c03.qxd 11/6/07 3:26 PM Page 108

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1 c03.qd 11/6/07 3:26 M age 108 eter ialobreski/laif/redu In man applications of mechanics, the sum of the forces acting on a bod is ero, and a state of equilibrium eists. This apparatus is designed to hold a car bod in equilibrium for a wide range of orientations during vehicle production.

2 c03.qd 11/6/07 3:26 M age EQUILIRIUM HTER UTLINE 3/1 Introduction SETIN EQUILIRIUM IN TW IMENSINS 3/2 Sstem Isolation and the Free-od iagram 3/3 Equilibrium onditions SETIN EQUILIRIUM IN THREE IMENSINS 3/4 Equilibrium onditions 3/5 hapter Review 3/1 INTRUTIN Statics deals primaril with the description of the force conditions necessar and sufficient to maintain the equilibrium of engineering structures. This chapter on equilibrium, therefore, constitutes the most important part of statics, and the procedures developed here form the basis for solving problems in both statics and dnamics. We will make continual use of the concepts developed in hapter 2 involving forces, moments, couples, and resultants as we appl the principles of equilibrium. When a bod is in equilibrium, the resultant of all forces acting on it is ero. Thus, the resultant force R and the resultant couple M are both ero, and we have the equilibrium equations R ΣF 0 M ΣM 0 (3/1) These requirements are both necessar and sufficient conditions for equilibrium. ll phsical bodies are three-dimensional, but we can treat man of them as two-dimensional when the forces to which the are subjected act in a single plane or can be projected onto a single plane. When this simplification is not possible, the problem must be treated as threedimensional. We will follow the arrangement used in hapter 2, and discuss in Section the equilibrium of bodies subjected to two-dimensional 109

3 c03.qd 11/6/07 3:26 M age hapter 3 Equilibrium force sstems and in Section the equilibrium of bodies subjected to three-dimensional force sstems. SETIN EQUILIRIUM IN TW IMENSINS 3/2 SYSTEM I SLTIN N THE F REE-Y IGRM efore we appl Eqs. 3/1, we must define unambiguousl the particular bod or mechanical sstem to be analed and represent clearl and completel all forces acting on the bod. mission of a force which acts on the bod in question, or inclusion of a force which does not act on the bod, will give erroneous results. mechanical sstem is defined as a bod or group of bodies which can be conceptuall isolated from all other bodies. sstem ma be a single bod or a combination of connected bodies. The bodies ma be rigid or nonrigid. The sstem ma also be an identifiable fluid mass, either liquid or gas, or a combination of fluids and solids. In statics we stud primaril forces which act on rigid bodies at rest, although we also stud forces acting on fluids in equilibrium. nce we decide which bod or combination of bodies to anale, we then treat this bod or combination as a single bod isolated from all surrounding bodies. This isolation is accomplished b means of the free-bod diagram, which is a diagraatic representation of the isolated sstem treated as a single bod. The diagram shows all forces applied to the sstem b mechanical contact with other bodies, which are imagined to be removed. If appreciable bod forces are present, such as gravitational or magnetic attraction, then these forces must also be shown on the free-bod diagram of the isolated sstem. nl after such a diagram has been carefull drawn should the equilibrium equations be written. ecause of its critical importance, we emphasie here that the free-bod diagram is the most important single step in the solution of problems in mechanics. efore attempting to draw a free-bod diagram, we must recall the basic characteristics of force. These characteristics were described in rt. 2/2, with primar attention focused on the vector properties of force. Forces can be applied either b direct phsical contact or b remote action. Forces can be either internal or eternal to the sstem under consideration. pplication of force is accompanied b reactive force, and both applied and reactive forces ma be either concentrated or distributed. The principle of transmissibilit permits the treatment of force as a sliding vector as far as its eternal effects on a rigid bod are concerned. We will now use these force characteristics to develop conceptual models of isolated mechanical sstems. These models enable us to

4 c03.qd 11/6/07 3:26 M age 111 rticle 3/2 Sstem Isolation and the Free-od iagram 111 write the appropriate equations of equilibrium, which can then be analed. Modeling the ction of Forces Figure 3/1 shows the coon tpes of force application on mechanical sstems for analsis in two dimensions. Each eample shows the force eerted on the bod to be isolated, b the bod to be removed. Newton s third law, which notes the eistence of an equal and opposite reaction to ever action, must be carefull observed. The force eerted on the bod in question b a contacting or supporting member is alwas in the sense to oppose the movement of the isolated bod which would occur if the contacting or supporting bod were removed. MELING THE TIN F FRES IN TW-IMENSINL NLYSIS Tpe of ontact and Force rigin ction on od to e Isolated 1. Fleible cable, belt, chain, or rope Weight of cable negligible Weight of cable not negligible θ θ T T θ θ Force eerted b a fleible cable is alwas a tension awa from the bod in the direction of the cable. 2. Smooth surfaces N ontact force is compressive and is normal to the surface. 3. Rough surfaces R F N Rough surfaces are capable of supporting a tangential compo-nent F (frictional force) as well as a normal component N of the resultant 4. Roller support N Roller, rocker, or ball support transmits a compressive force normal to the supporting surface. N 5. Freel sliding guide N N ollar or slider free to move along smooth guides; can support force normal to guide onl. Figure 3/1

5 c03.qd 11/6/07 3:26 M age hapter 3 Equilibrium MELING THE TIN F FRES IN TW-IMENSINL NLYSIS (cont.) Tpe of ontact and Force rigin 6. in connection ction on od to e Isolated in free to turn freel hinged pin connection is capable θ of supporting a force in an direction in the R plane normal to the R R pin ais. We ma either show two in not free to turn components R and R or a magnitude R and direction θ. pin R not free to turn also M supports a couple M. R 7. uilt-in or fied support or Weld F M V built-in or fied support is capable of supporting an aial force F, a transverse force V (shear force), and a couple M (bending moment) to prevent rotation. 8. Gravitational attraction m G W = mg The resultant of gravitational attraction on all elements of a bod of mass m is the weight W = mg and acts toward the center of the earth through the center mass G. 9. Spring action Neutral position F Linear F F = k Nonlinear F Hardening Softening F Spring force is tensile if spring is stretched and compressive if compressed. For a linearl elastic spring the stiffness k is the force required to deform the spring a unit distance. Figure 3/1, continued In Fig. 3/1, Eample 1 depicts the action of a fleible cable, belt, rope, or chain on the bod to which it is attached. ecause of its fleibilit, a rope or cable is unable to offer an resistance to bending, shear, or compression and therefore eerts onl a tension force in a direction tangent to the cable at its point of attachment. The force eerted b the cable on the bod to which it is attached is alwas awa from the bod. When the tension T is large compared with the weight of the cable, we ma assume that the cable forms a straight line. When the cable weight is not negligible compared with its tension, the sag of the cable becomes important, and the tension in the cable changes direction and magnitude along its length. When the smooth surfaces of two bodies are in contact, as in Eample 2, the force eerted b one on the other is normal to the tangent to the surfaces and is compressive. lthough no actual surfaces are per-

6 c03.qd 11/6/07 3:26 M age 113 rticle 3/2 Sstem Isolation and the Free-od iagram 113 fectl smooth, we can assume this to be so for practical purposes in man instances. When mating surfaces of contacting bodies are rough, as in Eample 3, the force of contact is not necessaril normal to the tangent to the surfaces, but ma be resolved into a tangential or frictional component F and a normal component N. Eample 4 illustrates a number of forms of mechanical support which effectivel eliminate tangential friction forces. In these cases the net reaction is normal to the supporting surface. Eample 5 shows the action of a smooth guide on the bod it supports. There cannot be an resistance parallel to the guide. Eample 6 illustrates the action of a pin connection. Such a connection can support force in an direction normal to the ais of the pin. We usuall represent this action in terms of two rectangular components. The correct sense of these components in a specific problem depends on how the member is loaded. When not otherwise initiall known, the sense is arbitraril assigned and the equilibrium equations are then written. If the solution of these equations ields a positive algebraic sign for the force component, the assigned sense is correct. negative sign indicates the sense is opposite to that initiall assigned. If the joint is free to turn about the pin, the connection can support onl the force R. If the joint is not free to turn, the connection can also support a resisting couple M. The sense of M is arbitraril shown here, but the true sense depends on how the member is loaded. Eample 7 shows the resultants of the rather comple distribution of force over the cross section of a slender bar or beam at a build-in or fied support. The sense of the reactions F and V and the bending couple M in a given problem depends, of course, on how the member is loaded. ne of the most coon forces is that due to gravitational attraction, Eample 8. This force affects all elements of mass in a bod and is, therefore, distributed throughout it. The resultant of the gravitational forces on all elements is the weight W mg of the bod, which passes through the center of mass G and is directed toward the center of the earth for earthbound structures. The location of G is frequentl obvious from the geometr of the bod, particularl where there is setr. When the location is not readil apparent, it must be determined b eperiment or calculations. Similar remarks appl to the remote action of magnetic and electric forces. These forces of remote action have the same overall effect on a rigid bod as forces of equal magnitude and direction applied b direct eternal contact. Eample 9 illustrates the action of a linear elastic spring and of a nonlinear spring with either hardening or softening characteristics. The force eerted b a linear spring, in tension or compression, is given b F k, where k is the stiffness of the spring and is its deformation measured from the neutral or undeformed position. The representations in Fig. 3/1 are not free-bod diagrams, but are merel elements used to construct free-bod diagrams. Stud these nine conditions and identif them in the problem work so that ou can draw the correct free-bod diagrams. nother view of the car-bod lifting device shown in the chapter-opening photograph. eter ialobreski/laif/redu

7 c03.qd 11/6/07 3:26 M age hapter 3 Equilibrium onstruction of Free-od iagrams The full procedure for drawing a free-bod diagram which isolates a bod or sstem consists of the following steps. Step 1. ecide which sstem to isolate. The sstem chosen should usuall involve one or more of the desired unknown quantities. Step 2. Net isolate the chosen sstem b drawing a diagram which represents its complete eternal boundar. This boundar defines the isolation of the sstem from all other attracting or contacting bodies, which are considered removed. This step is often the most crucial of all. Make certain that ou have completel isolated the sstem before proceeding with the net step. Step 3. Identif all forces which act on the isolated sstem as applied b the removed contacting and attracting bodies, and represent them in their proper positions on the diagram of the isolated sstem. Make a sstematic traverse of the entire boundar to identif all contact forces. Include bod forces such as weights, where appreciable. Represent all known forces b vector arrows, each with its proper magnitude, direction, and sense indicated. Each unknown force should be represented b a vector arrow with the unknown magnitude or direction indicated b smbol. If the sense of the vector is also unknown, ou must arbitraril assign a sense. The subsequent calculations with the equilibrium equations will ield a positive quantit if the correct sense was assumed and a negative quantit if the incorrect sense was assumed. It is necessar to be consistent with the assigned characteristics of unknown forces throughout all of the calculations. If ou are consistent, the solution of the equilibrium equations will reveal the correct senses. Step 4. Show the choice of coordinate aes directl on the diagram. ertinent dimensions ma also be represented for convenience. Note, however, that the free-bod diagram serves the purpose of focusing attention on the action of the eternal forces, and therefore the diagram should not be cluttered with ecessive etraneous information. learl distinguish force arrows from arrows representing quantities other than forces. For this purpose a colored pencil ma be used. ompletion of the foregoing four steps will produce a correct freebod diagram to use in appling the governing equations, both in statics and in dnamics. e careful not to omit from the free-bod diagram certain forces which ma not appear at first glance to be needed in the calculations. It is onl through complete isolation and a sstematic representation of all eternal forces that a reliable accounting of the effects of all applied and reactive forces can be made. Ver often a force which at first glance ma not appear to influence a desired result does indeed have an influence. Thus, the onl safe procedure is to include on the free-bod diagram all forces whose magnitudes are not obviousl negligible.

8 c03.qd 11/6/07 3:26 M age 115 rticle 3/2 Sstem Isolation and the Free-od iagram 115 The free-bod method is etremel important in mechanics because it ensures an accurate definition of a mechanical sstem and focuses attention on the eact meaning and application of the force laws of statics and dnamics. Review the foregoing four steps for constructing a freebod diagram while studing the sample free-bod diagrams shown in Fig. 3/2 and the Sample roblems which appear at the end of the net article. Eamples of Free-od iagrams Figure 3/2 gives four eamples of mechanisms and structures together with their correct free-bod diagrams. imensions and magnitudes are omitted for clarit. In each case we treat the entire sstem as SMLE FREE-Y IGRMS Mechanical Sstem Free-od iagram of Isolated od 1. lane truss Weight of truss assumed negligible compared with 2. antilever beam F 3 F 2 F 1 V F 3 F 2 F 1 3. eam Mass m F M W = mg Smooth surface contact at. Mass m M N W = mg M 4. Rigid sstem of interconnected bodies analed as a single unit Weight of mechanism neglected m W = mg Figure 3/2

9 c03.qd 11/6/07 3:26 M age hapter 3 Equilibrium a single bod, so that the internal forces are not shown. The characteristics of the various tpes of contact forces illustrated in Fig. 3/1 are used in the four eamples as the appl. In Eample 1 the truss is composed of structural elements which, taken all together, constitute a rigid framework. Thus, we ma remove the entire truss from its supporting foundation and treat it as a single rigid bod. In addition to the applied eternal load, the free-bod diagram must include the reactions on the truss at and. The rocker at can support a vertical force onl, and this force is transmitted to the structure at (Eample 4 of Fig. 3/1). The pin connection at (Eample 6 of Fig. 3/1) is capable of suppling both a horiontal and a vertical force component to the truss. If the total weight of the truss members is appreciable compared with and the forces at and, then the weights of the members must be included on the free-bod diagram as eternal forces. In this relativel simple eample it is clear that the vertical component must be directed down to prevent the truss from rotating clockwise about. lso, the horiontal component will be to the left to keep the truss from moving to the right under the influence of the horiontal component of. Thus, in constructing the free-bod diagram for this simple truss, we can easil perceive the correct sense of each of the components of force eerted on the truss b the foundation at and can, therefore, represent its correct phsical sense on the diagram. When the correct phsical sense of a force or its component is not easil recognied b direct observation, it must be assigned arbitraril, and the correctness of or error in the assignment is determined b the algebraic sign of its calculated value. In Eample 2 the cantilever beam is secured to the wall and subjected to three applied loads. When we isolate that part of the beam to the right of the section at, we must include the reactive forces applied to the beam b the wall. The resultants of these reactive forces are shown acting on the section of the beam (Eample 7 of Fig. 3/1). vertical force V to counteract the ecess of downward applied force is shown, and a tension F to balance the ecess of applied force to the right must also be included. Then, to prevent the beam from rotating about, a counterclockwise couple M is also required. The weight mg of the beam must be represented through the mass center (Eample 8 of Fig. 3/1). In the free-bod diagram of Eample 2, we have represented the somewhat comple sstem of forces which actuall act on the cut section of the beam b the equivalent force couple sstem in which the force is broken down into its vertical component V (shear force) and its horiontal component F (tensile force). The couple M is the bending moment in the beam. The free-bod diagram is now complete and shows the beam in equilibrium under the action of si forces and one couple. In Eample 3 the weight W mg is shown acting through the center of mass of the beam, whose location is assumed known (Eample 8 of Fig. 3/1). The force eerted b the corner on the beam is normal to the smooth surface of the beam (Eample 2 of Fig. 3/1). To perceive this action more clearl, visualie an enlargement of the contact point, which would appear somewhat rounded, and consider the force eerted b this rounded corner on the straight surface of the beam, which is as-

10 c03.qd 11/6/07 3:26 M age 117 rticle 3/2 Sstem Isolation and the Free-od iagram 117 sumed to be smooth. If the contacting surfaces at the corner were not smooth, a tangential frictional component of force could eist. In addition to the applied force and couple M, there is the pin connection at, which eerts both an - and a -component of force on the beam. The positive senses of these components are assigned arbitraril. In Eample 4 the free-bod diagram of the entire isolated mechanism contains three unknown forces if the loads mg and are known. n one of man internal configurations for securing the cable leading from the mass m would be possible without affecting the eternal response of the mechanism as a whole, and this fact is brought out b the free-bod diagram. This hpothetical eample is used to show that the forces internal to a rigid assembl of members do not influence the values of the eternal reactions. We use the free-bod diagram in writing the equilibrium equations, which are discussed in the net article. When these equations are solved, some of the calculated force magnitudes ma be ero. This would indicate that the assumed force does not eist. In Eample 1 of Fig. 3/2, an of the reactions,, or can be ero for specific values of the truss geometr and of the magnitude, direction, and sense of the applied load. ero reaction force is often difficult to identif b inspection, but can be determined b solving the equilibrium equations. Similar coents appl to calculated force magnitudes which are negative. Such a result indicates that the actual sense is the opposite of the assumed sense. The assumed positive senses of and in Eample 3 and in Eample 4 are shown on the free-bod diagrams. The correctness of these assumptions is proved or disproved according to whether the algebraic signs of the computed forces are plus or minus when the calculations are carried out in an actual problem. The isolation of the mechanical sstem under consideration is a crucial step in the formulation of the mathematical model. The most important aspect to the correct construction of the all-important free-bod diagram is the clear-cut and unambiguous decision as to what is included and what is ecluded. This decision becomes unambiguous onl when the boundar of the free-bod diagram represents a complete traverse of the bod or sstem of bodies to be isolated, starting at some arbitrar point on the boundar and returning to that same point. The sstem within this closed boundar is the isolated free bod, and all contact forces and all bod forces transmitted to the sstem across the boundar must be accounted for. The following eercises provide practice with drawing free-bod diagrams. This practice is helpful before using such diagrams in the application of the principles of force equilibrium in the net article. Even comple pulle sstems such as the ones seen here are easil handled with a sstematic equilibrium analsis. igital Vision/Gett Images

11 c03.qd 11/6/07 3:26 M age hapter 3 Equilibrium FREE-Y IGRM EXERISES 3/ In each of the five following eamples, the bod to be isolated is shown in the left-hand diagram, and an incomplete free-bod diagram (F) of the isolated bod is shown on the right. dd whatever forces are necessar in each case to form a complete free-bod diagram. The weights of the bodies are negligible unless otherwise indicated. imensions and numerical values are omitted for simplicit. od Incomplete F 1. ell crank supporting mass m with pin support at. Fleible cable m T mg ull 2. ontrol lever appling torque to shaft at. F 3. oom, of negligible mass compared with mass m. oom hinged at and supported b hoisting cable at. m T mg 4. Uniform crate of mass m leaning against smooth vertical wall and supported on a rough horiontal surface. mg 5. Loaded bracket supported b pin connection at and fied pin in smooth slot at. Load L L Figure 3/

12 c03.qd 11/6/07 3:26 M age 119 Free-od iagram Eercises 119 3/ In each of the five following eamples, the bod to be isolated is shown in the left-hand diagram, and either a wrong or an incomplete free-bod diagram (F) is shown on the right. Make whatever changes or addi- tions are necessar in each case to form a correct and complete free-bod diagram. The weights of the bodies are negligible unless otherwise indicated. imensions and numerical values are omitted for simplicit. od Wrong or Incomplete F 1. Lawn roller of mass m being pushed up incline θ. 2. rbar lifting bod having smooth horiontal surface. ar rests on horiontal rough surface. θ R N mg N 3. Uniform pole of mass m being hoisted into position b winch. Horiontal supporting surface notched to prevent slipping of pole. Notch T R mg F 4. Supporting angle bracket for frame; pin joints. F F 5. ent rod welded to support at and subjected to two forces and couple. M M Figure 3/

13 c03.qd 11/6/07 3:26 M age hapter 3 Equilibrium 3/ raw a complete and correct free-bod diagram of each of the bodies designated in the statements. The weights of the bodies are significant onl if the mass is stated. ll forces, known and unknown, should be labeled. (Note: The sense of some reaction components cannot alwas be determined without numerical calculation.) 1. Uniform horiontal bar of mass m suspended b vertical cable at and supported b rough inclined surface at. 5. Uniform grooved wheel of mass m supported b a rough surface and b action of horiontal cable. m m 2. Wheel of mass m on verge of being rolled over curb b pull. 6. ar, initiall horiontal but deflected under load L. inned to rigid support at each end. L 3. Loaded truss supported b pin joint at and b cable at. 7. Uniform heav plate of mass m supported in vertical plane b cable and hinge. m L 4. Uniform bar of mass m and roller of mass m 0 taken together. Subjected to couple M and supported as shown. Roller is free to turn. m M 0 8. Entire frame, pulles, and contacting cable to be isolated as a single unit. m L Figure 3/

14 c03.qd 11/6/07 3:26 M age 121 rticle 3/3 Equilibrium onditions 121 3/3 EQUILIRIUM NITINS In rt. 3/1 we defined equilibrium as the condition in which the resultant of all forces and moments acting on a bod is ero. Stated in another wa, a bod is in equilibrium if all forces and moments applied to it are in balance. These requirements are contained in the vector equations of equilibrium, Eqs. 3/1, which in two dimensions ma be written in scalar form as ΣF 0 ΣF 0 ΣM 0 (3/2) The third equation represents the ero sum of the moments of all forces about an point on or off the bod. Equations 3/2 are the necessar and sufficient conditions for complete equilibrium in two dimensions. The are necessar conditions because, if the are not satisfied, there can be no force or moment balance. The are sufficient because once the are satisfied, there can be no imbalance, and equilibrium is assured. The equations relating force and acceleration for rigid-bod motion are developed in Vol. 2 namics from Newton s second law of motion. These equations show that the acceleration of the mass center of a bod is proportional to the resultant force ΣF acting on the bod. onsequentl, if a bod moves with constant velocit (ero acceleration), the resultant force on it must be ero, and the bod ma be treated as in a state of translational equilibrium. For complete equilibrium in two dimensions, all three of Eqs. 3/2 must hold. However, these conditions are independent requirements, and one ma hold without another. Take, for eample, a bod which slides along a horiontal surface with increasing velocit under the action of applied forces. The force equilibrium equations will be satisfied in the vertical direction where the acceleration is ero, but not in the horiontal direction. lso, a bod, such as a flwheel, which rotates about its fied mass center with increasing angular speed is not in rotational equilibrium, but the two force equilibrium equations will be satisfied. ategories of Equilibrium pplications of Eqs. 3/2 fall naturall into a number of categories which are easil identified. The categories of force sstems acting on bodies in two-dimensional equilibrium are suaried in Fig. 3/3 and are eplained further as follows. ategor 1, equilibrium of collinear forces, clearl requires onl the one force equation in the direction of the forces (-direction), since all other equations are automaticall satisfied. ategor 2, equilibrium of forces which lie in a plane (- plane) and are concurrent at a point, requires the two force equations onl, since the moment sum about, that is, about a -ais through, is necessaril ero. Included in this categor is the case of the equilibrium of a particle. ategor 3, equilibrium of parallel forces in a plane, requires the one force equation in the direction of the forces (-direction) and one moment equation about an ais (-ais) normal to the plane of the forces.

15 c03.qd 11/6/07 3:26 M age hapter 3 Equilibrium TEGRIES F EQUILIRIUM IN TW IMENSINS Force Sstem Free-od iagram Independent Equations 1. ollinear F 3 Σ F = 0 F 2 F 1 2. oncurrent at a point F 1 F 2 Σ F = 0 Σ F = 0 3. arallel F 4 F 3 F 1 Σ F = 0 ΣM = 0 F 2 F 3 F 4 4. General F F 2 1 M F 3 Σ F = 0 Σ F = 0 Σ M = 0 F 4 Figure 3/3 Two-force members Figure 3/4 ategor 4, equilibrium of a general sstem of forces in a plane (-), requires the two force equations in the plane and one moment equation about an ais (-ais) normal to the plane. Two- and Three-Force Members You should be alert to two frequentl occurring equilibrium situations. The first situation is the equilibrium of a bod under the action of two forces onl. Two eamples are shown in Fig. 3/4, and we see that for such a two-force member to be in equilibrium, the forces must be equal, opposite, and collinear. The shape of the member does not affect this simple requirement. In the illustrations cited, we consider the weights of the members to be negligible compared with the applied forces. The second situation is a three-force member, which is a bod under the action of three forces, Fig. 3/5a. We see that equilibrium requires the lines of action of the three forces to be concurrent. If the were not concurrent, then one of the forces would eert a resultant moment about the point of intersection of the other two, which would violate the requirement of ero moment about ever point. The onl eception occurs when the three forces are parallel. In this case we ma consider the point of concurrenc to be at infinit.

16 c03.qd 11/6/07 3:26 M age 123 rticle 3/3 Equilibrium onditions 123 The principle of the concurrenc of three forces in equilibrium is of considerable use in carring out a graphical solution of the force equations. In this case the polgon of forces is drawn and made to close, as shown in Fig. 3/5b. Frequentl, a bod in equilibrium under the action of more than three forces ma be reduced to a three-force member b a combination of two or more of the known forces. F 1 F 2 lternative Equilibrium Equations In addition to Eqs. 3/2, there are two other was to epress the general conditions for the equilibrium of forces in two dimensions. The first wa is illustrated in Fig. 3/6, parts (a) and (b). For the bod shown in Fig. 3/6a, if ΣM 0, then the resultant, if it still eists, cannot be a couple, but must be a force R passing through. If now the equation ΣF 0 holds, where the -direction is arbitrar, it follows from Fig. 3/6b that the resultant force R, if it still eists, not onl must pass through, but also must be perpendicular to the -direction as shown. Now, if ΣM 0, where is an point such that the line is not perpendicular to the -direction, we see that R must be ero, and thus the bod is in equilibrium. Therefore, an alternative set of equilibrium equations is ΣF 0 ΣM 0 ΣM 0 where the two points and must not lie on a line perpendicular to the -direction. third formulation of the equilibrium conditions ma be made for a coplanar force sstem. This is illustrated in Fig. 3/6, parts (c) and (d). gain, if ΣM 0 for an bod such as that shown in Fig. 3/6c, the resultant, if an, must be a force R through. In addition, if ΣM 0, the resultant, if one still eists, must pass through as shown in Fig. 3/6d. Such a force cannot eist, however, if ΣM 0, where is not F 3 (a) (b) Three-force member F 3 F 1 F 2 losed polgon satisfies ΣF = 0 Figure 3/5 ΣM = 0 satisfied ΣM = 0 Σ F = 0 satisfied R R (a) (b) ΣM = 0 satisfied R ΣM = 0 Σ M = 0 satisfied R (c) (d) Figure 3/6

17 c03.qd 11/6/07 3:26 M age hapter 3 Equilibrium collinear with and. Thus, we ma write the equations of equilibrium as ΣM 0 ΣM 0 ΣM 0 where,, and are an three points not on the same straight line. When equilibrium equations are written which are not independent, redundant information is obtained, and a correct solution of the equations will ield 0 0. For eample, for a general problem in two dimensions with three unknowns, three moment equations written about three points which lie on the same straight line are not independent. Such equations will contain duplicated information, and solution of two of them can at best determine two of the unknowns, with the third equation merel verifing the identit 0 0. onstraints and Statical eterminac The equilibrium equations developed in this article are both necessar and sufficient conditions to establish the equilibrium of a bod. However, the do not necessaril provide all the information required to calculate all the unknown forces which ma act on a bod in equilibrium. Whether the equations are adequate to determine all the unknowns depends on the characteristics of the constraints against possible movement of the bod provided b its supports. constraint we mean the restriction of movement. In Eample 4 of Fig. 3/1 the roller, ball, and rocker provide constraint normal to the surface of contact, but none tangent to the surface. Thus, a tangential force cannot be supported. For the collar and slider of Eample 5, constraint eists onl normal to the guide. In Eample 6 the fied-pin connection provides constraint in both directions, but offers no resistance to rotation about the pin unless the pin is not free to turn. The fied support of Eample 7, however, offers constraint against rotation as well as lateral movement. If the rocker which supports the truss of Eample 1 in Fig. 3/2 were replaced b a pin joint, as at, there would be one additional constraint beond those required to support an equilibrium configuration with no freedom of movement. The three scalar conditions of equilibrium, Eqs. 3/2, would not provide sufficient information to determine all four unknowns, since and could not be solved for separatel; onl their sum could be determined. These two components of force would be dependent on the deformation of the members of the truss as influenced b their corresponding stiffness properties. The horiontal reactions and would also depend on an initial deformation required to fit the dimensions of the structure to those of the foundation between and. Thus, we cannot determine and b a rigid-bod analsis. gain referring to Fig. 3/2, we see that if the pin in Eample 3 were not free to turn, the support could transmit a couple to the beam through the pin. Therefore, there would be four unknown supporting reactions acting on the beam, namel, the force at, the two components of force at, and the couple at. onsequentl the three inde-

18 c03.qd 11/6/07 3:26 M age 125 rticle 3/3 Equilibrium onditions 125 pendent scalar equations of equilibrium would not provide enough information to compute all four unknowns. rigid bod, or rigid combination of elements treated as a single bod, which possesses more eternal supports or constraints than are necessar to maintain an equilibrium position is called staticall indeterminate. Supports which can be removed without destroing the equilibrium condition of the bod are said to be redundant. The number of redundant supporting elements present corresponds to the degree of statical indeterminac and equals the total number of unknown eternal forces, minus the number of available independent equations of equilibrium. n the other hand, bodies which are supported b the minimum number of constraints necessar to ensure an equilibrium configuration are called staticall determinate, and for such bodies the equilibrium equations are sufficient to determine the unknown eternal forces. The problems on equilibrium in this article and throughout Vol. 1 Statics are generall restricted to staticall determinate bodies where the constraints are just sufficient to ensure a stable equilibrium configuration and where the unknown supporting forces can be completel determined b the available independent equations of equilibrium. We must be aware of the nature of the constraints before we attempt to solve an equilibrium problem. bod can be recognied as staticall indeterminate when there are more unknown eternal reactions than there are available independent equilibrium equations for the force sstem involved. It is alwas well to count the number of unknown variables on a given bod and to be certain that an equal number of independent equations can be written; otherwise, effort might be wasted in attempting an impossible solution with the aid of the equilibrium equations onl. The unknown variables ma be forces, couples, distances, or angles. dequac of onstraints In discussing the relationship between constraints and equilibrium, we should look further at the question of the adequac of constraints. The eistence of three constraints for a two-dimensional problem does not alwas guarantee a stable equilibrium configuration. Figure 3/7 shows four different tpes of constraints. In part a of the figure, point of the rigid bod is fied b the two links and cannot move, and the third link prevents an rotation about. Thus, this bod is completel fied with three adequate (proper) constraints. In part b of the figure, the third link is positioned so that the force transmitted b it passes through point where the other two constraint forces act. Thus, this configuration of constraints can offer no initial resistance to rotation about, which would occur when eternal loads were applied to the bod. We conclude, therefore, that this bod is incompletel fied under partial constraints. The configuration in part c of the figure gives us a similar condition of incomplete fiit because the three parallel links could offer no initial resistance to a small vertical movement of the bod as a result of eternal loads applied to it in this direction. The constraints in these two eamples are often termed improper. (a) omplete fiit dequate constraints (c) (b) Incomplete fiit artial constraints Incomplete fiit artial constraints 4 (d) Ecessive fiit Redundant constraint Figure 3/7

19 c03.qd 11/6/07 3:26 M age hapter 3 Equilibrium In part d of Fig. 3/7 we have a condition of complete fiit, with link 4 acting as a fourth constraint which is unnecessar to maintain a fied position. Link 4, then, is a redundant constraint, and the bod is staticall indeterminate. s in the four eamples of Fig. 3/7, it is generall possible b direct observation to conclude whether the constraints on a bod in twodimensional equilibrium are adequate (proper), partial (improper), or redundant. s indicated previousl, the vast majorit of problems in this book are staticall determinate with adequate (proper) constraints. pproach to Solving roblems The sample problems at the end of this article illustrate the application of free-bod diagrams and the equations of equilibrium to tpical statics problems. These solutions should be studied thoroughl. In the problem work of this chapter and throughout mechanics, it is important to develop a logical and sstematic approach which includes the following steps: 1. Identif clearl the quantities which are known and unknown. 2. Make an unambiguous choice of the bod (or sstem of connected bodies treated as a single bod) to be isolated and draw its complete free-bod diagram, labeling all eternal known and unknown but identifiable forces and couples which act on it. 3. hoose a convenient set of reference aes, alwas using righthanded aes when vector cross products are emploed. hoose moment centers with a view to simplifing the calculations. Generall the best choice is one through which as man unknown forces pass as possible. Simultaneous solutions of equilibrium equations are frequentl necessar, but can be minimied or avoided b a careful choice of reference aes and moment centers. 4. Identif and state the applicable force and moment principles or equations which govern the equilibrium conditions of the problem. In the following sample problems these relations are shown in brackets and precede each major calculation. 5. Match the number of independent equations with the number of unknowns in each problem. 6. arr out the solution and check the results. In man problems engineering judgment can be developed b first making a reasonable guess or estimate of the result prior to the calculation and then comparing the estimate with the calculated value.

20 c03.qd 11/6/07 3:26 M age 127 rticle 3/3 Equilibrium onditions 127 Sample roblem 3/1 etermine the magnitudes of the forces and T, which, along with the other three forces shown, act on the bridge-truss joint. Solution. The given sketch constitutes the free-bod diagram of the isolated section of the joint in question and shows the five forces which are in equilibrium kn T Solution I (scalar algebra). For the - aes as shown we have [ΣF 0] 8 T cos 40 sin kn 40 8 kn 0.766T [ΣF 0] T sin 40 cos T Simultaneous solution of Eqs. (a) and (b) produces (a) (b) Helpful Hints Since this is a problem of concurrent forces, no moment equation is necessar. T 9.09 kn 3.03 kn ns. Solution II (scalar algebra). To avoid a simultaneous solution, we ma use aes - with the first suation in the -direction to eliminate reference to T. Thus, [ΣF 0] cos 20 3 cos 40 8 sin sin kn ns. [ΣF 0] T 8 cos cos 40 3 sin sin 20 0 T 9.09 kn ns. The selection of reference aes to facilitate computation is alwas an important consideration. lternativel in this eample we could take a set of aes along and normal to the direction of and emplo a force suation normal to to eliminate it. Solution III (vector algebra). With unit vectors i and j in the - and -directions, the ero suation of forces for equilibrium ields the vector equation [ΣF 0] 8i (T cos 40 )i (T sin 40 )j 3j ( sin 20 )i ( cos 20 )j 16i 0 Equating the coefficients of the i- and j-terms to ero gives 8 T cos 40 sin T sin 40 3 cos 20 0 which are the same, of course, as Eqs. (a) and (b), which we solved above. Solution IV (geometric). The polgon representing the ero vector sum of the five forces is shown. Equations (a) and (b) are seen iediatel to give the projections of the vectors onto the - and -directions. Similarl, projections onto the - and -directions give the alternative equations in Solution II. graphical solution is easil obtained. The known vectors are laid off headto-tail to some convenient scale, and the directions of T and are then drawn to close the polgon. The resulting intersection at point completes the solution, thus enabling us to measure the magnitudes of T and directl from the drawing to whatever degree of accurac we incorporate in the construction. T kn 8 kn The known vectors ma be added in an order desired, but the must be added before the unknown vectors. 3 kn

21 c03.qd 11/6/07 3:26 M age hapter 3 Equilibrium Sample roblem 3/2 alculate the tension T in the cable which supports the 500-kg mass with the pulle arrangement shown. Each pulle is free to rotate about its bearing, and the weights of all parts are small compared with the load. Find the magnitude of the total force on the bearing of pulle. T θ = 30 Solution. The free-bod diagram of each pulle is drawn in its relative position to the others. We begin with pulle, which includes the onl known force. With the unspecified pulle radius designated b r, the equilibrium of moments about its center and the equilibrium of forces in the vertical direction require [ΣM 0] [ΣF 0] T 1 r T 2 r 0 T 1 T 2 500(9.81) 0 T 1 T 2 2T 1 500(9.81) From the eample of pulle we ma write the equilibrium of forces on pulle b inspection as T 3 T 4 T 2 / N T 1 T N 500 kg T 30 T 3 F T 4 F For pulle the angle 30 in no wa affects the moment of T about the center of the pulle, so that moment equilibrium requires T T 3 or T 1226 N Equilibrium of the pulle in the - and -directions requires ns. T T (9.81) N [ΣF 0] 1226 cos 30 F 0 F 1062 N Helpful Hint [ΣF 0] F 1226 sin [F 2 F F 2 ] F (1062) 2 (613) N F 613 N ns. learl the radius r does not influence the results. nce we have analed a simple pulle, the results should be perfectl clear b inspection. Sample roblem 3/3 The uniform 100-kg I-beam is supported initiall b its end rollers on the horiontal surface at and. means of the cable at it is desired to elevate end to a position 3 m above end. etermine the required tension, the reaction at, and the angle made b the beam with the horiontal in the elevated position. 6 m 2 m Solution. In constructing the free-bod diagram, we note that the reaction on the roller at and the weight are vertical forces. onsequentl, in the absence of other horiontal forces, must also be vertical. From Sample roblem 3/2 we see iediatel that the tension in the cable equals the tension applied to the beam at. Moment equilibrium about eliminates force R and gives [ΣM 0] (6 cos ) 981(4 cos ) N ns. Equilibrium of vertical forces requires [ΣF 0] 654 R R 327 N The angle depends onl on the specified geometr and is sin 3/ ns. ns. R 4 m Helpful Hint θ 2 m 2 m 100(9.81) N 3 m learl the equilibrium of this parallel force sstem is independent of.

22 c03.qd 11/6/07 3:26 M age 129 T = kn rticle 3/3 Equilibrium onditions 129 Sample roblem 3/4 etermine the magnitude T of the tension in the supporting cable and the magnitude of the force on the pin at for the jib crane shown. The beam is a standard 0.5-m I-beam with a mass of 95 kg per meter of length. lgebraic solution. The sstem is setrical about the vertical - plane through the center of the beam, so the problem ma be analed as the equilibrium of a coplanar force sstem. The free-bod diagram of the beam is shown in the figure with the pin reaction at represented in terms of its two rectangular components. The weight of the beam is 95(10 3 )(5) kn and acts through its center. Note that there are three unknowns,, and T, which ma be found from the three equations of equilibrium. We begin with a moment equation about, which eliminates two of the three unknowns from the equation. In appling the moment equation about, it is simpler to consider the mo- ments of the - and -components of T than it is to compute the perpendicular distance from T to. Hence, with the counterclockwise sense as positive we write [ΣM 0] from which Equating the sums of forces in the - and -directions to ero gives [ΣF 0] [ΣF 0] (T cos 25 )0.25 (T sin 25 )(5 0.12) 10( ) 4.66( ) cos 25 0 T kn sin [ 2 2 ] (17.77) 2 (6.37) kn kn 6.37 kn ns. ns. Graphical solution. The principle that three forces in equilibrium must be concurrent is utilied for a graphical solution b combining the two known vertical forces of 4.66 and 10 kn into a single kN force, located as shown on the modified free-bod diagram of the beam in the lower figure. The position of this resultant load ma easil be determined graphicall or algebraicall. The intersection of the kN force with the line of action of the unknown tension T defines the point of concurrenc through which the pin reaction must pass. The unknown magnitudes of T and ma now be found b adding the forces head-to-tail to form the closed equilibrium polgon of forces, thus satisfing their ero vector sum. fter the known vertical load is laid off to a convenient scale, as shown in the lower part of the figure, a line representing the given direction of the tension T is drawn through the tip of the kN vector. Likewise a line representing the direction of the pin reaction, determined from the concurrenc established with the free-bod diagram, is drawn through the tail of the kN vector. The intersection of the lines representing vectors T and establishes the magnitudes T and necessar to make the vector sum of the forces equal to ero. These magnitudes are scaled from the diagram. The - and -components of ma be constructed on the force polgon if desired m 0.5 m Helpful Hints 0.12 m = kn 5 m 4.66 kn 10 kn T kn kn kn 10 kn Free-bod diagram 1.5 m 25 The justification for this step is Varignon s theorem, eplained in rt. 2/4. e prepared to take full advantage of this principle frequentl. The calculation of moments in twodimensional problems is generall handled more simpl b scalar algebra than b the vector cross product r F. In three dimensions, as we will see later, the reverse is often the case. The direction of the force at could be easil calculated if desired. However, in designing the pin or in checking its strength, it is onl the magnitude of the force that matters. T 10 kn Graphical solution

23 c03.qd 11/6/07 3:26 M age hapter 3 Equilibrium RLEMS Introductor roblems 3/1 etermine the force required to maintain the 200-kg engine in the position for which 30. The diameter of the pulle at is negligible. ns N 3/3 carpenter carries a 6-kg uniform board as shown. What downward force does he feel on his shoulder at? ns. N 88.3 N 1.5 m 0.6 m 0.3 m 2 m 2 m roblem 3/3 roblem 3/1 200 kg 3/4 In the side view of a 70-kg television resting on a 24-kg cabinet, the mass centers are labeled G 1 and G 2. etermine the force reactions at and. (Note that the mass center of most televisions is located well forward because of the heav nature of the front portion of picture tubes.) 3/2 The mass center G of the 1400-kg rear-engine car is located as shown in the figure. etermine the normal force under each tire when the car is in equilibrium. State an assumptions. G 2 G G roblem 3/ roblem 3/4

24 c03.qd 11/6/07 3:26 M age 131 rticle 3/3 roblems 131 3/5 The roller stand is used to support portions of long boards as the are being cut on a table saw. If the board eerts a 25-N downward force on the roller, determine the vertical reactions at and. Note that the connection at is rigid, and that the feet and are fairl length horiontal tubes with a nonslip coating. ns. N 8.45 N, N N G 55 kg 36 kg 36 kg 5 m 1 m 4 m 7 m kg roblem 3/7 3/8 The 20-kg homogeneous smooth sphere rests on the two inclines as shown. etermine the contact forces at and. 355 roblem 3/ roblem 3/8 3/6 The 450-kg uniform I-beam supports the load shown. etermine the reactions at the supports. 5.6 m 2.4 m 3/9 54-kg crate resets on the 27-kg pickup tailgate. alculate the tension T in each of the two restraining cables, one of which is shown. The centers of gravit are at G 1 and G 2. The crate is located midwa between the two cables. ns. T 577 N 350 roblem 3/6 220 kg 3/7 alculate the force and moment reactions at the bolted base of the overhead traffic-signal assembl. Each traffic signal has a mass of 36 kg, while the masses of members and are 50 kg and 55 kg, respectivel. ns. 0, 1736 N, M 7460 N m W G 2 70 G roblem 3/9 300

25 c03.qd 11/6/07 3:26 M age hapter 3 Equilibrium 3/10 portable electric generator has a mass of 160 kg with mass center at G. etermine the upward force F necessar to reduce the normal force at to onehalf its nominal (F 0) value. 3/12 The device shown is designed to aid in the removal of pull-tab tops from cans. If the user eerts a 40-N force at, determine the tension T in the portion of the pull tab. F 10 G 78 F = 40N roblem 3/10 3/11 With what force magnitude T must the person pull on the cable in order to cause the scale to read 2000 N? The weights of the pulles and cables are negligible. State an assumptions. ns. T 581 N roblem 3/12 3/13 woodcutter wishes to cause the tree trunk to fall uphill, even though the trunk is leaning downhill. With the aid of the winch W, what tension T in the cable will be required? The 600-kg trunk has a center of gravit at G. The felling notch at is sufficientl large so that the resisting moment there is negligible. ns. T 401 N m 1.3 m 4 m G 10 Horiontal W 500 kg roblem 3/11 roblem 3/13

26 c03.qd 11/6/07 3:26 M age 133 rticle 3/3 roblems 133 3/14 To facilitate shifting the position of a lifting hook when it is not under load, the sliding hanger shown is used. The projections at and engage the flanges of a bo beam when a load is supported, and the hook projects through a horiontal slot in the beam. ompute the forces at and when the hook supports a 300-kg mass / N aial force is required to remove the pulle from its shaft. What force F must be eerted on the handle of each of the two prbars? Friction at the contact points and E is sufficient to prevent slipping; friction at the pulle contact points and F is negligible. F E kg roblem 3/14 F 5 F 3/15 Three cables are joined at the junction ring. etermine the tensions in cables and caused b the weight of the 30-kg clinder. ns. T 215 N, T 264 N roblem 3/16 3/17 The uniform beam has a mass of 50 kg per meter of length. ompute the reactions at the support. The force loads shown lie in a vertical plane. ns. 0.7 kn, 5.98 kn, M 9.12 kn m kn 4 kn m 1.4 kn kg 1.8 m 0.6 m m 0.6 m roblem 3/17 roblem 3/15

27 c03.qd 11/6/07 3:26 M age hapter 3 Equilibrium Representative roblems 3/18 pipe is being bent b the pipe bender as shown. If the hdraulic clinder applies a force of magnitude F 24 kn to the pipe at, determine the magnitude of the roller reactions at and. 3/20 etermine the reactions at and E if 500 N. What is the maimum value which ma have for static equilibrium? Neglect the weight of the structure compared with the applied loads N m E F 4 m 4 m roblem 3/20 roblem 3/18 3/19 The uniform 15-m pole has a mass of 150 kg and is supported b its smooth ends against the vertical walls and b the tension T in the vertical cable. ompute the reactions at and. ns. 327 N 3/21 While digging a small hole prior to planting a tree, a homeowner encounters rocks. If he eerts a horiontal 225-N force on the prbar as shown, what is the horiontal force eerted on rock? Note that a small ledge on rock supports a vertical force reaction there. Neglect friction at. omplete solutions (a) including and (b) ecluding the weight of the 18-kg prbar. ns. (a) F 1705 N, (b) F 1464 N 225 N T 10 m 5 m m 20 roblem 3/ roblem 3/21

28 c03.qd 11/6/07 3:26 M age 135 rticle 3/3 roblems 135 3/22 etermine the force required to begin rolling the uniform clinder of mass m over the obstruction of height h. 3/24 person holds a 30-kg suitcase b its handle as indicated in the figure. etermine the tension in each of the four identical links. 30(9.81) N r h roblem 3/ /23 35-N aial force at is required to open the springloaded plunger of the water nole. etermine the required force F applied to the handle at and the magnitude of the pin reaction at. Note that the plunger passes through a verticall-elongated hole in the handle at, so that negligible vertical force is transmitted there. ns. F N, 48.8 N 18 roblem 3/24 3/25 block placed under the head of the claw haer as shown greatl facilitates the etraction of the nail. If a 200-N pull on the handle is required to pull the nail, calculate the tension T in the nail and the magnitude of the force eerted b the haer head on the block. The contacting surfaces at are sufficientl rough to prevent slipping. ns. T 800 N, 755 N N F F roblem 3/ roblem 3/25

29 c03.qd 11/6/07 3:26 M age hapter 3 Equilibrium 3/26 The indicated location of the center of mass of the 1600-kg pickup truck is for the unladen condition. If a load whose center of mass is 400 behind the rear ale is added to the truck, determine the mass m L for which the normal forces under the front and rear wheels are equal. G W L 3/28 To test the validit of aerodnamic assumptions made in the design of the aircraft, its model is being tested in a wind tunnel. The support bracket is connected to a force and moment balance, which is eroed when there is no airflow. Under test conditions, the lift L, drag, and pitching moment M G act as shown. The force balance records the lift, drag, and a moment M. etermine M G in terms of L,, and M. d L irflow h G M G roblem 3/26 3/27 The wall-mounted 2.5-kg light fiture has its mass center at G. etermine the reactions at and and also calculate the moment supported b the adjustment thumbscrew at. (Note that the lightweight frame has about 250 of horiontal tubing, directed into and out of the paper, at both and.) ns N 32.0 N, 24.5 N M 2.45 N m W roblem 3/28 3/29 The chain binder is used to secure loads of logs, lumber, pipe, and the like. If the tension T 1 is 2 kn when 30, determine the force required on the lever and the corresponding tension T 2 for this position. ssume that the surface under is perfectl smooth. ns N, T kn 500 G 230 T T 2 roblem 3/27 roblem 3/29

30 c03.qd 11/6/07 3:26 M age 137 rticle 3/3 roblems 137 3/30 The device shown is designed to appl pressure when bonding laminate to each side of a countertop near an edge. If a 120-N force is applied to the handle, determine the force which each roller eerts on its corresponding surface. Humerus Triceps 120 N G Ulna Hand Load cell 30 roblem 3/32 roblem 3/30 3/31 The two light pulles are fastened together and form an integral unit. The are prevented from turning about their bearing at b a cable wound securel around the smaller pulle and fastened to point. alculate the magnitude R of the force supported b the bearing for the applied 2-kN load. ns. R 4.38 kn /33 person is performing slow arm curls with a 10-kg weight as indicated in the figure. The brachialis muscle group (consisting of the biceps and brachialis muscles) is the major factor in this eercise. etermine the magnitude F of the brachialis-musclegroup force and the magnitude E of the elbow joint reaction at point E for the forearm position shown in the figure. Take the dimensions shown to locate the effective points of application of the two muscle groups; these points are 200 directl above E and 50 directl to the right of E. Include the effect of the 1.5-kg forearm mass with mass center at point G. State an assumptions. ns. F 753 N, E 644 N Humerus iceps rachialis 2 kn roblem 3/ /32 In a procedure to evaluate the strength of the triceps muscle, a person pushes down on a load cell with the palm of his hand as indicated in the figure. If the load-cell reading is 160 N, determine the vertical tensile force F generated b the triceps muscle. The mass of the lower arm is 1.5 kg with mass center at G. State an assumptions. 200 E G roblem 3/33 Ulna Radius 10 kg

31 c03.qd 11/6/07 3:26 M age hapter 3 Equilibrium 3/34 woman is holding a 3.6-kg sphere in her hand with the entire arm held horiontall as shown in the figure. tensile force in the deltoid muscle prevents the arm from rotating about the shoulder joint ; this force acts at the 21 angle shown. etermine the force eerted b the deltoid muscle on the upper arm at and the - and -components of the force reaction at the shoulder joint. The mass of the upper arm is m U 1.9 kg, the mass of the lower arm is m L 1.1 kg, and the mass of the hand is m H 0.4 kg; all the corresponding weights act at the locations shown in the figure. Femur Fibula Tibia Quadriceps muscle atella atellar tendon 3.6(9.81) N eltoid muscle roblem 3/35 F W U W L W H 3/36 The elements of an on-off mechanism for a table lamp are shown in the figure. The electrical switch S requires a 4-N force in order to depress it. What corresponding force F must be eerted on the handle at? roblem 3/ /35 With his weight W equall distributed on both feet, a man begins to slowl rise from a squatting position as indicated in the figure. etermine the tensile force F in the patellar tendon and the magnitude of the force reaction at point, which is the contact area between the tibia and the femur. Note that the line of action of the patellar tendon force is along its midline. Neglect the weight of the lower leg. ns. F 2.25W, 2.67W 30 S 30 roblem 3/ F

32 c03.qd 11/6/07 3:27 M age 139 rticle 3/3 roblems 139 3/37 The uniform 18-kg bar is held in the position shown b the smooth pin at and the cable. etermine the tension T in the cable and the magnitude and direction of the eternal pin reaction at. ns. T 99.5 N, 246 N, 70.3 W from -ais 1.5 m 3/39 The eercise machine is designed with a lightweight cart which is mounted on small rollers so that it is free to move along the inclined ramp. Two cables are attached to the cart one for each hand. If the hands are together so that the cables are parallel and if each cable lies essentiall in a vertical plane, determine the force which each hand must eert on its cable in order to maintain an equilibrium position. The mass of the person is 70 kg, the ramp angle is 15, and the angle is 18. In addition, calculate the force R which the ramp eerts on the cart. ns N, R 691 N 60 β roblem 3/37 3/38 person attempts to move a 20-kg shop vacuum b pulling on the hose as indicated. What force F will cause the unit to tip clockwise if wheel is against an obstruction? G 1.2 m F roblem 3/39 3/40 The device shown is used to test automobile-engine valve springs. The torque wrench is directl connected to arm. The specification for the automotive intake-valve spring is that 370 N of force should reduce its length from 50 (unstressed length) to 42. What is the corresponding reading M on the torque wrench, and what force F eerted on the torque-wrench handle is required to produce this reading? Neglect the small effects of changes in the angular position of arm. θ F roblem 3/ roblem 3/40

33 c03.qd 11/6/07 3:27 M age hapter 3 Equilibrium 3/41 uring an engine test on the ground, a propeller thrust T 3000 N is generated on the 1800-kg airplane with mass center at G. The main wheels at are locked and do not skid; the small tail wheel at has no brake. ompute the percent change n in the normal forces at and as compared with their engine-off values. ns. n 32.6%, n 2.28% 3/43 The hook wrench or pin spanner is used to turn shafts and collars. If a moment of 80 N m is required to turn the 200--diameter collar about its center under the action of the applied force, determine the contact force R on the smooth surface at. Engagement of the pin at ma be considered to occur at the peripher of the collar. ns. R 1047 N G m T m 0.8 m roblem 3/43 roblem 3/41 3/42 rocker arm with rollers at and is shown in the position when the valve is open and the valve spring is full compressed. In this position, the spring force is 900 N. etermine the force which the rocker arm eerts on the camshaft. lso calculate the magnitude of the force supported b the rocker-arm shaft. 3/44 The doll shown is useful in the handling of large drums. etermine the force F necessar to hold a drum in the position shown. You ma neglect the weight of the doll in comparison with that of the 250-kg drum, whose center of mass is at G. There is sufficient friction to prevent slipping at the contact point F G 150 roblem 3/ roblem 3/44

34 c03.qd 11/6/07 3:27 M age 141 rticle 3/3 roblems 141 3/45 In sailing at a constant speed with the wind, the sailboat is driven b a 4-kN force against its mainsail and a 1.6-kN force against its stasail as shown. The total resistance due to fluid friction through the water is the force R. etermine the resultant of the lateral forces perpendicular to motion applied to the hull b the water. ns. M 9.6 kn m 5 8 N R 1.6 kn 1.5 m 3 m Wind Slip tube 4 kn roblem 3/47 roblem 3/45 3/46 Estimate the force F required to lift the rear tires of the race car off the ground. You ma assume that part of of the doll jack is horiontal. The mass of the car and the driver combined is 700 kg with mass center at G. The driver applies the brakes during the jacking. State an additional assumptions. G E F /48 The small sailboat ma be tipped at its moorings as shown to effect repairs below the waterline. ne attached rope is passed under the keel and secured to the dock. The other rope is attached to the mast and is used to tip the boat. The boat shown has a displacement (which equals the total mass) of 5000 kg with mass center at G. The metacenter M is the point on the centerline of the boat through which the vertical resultant of the buoant forces passes, and GM 0.8 m. alculate the tension T required to hold the boat in the position shown. 6 m T roblem 3/46 3/47 portion of the shifter mechanism for a manual car transmission is shown in the figure. For the 8-N force eerted on the shift knob, determine the corresponding force eerted b the shift link on the transmission (not shown). Neglect friction in the ball-and-socket joint at, in the joint at, and in the slip tube near support. Note that a soft rubber bushing at allows the slip tube to self-align with link. ns N M 90 G roblem 3/48

35 c03.qd 11/6/07 3:27 M age hapter 3 Equilibrium 3/49 torque (moment) of 24 N m is required to turn the bolt about its ais. etermine and the forces between the smooth hardened jaws of the wrench and the corners and of the heagonal head. ssume that the wrench fits easil on the bolt so that contact is made at corners and onl. ns. 200 N, 2870 N, 3070 N etail L imensions in millimeters 1 ir flow 125 roblem 3/51 roblem 3/49 3/50 etermine the moment M which must be applied to the shaft in order to hold the homogeneous hemisphere in an arbitrar angular position as measured b the angle. The radii of gear, gear, and the hemisphere are r, r, and r, respectivel. ssume the friction in all bearings to be negligible. 3/52 To test the deflection of the uniform 100-kg beam the 50-kg bo eerts a pull of 150 N on the rope rigged as shown. ompute the force supported b the pin at the hinge. 1.5 m r m 0.75 m 2.5 m 0.75 m M roblem 3/50 3/51 The car complete with driver has a mass of 815 kg and without the two airfoils has a 50% 50% front rear weight distribution at a certain speed at which there is no lift on the car. It is estimated that at this speed each of the airfoils 1 and 2 will generate 2 kn of downward force L and 250 N of drag force on the car. Specif the vertical reactions N and N under the two pairs of wheels at that speed when the airfoils are added. ssume that the addition of the airfoils does not affect the drag and erolift conditions of the car bod itself and that the engine has sufficient power for equilibrium at that speed. The weight of the airfoils ma be neglected. ns. N 5750 N (48.0%), N 6240 N (52.0%) roblem 3/52 3/53 etermine the eternal reactions at and F for the roof truss loaded as shown. The vertical loads represent the effect of the supported roofing materials, while the 400-N force represents a wind load. ns. 346 N, 1100 N, F 1100 N 400 N 250 N N N 500 N E G 10 m 250 N F roblem 3/53

36 c03.qd 11/6/07 3:27 M age 143 rticle 3/3 roblems 143 3/54 The member and sheave at together have a mass of 500 kg, with a combined center of mass at G. alculate the magnitude of the force supported b the pin connection at when the 3-kN load is applied. The collar at can provide support in the horiontal direction onl. 3 kn m 4.5 m G 1.5 m 0.5 m 1 m 1 m 1 m 3/56 The man pushes the lawn mower at a stead speed with a force which is parallel to the incline. The mass of the mower with attached grass bag is 50 kg with mass center at G. If 15, determine the normal forces N and N under each pair of wheels and. Neglect friction. ompare with the normal forces for the conditions of 0 and 0. θ G roblem 3/54 3/55 It is desired that a person be able to begin closing the van hatch from the open position shown with a 40-N vertical force. s a design eercise, determine the necessar force in each of the two hdraulic struts. The mass center of the 40-kg door is 37.5 directl below point. Treat the problem as two-dimensional. ns. F 803 N roblem 3/56 3/57 etermine the tension T in the turnbuckle for the pulle cable sstem in terms of the mass m of the bod which it supports. Neglect the mass of the pulles and cable. ns. T 2 7 mg Hinge ais Strut detail T m roblem 3/57 roblem 3/55

37 c03.qd 11/6/07 3:27 M age hapter 3 Equilibrium 3/58 The cargo door for an airplane of circular fuselage section consists of the uniform quarter-circular segment of mass m. detent in the hinge at holds the door open in the position shown. etermine the moment eerted b the hinge on the door. Hori. r 30 3/60 ertain elements of an in-refrigerator ice-cube maker are shown in the figure. ( cube has the form of a clindrical segment!) nce the cube freees and a small heater (not shown) forms a thin film of water between the cube and supporting surface, a motor rotates the ejector arm to remove the cube. If there are eight cubes and eight arms, determine the required torque M as a function of. The mass of eight cubes is 0.25 kg, and the center-ofmass distance r 0.55r. Neglect friction, and assume that the resultant of the distributed normal force acting on the cube passes through point. 8 losed position of roblem 3/58 3/59 ulle delivers a stead torque (moment) of 100 N m to a pump through its shaft at. The tension in the lower side of the belt is 600 N. The driving motor has a mass of 100 kg and rotates clockwise. s a design consideration, determine the magnitude R of the force on the supporting pin at. ns. R kn r G r = 37 M roblem 3/ N roblem 3/59

38 c03.qd 11/6/07 3:27 M age 145 rticle 3/4 Equilibrium onditions 145 SETIN EQUILIRIUM IN THREE IMENSINS 3/4 EQUILIRIUM NITINS We now etend our principles and methods developed for twodimensional equilibrium to the case of three-dimensional equilibrium. In rt. 3/1 the general conditions for the equilibrium of a bod were stated in Eqs. 3/1, which require that the resultant force and resultant couple on a bod in equilibrium be ero. These two vector equations of equilibrium and their scalar components ma be written as F 0 or F 0 F 0 F 0 M 0 or M 0 M 0 M 0 (3/3) The first three scalar equations state that there is no resultant force acting on a bod in equilibrium in an of the three coordinate directions. The second three scalar equations epress the further equilibrium requirement that there be no resultant moment acting on the bod about an of the coordinate aes or about aes parallel to the coordinate aes. These si equations are both necessar and sufficient conditions for complete equilibrium. The reference aes ma be chosen arbitraril as a matter of convenience, the onl restriction being that a right-handed coordinate sstem should be chosen when vector notation is used. The si scalar relationships of Eqs. 3/3 are independent conditions because an of them can be valid without the others. For eample, for a car which accelerates on a straight and level road in the -direction, Newton s second law tells us that the resultant force on the car equals its mass times its acceleration. Thus ΣF 0, but the remaining two force equilibrium equations are satisfied because all other acceleration components are ero. Similarl, if the flwheel of the engine of the accelerating car is rotating with increasing angular speed about the -ais, it is not in rotational equilibrium about this ais. Thus, for the flwheel alone, ΣM 0 along with ΣF 0, but the remaining four equilibrium equations for the flwheel would be satisfied for its mass-center aes. In appling the vector form of Eqs. 3/3, we first epress each of the forces in terms of the coordinate unit vectors i, j, and k. For the first equation, ΣF 0, the vector sum will be ero onl if the coefficients of i, j, and k in the epression are, respectivel, ero. These three sums, when each is set equal to ero, ield precisel the three scalar equations of equilibrium, ΣF 0, ΣF 0, and ΣF 0. For the second equation, ΣM 0, where the moment sum ma be taken about an convenient point, we epress the moment of each force as the cross product r F, where r is the position vector from to an point on the line of action of the force F. Thus ΣM Σ(r F) 0. When the coefficients of i, j, and k in the resulting moment equation are

39 c03.qd 11/6/07 3:27 M age hapter 3 Equilibrium set equal to ero, respectivel, we obtain the three scalar moment equations ΣM 0, ΣM 0, and ΣM 0. Free-od iagrams The suations in Eqs. 3/3 include the effects of all forces on the bod under consideration. We learned in the previous article that the free-bod diagram is the onl reliable method for disclosing all forces and moments which should be included in our equilibrium equations. In three dimensions the free-bod diagram serves the same essential purpose as it does in two dimensions and should alwas be drawn. We have our choice either of drawing a pictorial view of the isolated bod with all eternal forces represented or of drawing the orthogonal projections of the free-bod diagram. oth representations are illustrated in the sample problems at the end of this article. The correct representation of forces on the free-bod diagram requires a knowledge of the characteristics of contacting surfaces. These characteristics were described in Fig. 3/1 for two-dimensional problems, and their etension to three-dimensional problems is represented in Fig. 3/8 for the most coon situations of force transmission. The representations in both Figs. 3/1 and 3/8 will be used in three-dimensional analsis. The essential purpose of the free-bod diagram is to develop a reliable picture of the phsical action of all forces (and couples if an) acting on a bod. So it is helpful to represent the forces in their correct phsical sense whenever possible. In this wa, the free-bod diagram becomes a closer model to the actual phsical problem than it would be if the forces were arbitraril assigned or alwas assigned in the same mathematical sense as that of the assigned coordinate ais. For eample, in part 4 of Fig. 3/8, the correct sense of the unknowns R and R ma be known or perceived to be in the sense opposite to those of the assigned coordinate aes. Similar conditions appl to the sense of couple vectors, parts 5 and 6, where their sense b the righthand rule ma be assigned opposite to that of the respective coordinate direction. this time, ou should recognie that a negative answer for an unknown force or couple vector merel indicates that its phsical action is in the sense opposite to that assigned on the freebod diagram. Frequentl, of course, the correct phsical sense is not known initiall, so that an arbitrar assignment on the free-bod diagram become necessar. ategories of Equilibrium pplication of Eqs. 3/3 falls into four categories which we identif with the aid of Fig. 3/9. These categories differ in the number and tpe (force or moment) of independent equilibrium equations required to solve the problem. ategor 1, equilibrium of forces all concurrent at point, requires all three force equations, but no moment equations because the moment of the forces about an ais through is ero. ategor 2, equilibrium of forces which are concurrent with a line, requires all equations ecept the moment equation about that line, which is automaticall satisfied.

40 c03.qd 11/6/07 3:27 M age 147 rticle 3/4 Equilibrium onditions 147 MELING THE TIN F FRES IN THREE-IMENSINL NLYSIS Tpe of ontact and Force rigin 1. Member in contact with smooth surface, or ball-supported member ction on od to e Isolated Force must be normal to the surface and directed toward the member. 2. Member in contact with rough surface 3. Roller or wheel support with lateral constraint N N N F The possibilit eists for a force F tangent to the surface (friction force) to act on the member, as well as a normal force N. lateral force eerted b the guide on the wheel can eist, in addition to the normal force N. 4. all-and-socket joint R R R ball-and-socket joint free to pivot about the center of the ball can support a force R with all three components. 5. Fied connection (embedded or welded) R R M In addition to three components of force, a fied connection can support a couple M represented b its three components. M R M 6. Thrust-bearing support R M R R M Thrust bearing is capable of supporting aial force R as well as radial forces R and R. ouples M and M must, in some cases, be assumed ero in order to provide statical determinac. Figure 3/8

41 c03.qd 11/6/07 3:27 M age hapter 3 Equilibrium TEGRIES F EQUILIRIUM IN THREE IMENSINS Force Sstem Free-od iagram Independent Equations 1. oncurrent F at a point 1 F 2 ΣF = 0 F 5 F 4 F 3 ΣF = 0 ΣF = 0 2. oncurrent with a line F 2 F 1 ΣF = 0 ΣM = 0 ΣF = 0 ΣM = 0 F 3 ΣF = 0 F 5 F 4 3. arallel F 1 F2 ΣF = 0 ΣM = 0 ΣM = 0 F 5 F 3 F 4 4. General F 1 F 2 M ΣF = 0 ΣM = 0 F 4 ΣF = 0 ΣM = 0 ΣF = 0 ΣM = 0 F3 Figure 3/9 ategor 3, equilibrium of parallel forces, requires onl one force equation, the one in the direction of the forces (-direction as shown), and two moment equations about the aes ( and ) which are normal to the direction of the forces. ategor 4, equilibrium of a general sstem of forces, requires all three force equations and all three moment equations. The observations contained in these statements are generall quite evident when a given problem is being solved. onstraints and Statical eterminac The si scalar relations of Eqs. 3/3, although necessar and sufficient conditions to establish equilibrium, do not necessaril provide all of the information required to calculate the unknown forces acting in a three-dimensional equilibrium situation. gain, as we found with two dimensions, the question of adequac of information is decided b the

42 c03.qd 11/6/07 3:27 M age 149 rticle 3/4 Equilibrium onditions 149 characteristics of the constraints provided b the supports. n analtical criterion for determining the adequac of constraints is available, but it is beond the scope of this treatment.* In Fig. 3/10, however, we cite four eamples of constraint conditions to alert the reader to the problem. art a of Fig. 3/10 shows a rigid bod whose corner point is completel fied b the links 1, 2, and 3. Links 4, 5, and 6 prevent rotations about the aes of links 1, 2, and 3, respectivel, so that the bod is completel fied and the constraints are said to be adequate. art b of the figure shows the same number of constraints, but we see that the provide no resistance to a moment which might be applied about ais E. Here the bod is incompletel fied and onl partiall constrained. Similarl, in Fig. 3/10c the constraints provide no resistance to an unbalanced force in the -direction, so here also is a case of incomplete fiit with partial constraints. In Fig. 3/10d, if a seventh constraining link were imposed on a sstem of si constraints placed properl for complete fiit, more supports would be provided than would be necessar to establish the equilibrium position, and link 7 would be redundant. The bod would then be staticall indeterminate with such a seventh link in place. With onl a few eceptions, the supporting constraints for rigid bodies in equilibrium in this book are adequate, and the bodies are staticall determinate (a) omplete fiit dequate constraints E 6 (b) Incomplete fiit artial constraints (c) Incomplete fiit artial constraints Thomas Ebert/Laif/Redu The equilibrium of structural components such as these shell-like panels is an issue both during and after construction. This structure will be used to house airships. 7 (d) Ecessive fiit Redundant constraints Figure 3/10 *See the first author s Statics, 2nd Edition SI Version, 1975, rt. 16.

43 c03.qd 11/6/07 3:27 M age hapter 3 Equilibrium Sample roblem 3/5 The uniform 7-m steel shaft has a mass of 200 kg and is supported b a balland-socket joint at in the horiontal floor. The ball end rests against the smooth vertical walls as shown. ompute the forces eerted b the walls and the floor on the ends of the shaft. 7 m Solution. The free-bod diagram of the shaft is first drawn where the contact forces acting on the shaft at are shown normal to the wall surfaces. In addition to the weight W mg 200(9.81) 1962 N, the force eerted b the floor on the ball joint at is represented b its -, -, and -components. These components are shown in their correct phsical sense, as should be evident from the requirement that be held in place. The vertical position of is found from h 2, h 3 m. Right-handed coordinate aes are assigned as shown. Vector solution. We will use as a moment center to eliminate reference to the forces at. The position vectors needed to compute the moments about are where the mass center G is located halfwa between and. The vector moment equation gives [ΣM 0] r G 1i 3j 1.5k m and r 2i 6j 3k m r ( ) r G W 0 6 m 2 m h 6 m 3.5 m 3.5 m W = mg G 2 m Equating the coefficients of i, j, and k to ero and solving give 654 N and 1962 N The forces at are easil determined b [ΣF 0] (654 )i (1962 )j ( 1962 )k 0 and 654 N 1962 N 1962 N Finall (654) 2 (1962) 2 (1962) N ns. ns. Scalar solution. Evaluating the scalar moment equations about aes through parallel, respectivel, to the - and -aes, gives [ΣM 0] [ΣM 0] 1962(3) (1) 3 0 The force equations give, simpl, [ΣF 0] [ΣF 0] [ΣF 0] ( 2i 6j 3k) ( i j) ( i 3j 1.5k) ( 1962k) 0 i j k i j k ( )i (3 1962)j ( 2 6 )k N 654 N 654 N 1962 N 1962 N Helpful Hints We could, of course, assign all of the unknown components of force in the positive mathematical sense, in which case and would turn out to be negative upon computation. The freebod diagram describes the phsical situation, so it is generall preferable to show the forces in their correct phsical senses wherever possible. Note that the third equation merel checks the results of the first two equations. This result could be anticipated from the fact that an equilibrium sstem of forces concurrent with a line requires onl two moment equations (ategor 2 under ategories of Equilibrium). We observe that a moment sum about an ais through parallel to the -ais merel gives us 6 2 0, which serves onl as a check as noted previousl. lternativel we could have first obtained from ΣF 0 and then taken our moment equations about aes through to obtain and.

44 c03.qd 11/6/07 3:27 M age 151 rticle 3/4 Equilibrium onditions 151 Sample roblem 3/6 200-N force is applied to the handle of the hoist in the direction shown. The bearing supports the thrust (force in the direction of the shaft ais), while bearing supports onl radial load (load normal to the shaft ais). etermine the mass m which can be supported and the total radial force eerted on the shaft b each bearing. ssume neither bearing to be capable of supporting a moment about a line normal to the shaft ais. Solution. The sstem is clearl three-dimensional with no lines or planes of setr, and therefore the problem must be analed as a general space sstem of forces. scalar solution is used here to illustrate this approach, although a solution using vector notation would also be satisfactor. The free-bod diagram of the shaft, lever, and drum considered a single bod could be shown b a space view if desired, but is represented here b its three orthogonal projections. The 200-N force is resolved into its three components, and each of the three views shows two of these components. The correct directions of and ma be seen b inspection b observing that the line of action of the resultant of the two 70.7-N forces passes between and. The correct sense of the forces and cannot be determined until the magnitudes of the moments are obtained, so the are arbitraril assigned. The - projection of the bearing forces is shown in terms of the sums of the unknown - and -components. The addition of and the weight W mg completes the free-bod diagrams. It should be noted that the three views represent three two-dimensional problems related b the corresponding components of the forces N m Radial bearing Thrust bearing imensions in millimeters 70.7 N 70.7 N N N 70.7 N 70.7 N + + mg = 9.81m mg = 9.81m From the - projection Helpful Hints [ΣM 0] 100(9.81m) 250(173.2) 0 m 44.1 kg From the - projection [ΣM 0] (70.7) 250(70.7) 0 [ΣF 0] The - view gives [ΣM 0] (173.2) 250(44.1)(9.81) 0 ns N 35.4 N 520 N If the standard three views of orthographic projection are not entirel familiar, then review and practice them. Visualie the three views as the images of the bod projected onto the front, top, and end surfaces of a clear plastic bo placed over and aligned with the bod. We could have started with the - projection rather than with the - projection. [ΣF 0] (44.1)(9.81) 0 [ΣF 0] 70.7 N The total radial forces on the bearings become 86.8 N The - view could have followed iediatel after the - view since the determination of and ma be made after m is found. [ r 2 2 ] [ 2 2 ] r (35.4) 2 (86.8) N (35.4) 2 (520) N ns. ns. Without the assumption of ero moment supported b each bearing about a line normal to the shaft ais, the problem would be staticall indeterminate.

45 c03.qd 11/6/07 3:27 M age hapter 3 Equilibrium Sample roblem 3/7 The welded tubular frame is secured to the horiontal - plane b a balland-socket joint at and receives support from the loose-fitting ring at. Under the action of the 2-kN load, rotation about a line from to is prevented b the cable, and the frame is stable in the position shown. Neglect the weight of the frame compared with the applied load and determine the tension T in the cable, the reaction at the ring, and the reaction components at. 2.5 m 3 m 4.5 m 2 kn 6 m Solution. The sstem is clearl three-dimensional with no lines or planes of setr, and therefore the problem must be analed as a general space sstem of forces. The free-bod diagram is drawn, where the ring reaction is shown in terms of its two components. ll unknowns ecept T ma be eliminated b a moment sum about the line. The direction of is specified b the unit 1 vector n The moment of T about (4.5j 6k) 1 (3j 4k). 2 5 is the component in the direction of of the vector moment about the point and equals r 1 T n. Similarl the moment of the applied load F about is r 2 F n. With 46.2 m, the vector epressions for T, F, r 1, and r 2 are The moment equation now becomes [ΣM 0] T T (2i 2.5j 6k) 46.2 r 1 i 2.5j m ( i 2.5j) F 2j kn r 2 2.5i 6k m T 46.2 (2i 2.5j 6k) 1 (3j 4k) 5 (2.5i 6k) (2j) 1 (3j 4k) m 1 m E T F = 2 kn r 2 n r 1 n r 1 T n r 1 T ompletion of the vector operations gives and the components of T become ns. We ma find the remaining unknowns b moment and force suations as follows: [ΣM 0] [ΣM 0] [ΣF 0] [ΣF 0] [ΣF 0] 48T T kn T kn T 2.50 kn 2(2.5) (3) kn 4.5 2(6) 1.042(6) kn T 2.83 kn kn 3.04 kn kn ns. ns. ns. ns. ns. Helpful Hints The advantage of using vector notation in this problem is the freedom to take moments directl about an ais. In this problem this freedom permits the choice of an ais that eliminates five of the unknowns. Recall that the vector r in the epression r F for the moment of a force is a vector from the moment center to an point on the line of action of the force. Instead of r 1, an equall simple choice would be the vector l. The negative signs associated with the -components indicate that the are in the opposite direction to those shown on the free-bod diagram.

46 c03.qd 11/6/07 3:27 M age 153 rticle 3/4 roblems 153 RLEMS Introductor roblems 3/61 etermine the tensions in cables,, and. ns. T 569 N, T 376 N, T 467 N 3/63 The horiontal steel shaft has a mass of 480 kg and is suspended b a vertical cable from and b a second cable which lies in a vertical transverse plane and loops underneath the shaft. alculate the tensions T 1 and T 2 in the cables. ns. T N, T N 1.5 m 2 m 1.5 m 1.5 m 2.5 m 0.5 m 1.25 m T 2 T 2 T 1 3 m 120 kg roblem 3/61 3 m 4 m 1 m 3/62 uniform steel plate 360 square with a mass of 15 kg is suspended in the horiontal plane b the three vertical wires as shown. alculate the tension in each wire. roblem 3/63 3/64 Two steel I-beams, each with a mass of 100 kg, are welded together at right angles and lifted b vertical cables so that the beams remain in a horiontal plane. ompute the tension in each of the cables,, and. 0.9 m m m roblem 3/62 roblem 3/64

47 c03.qd 11/6/07 3:27 M age hapter 3 Equilibrium 3/65 The chain-supported portion of a light fiture is in the shape of part of a spherical shell. If the mass of the glass unit is m, determine the tension T in each of the three chains. ns. T 0.373mg r 3/67 The 600-kg industrial door is a uniform rectangular panel which rolls along the fied rail on its hanger-mounted wheels and. The door is maintained in a vertical plane b the floor-mounted guide roller, which bears against the bottom edge. For the position shown compute the horiontal side thrust on each of the wheels and, which must be accounted for in the design of the brackets. ns. 235 N, 58.9 N 4r m 3 m 0.9 m 3r 3 m m etail of oor Hanger 1.5 m roblem 3/65 3/66 n overhead view of a car is shown in the figure. Two different locations and are considered for a single jack. In each case, the entire right side of the car is lifted just off the ground. etermine the normal reaction forces at and and the vertical jacking force required for the case of each jacking location. onsider the 1600-kg car to be rigid. The mass center G is on the midline of the car. roblem 3/67 3/68 uniform steel ring 600 in diameter has a mass of 50 kg and is lifted b the three cables, each 500 long, attached at points,, and as shown. ompute the tension in each cable. G roblem 3/ roblem 3/68 3/69 three-legged stool is subjected to the load L as shown. determine the vertical force reaction under each leg. Neglect the weight of the stool. ns. N 0.533L, N N 0.233L

48 c03.qd 11/6/07 3:27 M age 155 rticle 3/4 roblems L Representative roblems 3/71 ne of the vertical walls supporting end of the 200-kg uniform shaft of Sample roblem 3/5 is turned through a 30 angle as shown here. End is still supported b the ball-and-socket connection in the horiontal - plane. alculate the magnitudes of the forces and R eerted on the ball end of the shaft b the vertical walls and, respectivel. ns N, R 755 N 120 imensions in millimeters roblem 3/69 3/70 etermine the compression in each of the three legs of the tripod subjected to the vertical 2-kN force. The weight of the legs is negligible compared with the applied load. Solve b using the force equilibrium equation ΣF 0. F = 2 kn 6 m 7 m roblem 3/ /72 The 9-m steel boom has a mass of 600 kg with center of mass at midlength. It is supported b a ball-andsocket joint at and the two cables under tensions T 1 and T 2. The cable which supports the 2000-kg load leads through a sheave (pulle) at and is secured to the vertical - plane at F. alculate the magnitude of the tension T 1. (Hint: Write a moment equation which eliminates all unknowns ecept T 1.) 2 m roblem 3/ E 2 m 6 m 6 m T 1 F F = 3 m 2 m 5 m 4 m T kg roblem 3/72

49 c03.qd 11/6/07 3:27 M age hapter 3 Equilibrium 3/73 The smooth homogeneous sphere rests in the 120 groove and bears against the end plate, which is normal to the direction of the groove. etermine the angle, measured from the horiontal, for which the reaction on each side of the groove equals the force supported b the end plate. ns. 30 3/75 The mass center of the 30-kg door is in the center of the panel. If the weight of the door is supported entirel b the lower hinge, calculate the magnitude of the total force supported b the hinge at. ns N End view of V-groove kg θ Horiontal 1640 roblem 3/73 3/74 The small tripod-like stepladder is useful for supporting one end of a walking board. If F denotes the magnitude of the downward load from such a board (not shown), determine the reaction at each of the three feet,, and. Neglect friction G roblem 3/75 3/76 s part of a check on its design, a lower -arm (part of an automobile suspension) is supported b bearings at and and subjected to the pair of 900-N forces at and. The suspension spring, not shown for clarit, eerts a force F S at E as shown, where E is in plane. etermine the magnitude F S of the spring force and the magnitudes F and F of the bearing forces at and which are perpendicular to the hinge ais. F E roblem 3/ F s E 900 N N roblem 3/76

50 c03.qd 11/6/07 3:27 M age 157 rticle 3/4 roblems 157 3/77 The rigid unit of post, bracket, and motor has a mass of 30 kg with its mass center at G located 300 from the vertical centerline of the post. The post is welded to the fied base at. The motor, which drives a machine through a fleible shaft, turns in the direction indicated and delivers a torque of 200 N m. In addition, a 200-N force is applied to the bracket as shown. etermine the vector epressions for the total force R and moment M applied to the post at b the supporting base. (aution: e careful to assign the torque (couple) which acts on the motor shaft in its correct sense consistent with Newton s third law.) ns. R 200i 294k N, M 61.7j 15k N m 3/79 The 220-kg V-8 engine is supported on an engine stand and rotated 90 from its upright position so that its center of gravit G is in the position shown. etermine the vertical reaction at each roller of the stand. Neglect the weight of the stand itself. ns. N 1270 N, N N, N 732 N G N G roblem 3/79 roblem 3/77 3/78 etermine the magnitudes of the force R and couple M eerted b the nut and bolt on the loaded bracket at to maintain equilibrium. 3/80 uring a test, the left engine of the twin-engine airplane is revved up and a 2-kN thrust is generated. The main wheels at and are braked in order to prevent motion. etermine the change (compared with the nominal values with both engines off) in the normal reaction forces at,, and. T = 2 kn kn kn 4 m 2.4 m 2 m 2.4 m roblem 3/80 roblem 3/78

51 c03.qd 11/6/07 3:27 M age hapter 3 Equilibrium 3/81 The 25-kg rectangular access door is held in the 90 open position b the single prop. etermine the force F in the prop and the magnitude of the force normal to the hinge ais in each of the small hinges and. ns. F N, n 80.6 N, n 95.4 N imensions in millimeters roblem 3/82 roblem 3/81 3/82 ne of the three landing pads for a proposed Mars lander is shown in the figure. s part of a design check on the distribution of force in the landing struts, compute the force in each of the struts,, and when the lander is resting on a horiontal surface on Mars. The arrangement is setrical with respect to the - plane. The mass of the lander is 600 kg. (ssume equal support b the pads and consult Table /2 in ppendi as needed.) 3/83 Gear drives the V-belt pulle at a constant speed. For the belt tensions shown calculate the gear-tooth force and the magnitudes of the total forces supported b the bearings at and. ns N, 83.3 N, 208 N N N roblem 3/83

52 c03.qd 11/6/07 3:27 M age 159 rticle 3/4 roblems 159 3/84 The spring of modulus k 900 N/m is stretched a distance 60 when the mechanism is in the position shown. alculate the force min required to initiate rotation about the hinge ais, and determine the corresponding magnitudes of the bearing forces which are perpendicular to. What is the normal reaction force at if min /2? 55 k = 900 N/m roblem 3/ /85 force of 200 N on the handle of the cable reel is required to wind up the underground cable as it comes from the manhole. The drum diameter is For the horiontal position of the crank handle shown, calculate the magnitudes of the bearing forces at and. Neglect the weight of the drum. ns N, 313 N 90 3/86 The shaft, lever, and handle are welded together and constitute a single rigid bod. Their combined mass is 28 kg with mass center at G. The assembl is mounted in bearings and, and rotation is prevented b link. etermine the forces eerted on the shaft b bearings and while the 30-N m couple is applied to the handle as shown. Would these forces change if the couple were applied to the shaft rather than to the handle? N m G 220 roblem 3/ /87 Each of the two legs of the welded frame has a mass of 50 kg. wire from to prevents the frame from rotating out of the horiontal plane about an ais through its bearing at and its ball-and-socket joint at. alculate the tension T in the wire and the magnitude of the total force supported b the connection at. ns. T 1201 N, 601 N 2 m 2 m 2 m 45 1 m 2 m imensions in millimeters E roblem 3/85 roblem 3/87

53 c03.qd 11/6/07 3:27 M age hapter 3 Equilibrium 3/88 onsider the rudder assembl of a radio-controlled model airplane. For the 15 position shown in the figure, the net pressure acting on the left side of the rectangular rudder area is p 4(10 5 ) N/ 2. etermine the required force in the control rod E and the horiontal components of the reactions at hinges and which are parallel to the rudder surface. ssume the aerodnamic pressure to be uniform. 3/90 homogeneous door of mass m, height h, and width w is leaned against a wall for painting. Small wooden strips are placed beneath corners,, and. There is negligible friction at, but friction at and is sufficient to prevent slipping. etermine the - and -components of the force reactions at and and the force normal to the wall at E w m h roblem 3/88 3/89 The rigid pole and cross-arms of rob. 2/105 are shown again here. etermine the tensions T E and T GF in the two supporting cables resulting from the 1.2-kN tension in cable. ssume the absence of an resisting moments on the base of the pole at about the - and -aes, but not about the -ais. ns. T E 4.30 kn, T GF 3.47 kn F 1 m imensions in millimeters 2 m G 1 m 1.5 m 1.5 m 3 m 3 m 1.5 m E T = 1.2 kn roblem 3/90 3/91 The upper ends of the vertical coil springs in the stock racecar can be moved up and down b means of a screw mechanism not shown. This adjustment permits a change in the downward force at each wheel as an optimum handling setup is sought. Initiall, scales indicate the normal forces to be 3600 N, 3600 N, 4500 N, and 4500 N at,,, and, respectivel. If the top of the right rear spring at is lowered so that the scale at reads an additional 450 N, determine the corresponding changes in the normal forces at,, and. Neglect the effects of the small attitude changes (pitch and roll angles) caused b the spring adjustment. The front wheels are the same distance apart as the rear wheels. ns. N 450 N, N 450 N N 450 N roblem 3/89

54 c03.qd 11/6/07 3:27 M age 161 rticle 3/4 roblems 161 Simplified spring detail plane and applied at the midpoint M of the boom. Neglect the weight of the boom. ns. T kn, T kn 3 m 20 T 1 10 m M 20 kn roblem 3/91 3/92 uniform bar of length b and mass m is suspended at its ends b two wires, each of length b, from points and in the horiontal plane a distance b apart. couple M is applied to the bar causing it to rotate about a vertical ais to the equilibrium position shown. erive an epression for the height h which it rises from its original equilibrium position where it hangs freel with no applied moment. What value of M is required to raise the bar the maimum amount b? b b 2 m 4 m 4 m roblem 3/93 3/94 rectangular sign over a store has a mass of 100 kg, with the center of mass in the center of the rectangle. The support against the wall at point ma be treated as a ball-and-socket joint. t corner support is provided in the -direction onl. alculate the tensions T 1 and T 2 in the supporting wires, the total force supported at, and the lateral force R supported at. T 2 2 m 1.5 m 1.5 m b M b T 1 T 2.5 m m roblem 3/92 1 m 1 m 3/93 The boom lies in the vertical - plane and is supported b a ball-and-socket joint at and b the two cables at. alculate the tension in each cable resulting from the 20-kN force acting in the horiontal 4 m roblem 3/94

55 c03.qd 11/6/07 3:27 M age hapter 3 Equilibrium 3/95 The uniform rectangular panel has a mass of 40 kg and is hinged at its corners and to the fied vertical surface. wire from E to keeps edges and horiontal. Hinge can support thrust along the hinge ais, whereas hinge supports force normal to the hinge ais onl. Find the tension T in the wire and the magnitude of the force supported b hinge. ns. T 277 N, N T E 3/96 The vertical plane containing the utilit cable turns 30 at the vertical pole. The tensions T 1 and T 2 are both 950 N. In order to prevent long-term leaning of the pole, gu wires and E are utilied. If the two gu wires are adjusted so as to carr equal tensions T which together reduce the moment at to ero, determine the net horiontal reaction at. etermine the required value of T. Neglect the weight of the pole. ns N, T 471 N T 2 = 950 N T 1 = 950 N roblem 3/95 15 E 30 = 9 m = 11 m = 13 m = 8 m E = 10 m roblem 3/96

56 c03.qd 11/6/07 3:27 M age 163 rticle 3/5 hapter Review 163 3/5 HTER R EVIEW In hapter 3 we have applied our knowledge of the properties of forces, moments, and couples studied in hapter 2 to solve problems involving rigid bodies in equilibrium. omplete equilibrium of a bod requires that the vector resultant of all forces acting on it be ero (ΣF 0) and the vector resultant of all moments on the bod about a point (or ais) also be ero (ΣM 0). We are guided in all of our solutions b these two requirements, which are easil comprehended phsicall. Frequentl, it is not the theor but its application which presents difficult. The crucial steps in appling our principles of equilibrium should be quite familiar b now. The are: 1. Make an unequivocal decision as to which sstem (a bod or collection of bodies) in equilibrium is to be analed. 2. Isolate the sstem in question from all contacting bodies b drawing its free-bod diagram showing all forces and couples acting on the isolated sstem from eternal sources. 3. bserve the principle of action and reaction (Newton s third law) when assigning the sense of each force. 4. Label all forces and couples, known and unknown. 5. hoose and label reference aes, alwas choosing a right-handed set when vector notation is used (which is usuall the case for threedimensional analsis). 6. heck the adequac of the constraints (supports) and match the number of unknowns with the number of available independent equations of equilibrium. When solving an equilibrium problem, we should first check to see that the bod is staticall determinate. If there are more supports than are necessar to hold the bod in place, the bod is staticall indeterminate, and the equations of equilibrium b themselves will not enable us to solve for all of the eternal reactions. In appling the equations of equilibrium, we choose scalar algebra, vector algebra, or graphical analsis according to both preference and eperience; vector algebra is particularl useful for man three-dimensional problems. The algebra of a solution can be simplified b the choice of a moment ais which eliminates as man unknowns as possible or b the choice of a direction for a force suation which avoids reference to certain unknowns. few moments of thought to take advantage of these simplifications can save appreciable time and effort. The principles and methods covered in hapter 2 and 3 constitute the most basic part of statics. The la the foundation for what follows not onl in statics but in dnamics as well.

57 c03.qd 11/6/07 3:27 M age hapter 3 Equilibrium REVIEW RLEMS 3/97 alculate the magnitude of the force supported b the pin at for the bell crank loaded and supported as shown. ns. 240 N 120 N 125 roblem 3/97 3/98 50-kg acrobat pedals her uniccle across the taut but slightl elastic cable. If the deflection at the center of the 18-m span is 75, determine the tension in the cable. Neglect the effects of the weights of the cable and uniccle. 20 N. m /100 The tool shown is used for straightening twisted members as wooden framing is completed. If the force 150 N is applied to the handle as shown, determine the normal forces applied to the installed stud at points and. Ignore friction. 360 roblem 3/100 3/101 The device shown in the figure is useful for lifting drwall panels into position prior to fastening to the stud wall. Estimate the magnitude of the force required to lift the 25-kg panel. State an assumptions. ns. 351 N 25 kg m 9 m roblem 3/ /99 The uniform 5-m bar with a mass of 100 kg is hinged at and prevented from rotating in the vertical plane beond the 30 position b the fied roller at. alculate the magnitude of the total force supported b the pin at. ns N 5 m 30 1 m 0.5 m 90 roblem 3/ /102 The designers of an aircraft landing-gear sstem wish to cause the forces in both struts and to act along their respective lengths. What angle should the specif for strut? The weights of all members are small compared with the forces which act on the sstem shown. Treat as twodimensional. roblem 3/99

58 c03.qd 11/6/07 3:27 M age 165 rticle 3/5 Review roblems 165 3/104 If the weight of the boom is negligible compared with the applied 30-kN load, determine the cable tensions T 1 and T 2 and the force acting at the ball joint at m m T 1 4 m 2 m 3 m T 2 roblem 3/ kn 3/103 freewa sign measuring 4 m b 2 m is supported b the single mast as shown. The sign, supporting framework, and mast together have a mass of 300 kg, with center of mass 3.3 m awa from the vertical centerline of the mast. When the sign is subjected to the direct blast of a 125-km/h wind, an average pressure difference of 700 a is developed between the front and back sides of the sign, with the resultant of the wind-pressure forces acting at the center of the sign. etermine the magnitudes of the force and moment reactions at the base of the mast. Such results would be instrumental in the design of the base. ns. R 6330 N, M 38.1 kn m 4 m 1.3 m roblem 3/104 3/105 Magnetic tape under a tension of 10 N at passes around the guide pulles and through the erasing head at at constant speed. s a result of a small amount of friction in the bearings of the pulles, the tape at E is under a tension of 11 N. etermine the tension T in the supporting spring at. The plate lies in a horiontal plane and is mounted on a precision needle bearing at. ns. T N m E 100 ll pulles have a radius of m roblem 3/105 roblem 3/103

59 c03.qd 11/6/07 3:27 M age hapter 3 Equilibrium 3/106 The curved arm and attached cables and support a power line which lies in the vertical - plane. The tangents to the power line at the insulator below make 15 angles with the horiontal -ais. If the tension in the power line at the insulator is 1.3 kn, calculate the total force supported b the bolt at on the pole bracket. The weight of the arm can be neglected compared with the other forces, and it can be assumed that the bolt at E supports horiontal force onl. 3/107 etermine the tension T required to hold the rectangular solid in the position shown. The 125-kg rectangular solid is homogeneous. Friction at is negligible. ns. T 1053 N 0.5 m 1.2 m 1.2 m 0.4 m 0.4 m 0.3 m, E m T 600 roblem 3/107 3/108 vertical force on the foot pedal of the bell crank is required to produce a tension T of 400 N in the vertical control rod. etermine the corresponding bearing reactions at and. E etail of arm attachment roblem 3/ T = 400 N roblem 3/108