the 400-Nand 800-N forces exerted by the connecting rodsand the 200-N m couple. Replace this loading systemby a force–couple system at point A. Show thatR is not perpendicular to M A . Then replace thisforce–couple system by a wrench. Determine themagnitude M of the moment of the wrench, themagnitude of the force R of the wrench, and the coordinatesof the point in the x-z plane throughwhich the line of action of the wrench passes.Ans. M 85.8 N m, R 1108 Nx 0.1158 m, z 0.478 my24°400 N200mm24°800 NBProblem 2/159z200 N·mA200mm200mm400 mm200mmxProblem 2/160

c02.qxd 10/29/07 1:38 PM Page 99Article 2/10 Chapter Review 992/10 CHAPTER R EVIEWIn Chapter 2 we have established the properties of forces, moments,and couples, and the correct procedures for representing their effects.Mastery of this material is essential for our study of equilibrium in thechapters which follow. Failure to correctly use the procedures of Chapter2 is a common cause of errors in applying the principles of equilibrium.When difficulties arise, you should refer to this chapter to be surethat the forces, moments, and couples are correctly represented.ForcesThere is frequent need to represent forces as vectors, to resolve asingle force into components along desired directions, and to combinetwo or more concurrent forces into an equivalent resultant force. Specifically,you should be able to:1. Resolve a given force vector into its components along given directions,and express the vector in terms of the unit vectors along agiven set of axes.2. Express a force as a vector when given its magnitude and informationabout its line of action. This information may be in the form oftwo points along the line of action or angles which orient the line ofaction.3. Use the dot product to compute the projection of a vector onto aspecified line and the angle between two vectors.4. Compute the resultant of two or more forces concurrent at a point.MomentsThe tendency of a force to rotate a body about an axis is describedby a moment (or torque), which is a vector quantity. We have seen thatfinding the moment of a force is often facilitated by combining the momentsof the components of the force. When working with moment vectorsyou should be able to:1. Determine a moment by using the moment-arm rule.2. Use the vector cross product to compute a moment vector in termsof a force vector and a position vector locating the line of action ofthe force.3. Utilize Varignon’s theorem to simplify the calculation of moments,in both scalar and vector forms.4. Use the triple scalar product to compute the moment of a force vectorabout a given axis through a given point.CouplesA couple is the combined moment of two equal, opposite, and noncollinearforces. The unique effect of a couple is to produce a pure twistor rotation regardless of where the forces are located. The couple isuseful in replacing a force acting at a point by a force–couple system at

c02.qxd 10/29/07 1:38 PM Page 100100 Chapter 2 Force Systemsa different point. To solve problems involving couples you should beable to:1. Compute the moment of a couple, given the couple forces and eithertheir separation distance or any position vectors locating their linesof action.2. Replace a given force by an equivalent force–couple system, and viceversa.ResultantsWe can reduce an arbitrary system of forces and couples to a singleresultant force applied at an arbitrary point, and a corresponding resultantcouple. We can further combine this resultant force and couple intoa wrench to give a single resultant force along a unique line of action,together with a parallel couple vector. To solve problems involving resultantsyou should be able to:1. Compute the magnitude, direction, and line of action of the resultantof a system of coplanar forces if that resultant is a force; otherwise,compute the moment of the resultant couple.2. Apply the principle of moments to simplify the calculation of themoment of a system of coplanar forces about a given point.3. Replace a given general force system by a wrench along a specificline of action.EquilibriumYou will use the preceding concepts and methods when you studyequilibrium in the following chapters. Let us summarize the concept ofequilibrium:1. When the resultant force on a body is zero (ΣF 0), the body is intranslational equilibrium. This means that its center of mass is eitherat rest or moving in a straight line with constant velocity.2. In addition, if the resultant couple is zero (ΣM 0), the body is inrotational equilibrium, either having no rotational motion or rotatingwith a constant angular velocity.3. When both resultants are zero, the body is in complete equilibrium.

c02.qxd 10/29/07 1:38 PM Page 101Article 2/10 Review Problems 101REVIEW PROBLEMS2/161 A die is being used to cut threads on a rod. If 60-Nforces are applied as shown, determine the magnitudeF of the equal forces exerted on the 6-mm rodby each of the four cutting surfaces so that their externaleffect on the rod is equivalent to that of thetwo 60-N forces.Ans. F 1200 N6 mm2/163 The control lever is subjected to a clockwise coupleof 80 N m exerted by its shaft at A and is designedto operate with a 200-N pull as shown. If the resultantof the couple and the force passes through A,determine the proper dimension x of the lever.Ans. x 266 mm200 N150 mmx20°60 N120mmProblem 2/1612/162 Using the principles of equilibrium to be developedin Chapter 3, you will soon be able to verify thatthe tension in cable AB is 85.8% of the weight ofthe cylinder of mass m, while the tension in cableAC is 55.5% of the suspended weight. Write eachtension force acting on point A as a vector if themass m is 60 kg.B120mm0.8 m 2 mC60 NProblem 2/1632/164 The blades of the portable fan generate a 4-Nthrust T as shown. Compute the moment M O ofthis force about the rear support point O. For comparison,determine the moment about O due to theweight of the motor–fan unit AB, whose weight of40 N acts at G.A15°100 mmGBAAir flowT = 4 N1.2 my250 mm50mmAmProblem 2/162xO50mm200 mmProblem 2/164

c02.qxd 10/29/07 1:38 PM Page 102102 Chapter 2 Force Systems2/165 For the angular position 60 of the crank OA,the gas pressure on the piston induces a compressiveforce P in the connecting rod along its centerlineAB. If this force produces a moment of 720N m about the crank axis O, calculate P.Ans. P 9.18 kN2/167 Represent the resultant of the three forces and coupleby a force–couple system located at point A.Ans. R 4.75 kN, M A 21.1 kN m CCW3 kN1.5 m3.5 m2.5 mB10 kN·mA5 kN160°2 m4 kNOA = 125 mmAB = 300 mmProblem 2/167AθO2/168 Reduce the given loading system to a force–couplesystem at point A. Then determine the distance xto the right of point A at which the resultant of thethree forces acts.800 N720 NProblem 2/1652/166 Calculate the moment M O of the 250-N force aboutthe base point O of the robot.A200 mm450 mm500 mm20°1200 NA60°400 mmC250 NProblem 2/168500 mmBO300 mmProblem 2/166

c02.qxd 10/29/07 1:38 PM Page 103Article 2/10 Review Problems 1032/169 The directions of rotation of the input shaft A andoutput shaft B of the worm-gear reducer are indicatedby the curved dashed arrows. An inputtorque (couple) of 80 N m is applied to shaft A inthe direction of rotation. The output shaft B suppliesa torque of 320 N m to the machine which itdrives (not shown). The shaft of the driven machineexerts an equal and opposite reacting torqueon the output shaft of the reducer. Determine theresultant M of the two couples which act on the reducerunit and calculate the direction cosine of Mwith respect to the x-axis.Ans. M 320i 80j N mcos x 0.970z2/171 When the pole OA is in the position shown, the tensionin cable AB is 3 kN. (a) Write the tension forceexerted on the small collar at point A as a vectorusing the coordinates shown. (b) Determine themoment of this force about point O and state themoments about the x-, y-, and z-axes. (c) Determinethe projection of this tension force onto line AO.Ans. (a) T AB 2.05i 1.432j 1.663k kN(b) M O 7.63i 10.90j kN m(c) T AO 2.69 kNOA = 10 mOB = 8 mOG = 6 mzGAy320 N·mB60°B80 N·mO35°150 mmxxAyProblem 2/171Problem 2/1692/170 Determine the moment of the force P about point A.x2/172 A force F acts along the line AB inside the rightcircular cylindrical shell as shown. The quantitiesr, h, , and F are known. Using the x-, y-, andz-coordinates shown, express F as a vector.yAhAbbbzB34PyzOFBθrxProblem 2/170Problem 2/172

c02.qxd 10/29/07 1:38 PM Page 104104 Chapter 2 Force Systems2/173 Three couples are formed by the three pairs ofequal and opposite forces. Determine the resultantM of the three couples.Ans. M 20i 6.77j 37.2k N m2/175 The combined action of the three forces on the baseat O may be obtained by establishing their resultantthrough O. Determine the magnitudes of Rand the accompanying couple M.Ans. R 10.93 kN, M 38.9 kN m180mm100 N80 N100 N 20°z4 kNx100mm200mm20°80 N45°30° y120 N2.8 m1.2 m1.2 m1.2 m45°120 NyO45°6 kNzx5 kNProblem 2/1732/174 During a drilling operation, the small robotic deviceis subjected to an 800-N force at point C asshown. Replace this force by an equivalent force–couple system at point O.z300 mm20°Problem 2/175*Computer-Oriented Problems*2/176 Four forces are exerted on the eyebolt as shown. Ifthe net effect on the bolt is a direct pull of 1200 Nin the y-direction, determine the necessary valuesof T and .AB480 NyT250 mmyCθ30°800 NOx30°F = 800 N720 N30°xProblem 2/174Problem 2/176

c02.qxd 10/29/07 1:38 PM Page 105Article 2/10 Review Problems 105*2/177 The force F is directed from A toward D and D isallowed to move from B to C as measured by thevariable s. Consider the projection of F onto lineEF as a function of s. In particular, determine andplot the fraction n of the magnitude F which isprojected as a function of s/d. Note that s/d variesfrom 0 to 22.A2dFCAns. n sDEB2d2 s d 15 s d 2 5 22 s ddF*2/179 With the cylindrical part P of weight 1500 N in itsgrip, the robotic arm pivots about O through therange 45 45 with the angle at A lockedat 120. Determine and plot (as a function of ) themoment at O due to the combined effects of theweight of part P, the 600-N weight of member OA(mass center at G 1 ), and the 250-N weight ofmember AB (mass center at G 2 ). The end grip isincluded as a part of member AB. The lengths L 1and L 2 are 900 mm and 600 mm, respectively.What is the maximum value of M O and at whatvalue of does this maximum occur?Ans. M O 1845 cos 975 cos(60 ) N m(M O ) max 2480 N m at 19.90OL 1—–2L 1—–2AL 2—– 2120° G 2G 1 BθL 2—– 2PProblem 2/177*2/178 The throttle-control lever OA rotates in the range0 90. An internal torsional return springexerts a restoring moment about O given by M K( /4), where K 500 N mm/rad and is inradians. Determine and plot as a function of thetension T required to make the net moment aboutO zero. Use the two values d 60 mm and d 160 mm and comment on the relative design merits.The effects of the radius of the pulley at B arenegligible.Problem 2/179θAB40 mm45° 40 mmOTMdProblem 2/178

c02.qxd 10/29/07 1:38 PM Page 106106 Chapter 2 Force Systems*2/180 A motor attached to the shaft at O causes the armOA to rotate over the range 0 180. The unstretchedlength of the spring is 0.65 m, and it cansupport both tension and compression. If the netmoment about O must be zero, determine and plotthe required motor torque M as a function of .y*2/182 The tension T in cable AB is maintained at aconstant value of 120 N. Determine the momentM O of this tension about point O over the range0 90. Plot the x-, y-, and z-components ofM O as functions of .TBM0.5 mA′′ θA′Ok = 600 N/m0.3 mAx800 mmy400 mmxBOAProblem 2/180*2/181 A flagpole with attached light triangular frame isshown here for an arbitrary position during itsraising. The 75-N tension in the erecting cable remainsconstant. Determine and plot the momentabout the pivot O of the 75-N force for the range0 90. Determine the maximum value ofthis moment and the elevation angle at which itoccurs; comment on the physical significance ofthe latter. The effects of the diameter of the drumat D may be neglected.1350 sin ( 60)Ans. M O 45 36 cos ( 60) k N m(M O ) max 225 N m at 60θ800 mmProblem 2/182A′zCA3 mBD3 mO3 mθ6 mProblem 2/181

c02.qxd 10/29/07 1:38 PM Page 107Article 2/10 Review Problems 107*2/183 The arm AB rotates in the range 0 180,and the spring is unstretched when 90. Determineas a function of the moment about O ofthe spring force as applied at B. Plot the threescalar components of M O , and state the maximumabsolute value of each component.1.5 0.5 sin 1Ans. M O [674i 337 cos k] N m1.5 0.5 sin (M Ox) max = 123.7 N m at 0, 180(M Oz) max = 61.8 N m at 0, 180*2/184 The rectangular plate is tilted about its loweredge by a cable tensioned at a constant 600 N. Determineand plot the moment of this tensionabout the lower edge AB of the plate for the range0 90.z5 mT = 600 NyCz3 mA360 mmO5 mθA4 mBBxProblem 2/184k = 2.6 kN/m720 mmxθODyC360 mmProblem 2/183