,G &$ru d M',,: .;lt:.,,-

l;i::-.,:*, *j:'1.,;* : = l " ';.g-i:: .;

,G &$ru d M',,: .;lt:.,,-

l;i::-.,:*, *j:'1.,;* : = l " ';.g-i:: .;

2. The i.ntem,al forcas are the forces which hold together the parforming the rigid body. If the rigid body is structurally ticles composedof severalparts, the forces holding the component partJ together are alsode{ined asinternal forces.Internal forces will be consideredin Chaps. 6 and 7. As an example of external forces, let us consider the forces acting on a disabled truck that three people are pulling forward by *"ans of a rope attachedto the front bumper (Fig. 3.1). The external forcesactingon the truck are shownin afree-bodydiagram (Fig. 3.2). Let us first consider rhe roeight of the truck. Although it embodies the eflect of tlre earths pull"on each of the particles Formingthe Fig. 3.1 truck, the weight can be represented by the single force W' The point of application of this force, i.e., the point at which the force acts, is de{ined as the center of graoity of the truck. It will be seen in Chap. 5 how centers of gravity can be determined. The weight W tends to make the truck move vertically downward. In fact, it would actuallycausethe truck to move downward,i.e., to fall, if it were not for the presenceof the ground. the ground opposesthe downward motion of the truck by means of the reactions R1 and R2. These forces are exerted by the ground on the truck and must therefore Fis.3.2 be included among the external forces acting on the truck. The people p"lling on the rope exeft the for"" F. The point of applicationof F is on the front bumper. The force F tends to make the truck move forward in a straight line and does actually make it this motion. (Rolling resistance move, sinceno externalforce opposes neglectedhere for simplicity.) This forward motion of the has been truck, during which each straight line keeps its original orientation (the floor of the truck remainshorizontal,and the walls remain vertical), is known as atranslation..Other forces might causethe truck to move differently. For example, the force exerted by a jack placed under the front axlewould causethe truck to pivot about its rear axle. Such a motion is a rotation It can be concluded,therefore, that each impart of the external forces acting on a rigid body can,if unopposed, to the rigid body a motion of translationor rotation, or both.

of 3.3 Principle Tronsmissibility. Forces Equivolenl

75

OF 3.3 PRINCIPLETRANSMISSIBITIW. FORCES EQUIVATENTThe pri.nciple of transmisslbllltg statesthat the conditions of equilibrium or motion of a rigid body will remain unchangedif a force F acting at a given point of the rigid body is replaced by a force F' of the samemagnitudeand samedirection,but actingat a different point, line of action (Fig. 3.3). proaidzd that the tuo forces haue the sam.e The two forces F and F' have the sameeffect on the rigid body and are said to be equir:alent.This principle, which statesthat the action along its line of action, is based on of a force may be transn'aitted experimental evidence. It cannot be derived from the properties esiablishedso far in this text and must therefore be acceptedas an experimentallaw. However,asyou will seein Sec,16.5,the principle_ of transmissibilitycan be derived from the study of the dlnamics of rigid bodies, bui this study requires the introduction of Newton's Fig. 3.3

2. The i.ntem,al forcas are the forces which hold together the parforming the rigid body. If the rigid body is structurally ticles composedof severalparts, the forces holding the component partJ together are alsode{ined asinternal forces.Internal forces will be consideredin Chaps. 6 and 7. As an example of external forces, let us consider the forces acting on a disabled truck that three people are pulling forward by *"ans of a rope attachedto the front bumper (Fig. 3.1). The external forcesactingon the truck are shownin afree-bodydiagram (Fig. 3.2). Let us first consider rhe roeight of the truck. Although it embodies the eflect of tlre earths pull"on each of the particles Formingthe Fig. 3.1 truck, the weight can be represented by the single force W' The point of application of this force, i.e., the point at which the force acts, is de{ined as the center of graoity of the truck. It will be seen in Chap. 5 how centers of gravity can be determined. The weight W tends to make the truck move vertically downward. In fact, it would actuallycausethe truck to move downward,i.e., to fall, if it were not for the presenceof the ground. the ground opposesthe downward motion of the truck by means of the reactions R1 and R2. These forces are exerted by the ground on the truck and must therefore Fis.3.2 be included among the external forces acting on the truck. The people p"lling on the rope exeft the for"" F. The point of applicationof F is on the front bumper. The force F tends to make the truck move forward in a straight line and does actually make it this motion. (Rolling resistance move, sinceno externalforce opposes neglectedhere for simplicity.) This forward motion of the has been truck, during which each straight line keeps its original orientation (the floor of the truck remainshorizontal,and the walls remain vertical), is known as atranslation..Other forces might causethe truck to move differently. For example, the force exerted by a jack placed under the front axlewould causethe truck to pivot about its rear axle. Such a motion is a rotation It can be concluded,therefore, that each impart of the external forces acting on a rigid body can,if unopposed, to the rigid body a motion of translationor rotation, or both.

of 3.3 Principle Tronsmissibility. Forces Equivolenl

75

OF 3.3 PRINCIPLETRANSMISSIBITIW. FORCES EQUIVATENTThe pri.nciple of transmisslbllltg statesthat the conditions of equilibrium or motion of a rigid body will remain unchangedif a force F acting at a given point of the rigid body is replaced by a force F' of the samemagnitudeand samedirection,but actingat a different point, line of action (Fig. 3.3). proaidzd that the tuo forces haue the sam.e The two forces F and F' have the sameeffect on the rigid body and are said to be equir:alent.This principle, which statesthat the action along its line of action, is based on of a force may be transn'aitted experimental evidence. It cannot be derived from the properties esiablishedso far in this text and must therefore be acceptedas an experimentallaw. However,asyou will seein Sec,16.5,the principle_ of transmissibilitycan be derived from the study of the dlnamics of rigid bodies, bui this study requires the introduction of Newton's Fig. 3.3

can be added according to the mles of Chap. 2, and, as these forces are equal and opposite, their sum is equal to zero. Thus, in terms of the eiternal behavior of the bar, the original system of forces shown

of 3.4 VectorProduct Two Vectors

77

to in Fig. 3.5ais equivalent no forceat"all(nig. S.5c).Consider now the two equal and opposite forces P1 and P2 acting on the bar AB as shown in Fig. 3.5d. The force P2 can be replacedby a force Pj havingthe samemagnitude,the samedirection, and the sameline of actionbut actinqat B insteadof at A (Fie. 3.5e). The forcesP1and Pi can then be added.and their sum is ugiin ,"to (Fig. 3.5/). From the point of view of the mechanics rigid bodies, of the systemsshown in Fig. 3.5a and d are thus equivalent. But the intem.alforces and deformafionsproduced by the two systemsare clearly different. The bar of Fig. 3.5a is in tension and, if not absoIutely rigid, will increasein length slightly; the bar of Fig. 3.5d is in and, if not absolutelyrigid, will decreasein length compresision slightly.Thus, while the principle of transmissibilitymay be used freely to determine the conditions of motion or equilibrium of rigid bodiesand to computethe externalforcesactingon thesebodies,it or shouldbe avoided, at leastusedwith care,in determininginternal forces and deformations.

PRODUCT TWOVECTORS OF 3.4 VECTORof In order to gain a better understanding the effect of a force on a the conceptof a momentof a force about a rigid body,u ,re* "otr""pt, point, will be introducedat this time. This conceptwill be more clearly if understood,and appliedmore effectively, we first add to the mathematicaltools at our disposalthe aectorproduct of two vectors. The vector product of two vectors P and Q is defined as the vector V which satisffesthe following conditions. to l. The line of actionof V is perpendicular the plane containing P and Q (Fig. 3.6a). of 2. The magnitudeof V is the product of the magnitudes P and and of the sine of the angle g formed by P and Q (the meaQ sure of which will alwaysbe 180" or less);we thus have V:PQsin0

(3.r)

(a)

3. The direction of V is obtained from the rieht-hand de, Close your right hand and hold it so that yout lingers are curled in the samesenseas the rotation through 0 which brings the vector P in line with the vector Q; your thumb will then indicate the direction of the vector V (Fig. 3.6b). Note that if P and Q they should iirst Fiq.3.6 do not have a commonpoint of application, be redrawn from the samepoint. The three vectors P, Q, and V-taken in that order-are said to form a rieht-handed triad.I

t:I (b)

iWe should note ihat the r, y, and : axes used in Chap. 2 forn-r a right-handed systerr of orthogonal a.xesand that the unit vectors i, j, k defined in Sec. 2.I2 lbrm a right-handed orthogonal triad.

can be added according to the mles of Chap. 2, and, as these forces are equal and opposite, their sum is equal to zero. Thus, in terms of the eiternal behavior of the bar, the original system of forces shown

of 3.4 VectorProduct Two Vectors

77

to in Fig. 3.5ais equivalent no forceat"all(nig. S.5c).Consider now the two equal and opposite forces P1 and P2 acting on the bar AB as shown in Fig. 3.5d. The force P2 can be replacedby a force Pj havingthe samemagnitude,the samedirection, and the sameline of actionbut actinqat B insteadof at A (Fie. 3.5e). The forcesP1and Pi can then be added.and their sum is ugiin ,"to (Fig. 3.5/). From the point of view of the mechanics rigid bodies, of the systemsshown in Fig. 3.5a and d are thus equivalent. But the intem.alforces and deformafionsproduced by the two systemsare clearly different. The bar of Fig. 3.5a is in tension and, if not absoIutely rigid, will increasein length slightly; the bar of Fig. 3.5d is in and, if not absolutelyrigid, will decreasein length compresision slightly.Thus, while the principle of transmissibilitymay be used freely to determine the conditions of motion or equilibrium of rigid bodiesand to computethe externalforcesactingon thesebodies,it or shouldbe avoided, at leastusedwith care,in determininginternal forces and deformations.

PRODUCT TWOVECTORS OF 3.4 VECTORof In order to gain a better understanding the effect of a force on a the conceptof a momentof a force about a rigid body,u ,re* "otr""pt, point, will be introducedat this time. This conceptwill be more clearly if understood,and appliedmore effectively, we first add to the mathematicaltools at our disposalthe aectorproduct of two vectors. The vector product of two vectors P and Q is defined as the vector V which satisffesthe following conditions. to l. The line of actionof V is perpendicular the plane containing P and Q (Fig. 3.6a). of 2. The magnitudeof V is the product of the magnitudes P and and of the sine of the angle g formed by P and Q (the meaQ sure of which will alwaysbe 180" or less);we thus have V:PQsin0

(3.r)

(a)

3. The direction of V is obtained from the rieht-hand de, Close your right hand and hold it so that yout lingers are curled in the samesenseas the rotation through 0 which brings the vector P in line with the vector Q; your thumb will then indicate the direction of the vector V (Fig. 3.6b). Note that if P and Q they should iirst Fiq.3.6 do not have a commonpoint of application, be redrawn from the samepoint. The three vectors P, Q, and V-taken in that order-are said to form a rieht-handed triad.I

t:I (b)

iWe should note ihat the r, y, and : axes used in Chap. 2 forn-r a right-handed systerr of orthogonal a.xesand that the unit vectors i, j, k defined in Sec. 2.I2 lbrm a right-handed orthogonal triad.

note that the left-hand member of Eq. (3.5) can be replaced by P x Q' and that, similarly,the vector products P x Qr and P x Qn can respectivelybe replaced by P x Qi and P x Qi. Thus, the relation to be proved can be written in the form

in Exoressed Terms 3.5 VectorProducts of Rectongulor Components

79

PxQ':PxQi+PxQi

(3.5')

We now observe that P x Q' can be obtained from Q' by multiplylng this vector by the scalarP and rotating it counterclockwise through 90" in the ar plane (Fig. 3.9b); the other two vector

,l

la)

Fis. 3.9 productsin (3.5') can be obtainedin the samemannerfrom Qi and Qi, respectively.Now, since the projection of a parallelogramonto an arbitrary plane is a parallelogram,the projection Q' of the sum Q of Qr and Q2 must be the sum of the projectionsQi and Qi of Q1 and Qz on the sameplane (Fig. 3.9a).This relationbetweenthe vectors Q', Qi, and Qi will still hold after the three vectors have been multiplied by the scalarP and rotated through 90' (Fig. 3.9b). Thus, the relation (3.5') has been proved, and we can now be sure that the distributive property holds for vector products. A third property, the associativeproperty, does not apply to vector products; we have in general

(PxQ) S+Px(QxS) x

(3.6) z\a)

PRODUCTS EXPRESSEDTERMS IN 3.5 VECTOR OF RECTANGULAR COMPONENTSLet us now determine the vector product of any two of the unit vectorsi, j, and k, which were de{inLd in Chap. 2. Consider first the product i x j (Fig. 3.10a). Since both vectors have a magnitude equal to I and since they are at a right angle to each other, their vector product will alsobe a unit vector. This unit vector must be k, rule given on page 77 that the product j x i will be equal to -k (Fig. 3.f0b). Finally, it should be observedthat the vector product

l(

(b)

Fis.3.lO

note that the left-hand member of Eq. (3.5) can be replaced by P x Q' and that, similarly,the vector products P x Qr and P x Qn can respectivelybe replaced by P x Qi and P x Qi. Thus, the relation to be proved can be written in the form

in Exoressed Terms 3.5 VectorProducts of Rectongulor Components

79

PxQ':PxQi+PxQi

(3.5')

We now observe that P x Q' can be obtained from Q' by multiplylng this vector by the scalarP and rotating it counterclockwise through 90" in the ar plane (Fig. 3.9b); the other two vector

,l

la)

Fis. 3.9 productsin (3.5') can be obtainedin the samemannerfrom Qi and Qi, respectively.Now, since the projection of a parallelogramonto an arbitrary plane is a parallelogram,the projection Q' of the sum Q of Qr and Q2 must be the sum of the projectionsQi and Qi of Q1 and Qz on the sameplane (Fig. 3.9a).This relationbetweenthe vectors Q', Qi, and Qi will still hold after the three vectors have been multiplied by the scalarP and rotated through 90' (Fig. 3.9b). Thus, the relation (3.5') has been proved, and we can now be sure that the distributive property holds for vector products. A third property, the associativeproperty, does not apply to vector products; we have in general

(PxQ) S+Px(QxS) x

(3.6) z\a)

PRODUCTS EXPRESSEDTERMS IN 3.5 VECTOR OF RECTANGULAR COMPONENTSLet us now determine the vector product of any two of the unit vectorsi, j, and k, which were de{inLd in Chap. 2. Consider first the product i x j (Fig. 3.10a). Since both vectors have a magnitude equal to I and since they are at a right angle to each other, their vector product will alsobe a unit vector. This unit vector must be k, rule given on page 77 that the product j x i will be equal to -k (Fig. 3.f0b). Finally, it should be observedthat the vector product

l(

(b)

Fis.3.lO

3.6 MOMENT A FORCE OF ABOUT POINT ALet us now considera force F actingon a rigid body (Fig. 3.12a).As we by know,the force F is represented a vectorwhich definesits magnitude the anddirection.However, effectof the force on the rigid body depends also upon its point of applicationA. The position of A can be conveniently defined by the vector r which joins the ffxed referencepoint O with A; this vector is known as the position aector of A. t The position vector r and the force F deiine the plane shownin Fig. 3.72a. We will define the mom.ent F about O as the vector producr of of r and F: Mo:rXF (3.11)

3.6 Momenfof o Forceobout o Point

8l

Accordingto the de{initionof the vectorproduct givenin Sec.3.4, the moment Mp must be pelpendicularto the plane containingO and the force F. The senseof Me is defined by the senseof the rotation which will bring the vector r in line with the vector F; this rotation will be observedas counterclockuise an observerlocated at the tip of by Me. Anotherway of de{iningthe sense Me is furnishedby a variation of of the right-hand rule: Closeyour right hand and hold it so that your fingers are curled in the senseof the rotation that F would impart to ttrJrigra body abouta fr-xed directedalongthe line of actionbf Uo, F i g . 3 . 1 2 axis your thumb will indicatethe senseof the moment Me (Fig. 3.12b). Finally, denoting by 0 the angle between the lines of action of the position vector r and the force F, we find that the magnitude of the moment of F about O is Ms: rF sin? : Fd

la)

,-t(b)

(3.r2)

where d represents the pelpendicular distance from O to the line of action of F. Since the tendency of a force F to make a rigid body rotate about a fixed aris pelpendicular to the force depends upon the distance of F from that axis as well as upon the magnitude of F, we note that the magnitude of M6 measures the tendency of the force F to make the rigid bod:t1rotate about a fixed axis directed alongMs. In the SI system of units, where a force is expressedin newtons (N) and a distance in meters (m), the moment of a force is expressed in newton-meters (N ' m). In the U.S. customary system of units, where a force is erpressed in pounds and a distance in feet or inches, the moment of a force is ermressedin lb . ft or lb . ln. We can observe that although the moment M6 of a force about a point depends upon the *ugt-tit.rd". the line of action, and the sense of the force, it does nof depend upon the actual position of the point of application of the force along its line of action. Conversely, the moment Me of a force F does not characterize the position of the point of application of F.lWe can easily verify that position vectors obey the law of vector addition and, thus, are trrly vectors. Consider, for example, the position vectors r and r' ofA with respect to two reference points O and O' and the position vector s of O with respect to O' (Fig. 3.40a, Sec. 3.16). We verifz that the position vector r' : O'A can be obtained from the position vectors s = Oj 6 and , = Oi by applying the triangle rule for the addition ofvectors.

3.6 MOMENT A FORCE OF ABOUT POINT ALet us now considera force F actingon a rigid body (Fig. 3.12a).As we by know,the force F is represented a vectorwhich definesits magnitude the anddirection.However, effectof the force on the rigid body depends also upon its point of applicationA. The position of A can be conveniently defined by the vector r which joins the ffxed referencepoint O with A; this vector is known as the position aector of A. t The position vector r and the force F deiine the plane shownin Fig. 3.72a. We will define the mom.ent F about O as the vector producr of of r and F: Mo:rXF (3.11)

3.6 Momenfof o Forceobout o Point

8l

Accordingto the de{initionof the vectorproduct givenin Sec.3.4, the moment Mp must be pelpendicularto the plane containingO and the force F. The senseof Me is defined by the senseof the rotation which will bring the vector r in line with the vector F; this rotation will be observedas counterclockuise an observerlocated at the tip of by Me. Anotherway of de{iningthe sense Me is furnishedby a variation of of the right-hand rule: Closeyour right hand and hold it so that your fingers are curled in the senseof the rotation that F would impart to ttrJrigra body abouta fr-xed directedalongthe line of actionbf Uo, F i g . 3 . 1 2 axis your thumb will indicatethe senseof the moment Me (Fig. 3.12b). Finally, denoting by 0 the angle between the lines of action of the position vector r and the force F, we find that the magnitude of the moment of F about O is Ms: rF sin? : Fd

la)

,-t(b)

(3.r2)

where d represents the pelpendicular distance from O to the line of action of F. Since the tendency of a force F to make a rigid body rotate about a fixed aris pelpendicular to the force depends upon the distance of F from that axis as well as upon the magnitude of F, we note that the magnitude of M6 measures the tendency of the force F to make the rigid bod:t1rotate about a fixed axis directed alongMs. In the SI system of units, where a force is expressedin newtons (N) and a distance in meters (m), the moment of a force is expressed in newton-meters (N ' m). In the U.S. customary system of units, where a force is erpressed in pounds and a distance in feet or inches, the moment of a force is ermressedin lb . ft or lb . ln. We can observe that although the moment M6 of a force about a point depends upon the *ugt-tit.rd". the line of action, and the sense of the force, it does nof depend upon the actual position of the point of application of the force along its line of action. Conversely, the moment Me of a force F does not characterize the position of the point of application of F.lWe can easily verify that position vectors obey the law of vector addition and, thus, are trrly vectors. Consider, for example, the position vectors r and r' ofA with respect to two reference points O and O' and the position vector s of O with respect to O' (Fig. 3.40a, Sec. 3.16). We verifz that the position vector r' : O'A can be obtained from the position vectors s = Oj 6 and , = Oi by applying the triangle rule for the addition ofvectors.

THEOREM 3.7 VARIGNON'SThe distributive property of vector products can be used to determine the moment of the resultant of several concurrentforces. Il forcesF1, F2, . , . are appliedat the samepoint A (Fig. 3.1a), several if we denote by r the position vector of A, it follows immediately and from Eq. (3.5)of Sec.3.4 that r x ( F 1+ F 2 + " ' ) : r X F 1 * r x F 2 * (3.14)

3.8 Rectongulor of Components the Momenl of o Force

83

about a gi.oen In words,the mom.ent point O of the resultantof seaeral conanrrentforces is equal to the sum of the mamentsof the aarious the same point O. This property, whlch was originally forces abcvut z by established the French mathematicianVarignon (1654-1722)long before the introduction of vector algebra, is known as Yarignon's Fis.3.14 theorem. The relation (3.14)makesit possibleto replacethe direct determination of the moment of a force F by the determination of the moments of two or more comDonentforces. As vou will see in the into components F Le parallel next section, \rill generally resolued However,it may be more expeditiousin some to the coordinatea,xes. instancesto resolveF into componentswhich are not parallel to the coordinateaxes(seeSampleProb. 3.3).

MOMENT COMPONENTS THE OF 3.8 RECTANGULAR A FORCE OFIn general, the determination of the moment of a force in space will be considerably simplified if the force and the position vector of its point of application are resolved into rectangular r, y, and a components. Consider, for example, the moment M6 about O of a force F whose components are F*, Fo, and F" and whlch is applied at a point A of coordinates r, !, and z (Fig, 3.15). Observing that the components of the position vector r are respectively equal to the coordinates r, y, and z of the point A, we write

r:ri+yj+zk F:F,i+Fyj+F,k for Substituting r and F from (3.15)and (3.16)intoM O : r X F

(3.15) (3.r6)z

(3.u)

F i g .3 . 1 5

and recalling the results obtained in Sec. 3.5, we write the moment Me of F about O in the form

Mo:M,i+Mvj+M"k

(3.17)

where the components M,, M* and M" are defined by the relations

M, : UF^- zFv Mv : zFx - xFz M" : xF, _ yF,

(3.r8)

THEOREM 3.7 VARIGNON'SThe distributive property of vector products can be used to determine the moment of the resultant of several concurrentforces. Il forcesF1, F2, . , . are appliedat the samepoint A (Fig. 3.1a), several if we denote by r the position vector of A, it follows immediately and from Eq. (3.5)of Sec.3.4 that r x ( F 1+ F 2 + " ' ) : r X F 1 * r x F 2 * (3.14)

3.8 Rectongulor of Components the Momenl of o Force

83

about a gi.oen In words,the mom.ent point O of the resultantof seaeral conanrrentforces is equal to the sum of the mamentsof the aarious the same point O. This property, whlch was originally forces abcvut z by established the French mathematicianVarignon (1654-1722)long before the introduction of vector algebra, is known as Yarignon's Fis.3.14 theorem. The relation (3.14)makesit possibleto replacethe direct determination of the moment of a force F by the determination of the moments of two or more comDonentforces. As vou will see in the into components F Le parallel next section, \rill generally resolued However,it may be more expeditiousin some to the coordinatea,xes. instancesto resolveF into componentswhich are not parallel to the coordinateaxes(seeSampleProb. 3.3).

MOMENT COMPONENTS THE OF 3.8 RECTANGULAR A FORCE OFIn general, the determination of the moment of a force in space will be considerably simplified if the force and the position vector of its point of application are resolved into rectangular r, y, and a components. Consider, for example, the moment M6 about O of a force F whose components are F*, Fo, and F" and whlch is applied at a point A of coordinates r, !, and z (Fig, 3.15). Observing that the components of the position vector r are respectively equal to the coordinates r, y, and z of the point A, we write

r:ri+yj+zk F:F,i+Fyj+F,k for Substituting r and F from (3.15)and (3.16)intoM O : r X F

(3.15) (3.r6)z

(3.u)

F i g .3 . 1 5

and recalling the results obtained in Sec. 3.5, we write the moment Me of F about O in the form

Mo:M,i+Mvj+M"k

(3.17)

where the components M,, M* and M" are defined by the relations

M, : UF^- zFv Mv : zFx - xFz M" : xF, _ yF,

(3.r8)

A

pROSLrt$3.I sAlvtPLEA 1001bvertical force is applied to the end of a lever which is attachedto a shaft at O. Determine (a) the moment of the 1001bforce about O; (b) tlie horizontal force applied at A which createsthe samemoment about O; (c) the smallest force applied at A which createsthe samemoment about O; (d) how far from the shaft a240-Ibvertical force must act to createthe samemoment about O; (e) whether any one of the forcesobtainedin partsb, c, andd is equivalentto the oriqinalforce.

s(}LUTroru$. fltarflent cbout 0. The perpendicular distance from O to the line of action ol the 100-lb force is d : (24 in.) cos 60" : 12 in. The magnitude of the moment about O of the 1001b force is Mo : Fd: (100lbXr2 in.) : 12oolb ' in.

Since the force tends to rotate the lever clockvise about O, the moment will be representedby a vector M6 perpendicularto the plane of the figure and pointing into the paper. We expressthis fact by writing bIo : l2l]0 llr ' i1. / ';': &- lisrizanfal Force" In this case,we have d : (24 in.) sin 60" : 20.8 in. Since the moment about O must be 1200 lb ' in., we write Mo: FcJ 1200lb ' in. : F(20.8in.) F : 57.7lb

E * ,17,7lb *+

'.i

e. $malle*t Forc*. Since Mp : Fd, the smallestvalue of F occurs when d is maximum. We choose tl-re force perpendicular to OA and note that d : 2a in.; thus Mo : F d 1200 lb ' in. = F(24 in,): Dt, lb

F = 5(l lir =d3i)-

-Fl

,/f,/$tun,

'E!

B /!.i

ftf t'/ i

6{d lt!

d- ?40-lb' V*rticEl force. but

In this caseM6 : Pd yields d. : 5 in. 0S : ttl in. '.1:

1200 lb . in. : (240 tb)d OB cos 60" : cl

nr.,1l = ' I o i't/ffo'i, P --'l i \ F-d

iit'

e. None of the forces consideredin parts b, c, and d is equivalent to the original 1001b force. Although they have the same moment about O, they have different r and y components. In other words, although each force tends to rotate the shaft in the same manner, each causesthe lever to pull on the shaft in a dilferent way.

85

A

pROSLrt$3.I sAlvtPLEA 1001bvertical force is applied to the end of a lever which is attachedto a shaft at O. Determine (a) the moment of the 1001bforce about O; (b) tlie horizontal force applied at A which createsthe samemoment about O; (c) the smallest force applied at A which createsthe samemoment about O; (d) how far from the shaft a240-Ibvertical force must act to createthe samemoment about O; (e) whether any one of the forcesobtainedin partsb, c, andd is equivalentto the oriqinalforce.

s(}LUTroru$. fltarflent cbout 0. The perpendicular distance from O to the line of action ol the 100-lb force is d : (24 in.) cos 60" : 12 in. The magnitude of the moment about O of the 1001b force is Mo : Fd: (100lbXr2 in.) : 12oolb ' in.

Since the force tends to rotate the lever clockvise about O, the moment will be representedby a vector M6 perpendicularto the plane of the figure and pointing into the paper. We expressthis fact by writing bIo : l2l]0 llr ' i1. / ';': &- lisrizanfal Force" In this case,we have d : (24 in.) sin 60" : 20.8 in. Since the moment about O must be 1200 lb ' in., we write Mo: FcJ 1200lb ' in. : F(20.8in.) F : 57.7lb

E * ,17,7lb *+

'.i

e. $malle*t Forc*. Since Mp : Fd, the smallestvalue of F occurs when d is maximum. We choose tl-re force perpendicular to OA and note that d : 2a in.; thus Mo : F d 1200 lb ' in. = F(24 in,): Dt, lb

F = 5(l lir =d3i)-

-Fl

,/f,/$tun,

'E!

B /!.i

ftf t'/ i

6{d lt!

d- ?40-lb' V*rticEl force. but

In this caseM6 : Pd yields d. : 5 in. 0S : ttl in. '.1:

1200 lb . in. : (240 tb)d OB cos 60" : cl

nr.,1l = ' I o i't/ffo'i, P --'l i \ F-d

iit'

e. None of the forces consideredin parts b, c, and d is equivalent to the original 1001b force. Although they have the same moment about O, they have different r and y components. In other words, although each force tends to rotate the shaft in the same manner, each causesthe lever to pull on the shaft in a dilferent way.

85

PROB[f;fiiI3.4 SA'YTPLEA rectangular by plateis suppofied brackets A andB andby a wire CD. atKnowing that the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire on point C.

50r"uflsF{The moment Ma about A of the force F exerted by the wire on point C is obtained by forming the vector productMa:166XF where r67a is the vector drawn from A to C,

{1}

,",o: Ed : (0.3m)i + (0,08 m)k

i:)

and F is the 200-N force directedalongCD. Introducingthe unit vector It: citcn, we writex.

F : F,I : (200N) cD

Cts

t3i

components, have we Resolving vectorC-d into rectangular the c n : - ( 0 . 3 m ) i+ ( 0 . 2 4 m ) j ( 0 . 3 2 m ) k c D : 0 . 5 0 m into Substituting (3), we obtain : 200N . m)i - (0.32 m)k] [-(0.3 m)i + (0.24 " -(1 : ";fr 2 0 N ) i + ( e 6N ) j - ( 1 2 8 ) k N

(28.8 N'n)j

r4j for the Substituting ra^ andF from (2) and (4) into (1) andrecalling (3.7)of Sec. we obtain 3.5, relations x x Ma: 1616 F: (0.3i+ 0.08k) (-120i + 96j - l28k) : (0.3xe6)k (0.3x-128x-j) + (0.08)(-120)j (0.08)(e6x-i) + + } M , r : - i 7 . 6 8 f " r r i ) i + i ! E . 8 N . m X+ i 2 8 . 8 h i . i n ) k " . Alternafive SalufiE*. As indicatedin Sec.3.8, the momentMa can beexpressedin the form of a determinant:

( 2 b . 8N . m l k

: M,1

I

-t a r lx1

jF , r

kF "

l

i

j

k l0.08 -128 | I

Ac-Ae zc-:ol:

1

4

0.3 0 -120 96 l

M . { : - ( 7 . 6 S N ' r n ) i + ( 2 6 . 8 N . n i ) j+ { 2 E . 8 N . n r ) k . ; i ; a

87

PROB[f;fiiI3.4 SA'YTPLEA rectangular by plateis suppofied brackets A andB andby a wire CD. atKnowing that the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire on point C.

50r"uflsF{The moment Ma about A of the force F exerted by the wire on point C is obtained by forming the vector productMa:166XF where r67a is the vector drawn from A to C,

{1}

,",o: Ed : (0.3m)i + (0,08 m)k

i:)

and F is the 200-N force directedalongCD. Introducingthe unit vector It: citcn, we writex.

F : F,I : (200N) cD

Cts

t3i

components, have we Resolving vectorC-d into rectangular the c n : - ( 0 . 3 m ) i+ ( 0 . 2 4 m ) j ( 0 . 3 2 m ) k c D : 0 . 5 0 m into Substituting (3), we obtain : 200N . m)i - (0.32 m)k] [-(0.3 m)i + (0.24 " -(1 : ";fr 2 0 N ) i + ( e 6N ) j - ( 1 2 8 ) k N

(28.8 N'n)j

r4j for the Substituting ra^ andF from (2) and (4) into (1) andrecalling (3.7)of Sec. we obtain 3.5, relations x x Ma: 1616 F: (0.3i+ 0.08k) (-120i + 96j - l28k) : (0.3xe6)k (0.3x-128x-j) + (0.08)(-120)j (0.08)(e6x-i) + + } M , r : - i 7 . 6 8 f " r r i ) i + i ! E . 8 N . m X+ i 2 8 . 8 h i . i n ) k " . Alternafive SalufiE*. As indicatedin Sec.3.8, the momentMa can beexpressedin the form of a determinant:

( 2 b . 8N . m l k

: M,1

I

-t a r lx1

jF , r

kF "

l

i

j

k l0.08 -128 | I

Ac-Ae zc-:ol:

1

4

0.3 0 -120 96 l

M . { : - ( 7 . 6 S N ' r n ) i + ( 2 6 . 8 N . n i ) j+ { 2 E . 8 N . n r ) k . ; i ; a

87

3"r A foot valve for a pneumatic systemis hinged at B. Knowing thatd. : 28o, determine the moment of the l6-N force about point B by resolving the force into horizontal and vertical compon;nts.

3"? A foot valve for a pneumatic systemis hinged at B. Knowing thata : 28o,determine the moment of the 16-N force about point B by resolvingthe force into componentsalong ABC arrdin a direction perpendicular ABC. to

3.3 A 300-N force is applied at A as shown.Determine (a) the momentof the 300-N force about D, (b) the smallest force applied at B that createsthe samemoment about D.

Fig. P3.I ond P3.2

125 mm

{C B

I I

Fig. P3.3 ond P3.4 3-4 A 300-N forceis appliedat A asshown.Determine(a) the moment of the 300-N force about D, (b) the magnitude and senseof the horizontal force applied at C that createsthe same moment about D, (c) the smallestforce applied at C that createsthe samemoment about D. 3,5 An 8lb force P is applied to a shift lever. Determine the moment of P about B when a is equal to 25o. 3.8 For the shift lever shown, determine the magnitude and the direction of the smallestforce P that has a 2101b . in. cloclovisemoment about B. 3.7 An l1-lb force P is applied to a shift lever. The moment of P about B is clocloviseand has a masnitude of 250 Ib . in. Determine the value of a. 3"* It is known that a vertical force of 200 lb is recuired to remove the nail at C from the board. As the nail first starts rroving, determine (a) the moment about B of the force exerled on the nail, (b) the magnitude of the force P that createsthe same moment about B if a : 10". (c) the smallest force P that creates the same moment about B. Fig. P3.8

Fig. P3.5, P3.6, snd P3.7

3"r A foot valve for a pneumatic systemis hinged at B. Knowing thatd. : 28o, determine the moment of the l6-N force about point B by resolving the force into horizontal and vertical compon;nts.

3"? A foot valve for a pneumatic systemis hinged at B. Knowing thata : 28o,determine the moment of the 16-N force about point B by resolvingthe force into componentsalong ABC arrdin a direction perpendicular ABC. to

3.3 A 300-N force is applied at A as shown.Determine (a) the momentof the 300-N force about D, (b) the smallest force applied at B that createsthe samemoment about D.

Fig. P3.I ond P3.2

125 mm

{C B

I I

Fig. P3.3 ond P3.4 3-4 A 300-N forceis appliedat A asshown.Determine(a) the moment of the 300-N force about D, (b) the magnitude and senseof the horizontal force applied at C that createsthe same moment about D, (c) the smallestforce applied at C that createsthe samemoment about D. 3,5 An 8lb force P is applied to a shift lever. Determine the moment of P about B when a is equal to 25o. 3.8 For the shift lever shown, determine the magnitude and the direction of the smallestforce P that has a 2101b . in. cloclovisemoment about B. 3.7 An l1-lb force P is applied to a shift lever. The moment of P about B is clocloviseand has a masnitude of 250 Ib . in. Determine the value of a. 3"* It is known that a vertical force of 200 lb is recuired to remove the nail at C from the board. As the nail first starts rroving, determine (a) the moment about B of the force exerled on the nail, (b) the magnitude of the force P that createsthe same moment about B if a : 10". (c) the smallest force P that creates the same moment about B. Fig. P3.8

Fig. P3.5, P3.6, snd P3.7

3.14 A mechanicuses a piece of pipe AB as a lever when tightening an alternator belt. When he pushes down at A, a force of 485 N is exertedon the alternatorat B. Determine the moment of that force about bolt C if its line of action passes through O.

9l

120mm

Fig. P3.14

x 3 . f 5 F o r m t h e v e c t o r p r o d u c t s BC a n d B ' X C , w h e r e : B ' , a n d B use the results obtained to prove the identity s i n a c o s B : j s i n( a + 9 1 + j s i n( a - B ) . 3.f6 A line passes throughthe points (20 m, 16 m) and (-1 m, -4 m). Determine the perpendiculardistanced from the line to the origin O of the systemof coordinates.

3.17 The vectorsP and Q are two adjacent sidesof a parallelogram. Determine the area of the parallelogram when (a) P : -7i + 3j - 3k andg : 2i + 2j+ 5k, (b) P : 6i - 5j - 2kandQ : -2i + 5j-k. 3,tg A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) i + 2j - Skand 4i - 7j - lk,(b) 3i - 3j + 2k and -2i * 6j - 4k.

Fig. P3.15

3.19 Determine the moment about the oriein O of the force F : 4i + 5j - 3k that acts at a point A. Assurnethat the position vector of A is (a) r : 2i - 3j + 4k, (b) : 2i + 2.5j - 1.5k, (c) r : 2i + " 5j + 6k. 3,2O Determine the moment about the origin O of the force F = -Asurn" -2i + 3j + 5k that acts at a point A. that the position vectorof A is (a) r : i * j + k, (b), :2i + 3j - 5k, (c) r : -4i+6j+10k. 3.21 A 200-N force is applied as shown to the bracketABC. Determine t]le moment of the force about A.

Fig. P3.21

3.14 A mechanicuses a piece of pipe AB as a lever when tightening an alternator belt. When he pushes down at A, a force of 485 N is exertedon the alternatorat B. Determine the moment of that force about bolt C if its line of action passes through O.

9l

120mm

Fig. P3.14

x 3 . f 5 F o r m t h e v e c t o r p r o d u c t s BC a n d B ' X C , w h e r e : B ' , a n d B use the results obtained to prove the identity s i n a c o s B : j s i n( a + 9 1 + j s i n( a - B ) . 3.f6 A line passes throughthe points (20 m, 16 m) and (-1 m, -4 m). Determine the perpendiculardistanced from the line to the origin O of the systemof coordinates.

3.17 The vectorsP and Q are two adjacent sidesof a parallelogram. Determine the area of the parallelogram when (a) P : -7i + 3j - 3k andg : 2i + 2j+ 5k, (b) P : 6i - 5j - 2kandQ : -2i + 5j-k. 3,tg A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) i + 2j - Skand 4i - 7j - lk,(b) 3i - 3j + 2k and -2i * 6j - 4k.

Fig. P3.15

3.19 Determine the moment about the oriein O of the force F : 4i + 5j - 3k that acts at a point A. Assurnethat the position vector of A is (a) r : 2i - 3j + 4k, (b) : 2i + 2.5j - 1.5k, (c) r : 2i + " 5j + 6k. 3,2O Determine the moment about the origin O of the force F = -Asurn" -2i + 3j + 5k that acts at a point A. that the position vectorof A is (a) r : i * j + k, (b), :2i + 3j - 5k, (c) r : -4i+6j+10k. 3.21 A 200-N force is applied as shown to the bracketABC. Determine t]le moment of the force about A.

Fig. P3.21

3.26

A small boat hangs from two davits, one of which is shown in the

figure.The tension line ABAD is 82 lb. Determine moment in the aboutC of the resultant forceRo exerted the davitat A. on

93

Fig. P3.26 3.27 3.28 In Prob. 3.22, determine the pelpendicular distancefrom point O to cableAB. In Prob.3.22, determine the pelpendicular distancefrom point O to cableBC.

3.29 InProb.3.24, determine the perpendicular distancefrom point D to a line drawn through points A and B. 3.3O In Prob. 3.24, determine the perpendicular distancefrom point C to a line drawn through points A and B. 3.31 In Prob. 3.25, determine the pelpendicular distancefrom point A to portion DE of cableDEF. 3.32 3.33 3.34 In Prob. 3.25, determine the perpendicular distancefrom point A to a line drawn through points C and G. In Prob. 3.26, determine the perpendicular distancefrom point C to portion AD of the line ABAD. Determine the value of a that minimizes the perpendicular distance from point C to a section of pipeline that passesthrough ooints A and B.

Fis. P3.34

3.26

A small boat hangs from two davits, one of which is shown in the

figure.The tension line ABAD is 82 lb. Determine moment in the aboutC of the resultant forceRo exerted the davitat A. on

93

Fig. P3.26 3.27 3.28 In Prob. 3.22, determine the pelpendicular distancefrom point O to cableAB. In Prob.3.22, determine the pelpendicular distancefrom point O to cableBC.

3.29 InProb.3.24, determine the perpendicular distancefrom point D to a line drawn through points A and B. 3.3O In Prob. 3.24, determine the perpendicular distancefrom point C to a line drawn through points A and B. 3.31 In Prob. 3.25, determine the pelpendicular distancefrom point A to portion DE of cableDEF. 3.32 3.33 3.34 In Prob. 3.25, determine the perpendicular distancefrom point A to a line drawn through points C and G. In Prob. 3.26, determine the perpendicular distancefrom point C to portion AD of the line ABAD. Determine the value of a that minimizes the perpendicular distance from point C to a section of pipeline that passesthrough ooints A and B.

Fis. P3.34

definition of the scalarproduct it follows that the scalarproducts of the unit vectors are either zero or one.

3.9 Scolor Productof Two Vectors

95

i . i : i ; . i ' j : 0

j . j : 1 j . k : 0

k . k : I k . i : 0

(3.2e)

Thus,the expression obtained P . Q reduces for to

, P;',:.E.,f*;. *-,,ffi,1 f.18a'In the particular casewhen P and Q are equal, we note that

(3.30)

P.P: ri + rl + P!: pz

(3.31)

Appliccfionsl. Angleformed by tuo giaen oectors. Let two vectors be given in terms of their components: P:P"i+Psi+P"k

Q: Q"i+ Qai+ Q"kTo determine the angle formed by the two vectors,we equate the expressions obtained in (3.24) and (3.30) for their scalar product and write PQ cos 0 : P,Q, + PaQa* P,Q" Solving for cos 0, we have

P,Q,rPvQa+P^Q^cos0:

PQ

(3.32)

2. Projecti,onof a oector on a gi,oenaxi,s. Consider a vector Pforming an angle 0 with an axis, or directed line, OL (Fig. 3.21). The projection of P on the axi.sOL is defined as the scalar Fig. 3.2r Por,:Pcos0 (3.33) We note that the projection Pe1 is equal in absolute value to the length of the segmentOA; it will be positive if OA has the same senseas the axis OL, that is, if 0 is acute, and negative otherwise. If P and OL are at a right angle, the projection of P on OL is zero. Consider now a vector Q directed along OL and of the same senseas OL (Fig. 3.22). The scalarproduct of P and Q can be expressed as P .Q: P Q c o s0 : P o r Q

definition of the scalarproduct it follows that the scalarproducts of the unit vectors are either zero or one.

3.9 Scolor Productof Two Vectors

95

i . i : i ; . i ' j : 0

j . j : 1 j . k : 0

k . k : I k . i : 0

(3.2e)

Thus,the expression obtained P . Q reduces for to

, P;',:.E.,f*;. *-,,ffi,1 f.18a'In the particular casewhen P and Q are equal, we note that

(3.30)

P.P: ri + rl + P!: pz

(3.31)

Appliccfionsl. Angleformed by tuo giaen oectors. Let two vectors be given in terms of their components: P:P"i+Psi+P"k

Q: Q"i+ Qai+ Q"kTo determine the angle formed by the two vectors,we equate the expressions obtained in (3.24) and (3.30) for their scalar product and write PQ cos 0 : P,Q, + PaQa* P,Q" Solving for cos 0, we have

P,Q,rPvQa+P^Q^cos0:

PQ

(3.32)

2. Projecti,onof a oector on a gi,oenaxi,s. Consider a vector Pforming an angle 0 with an axis, or directed line, OL (Fig. 3.21). The projection of P on the axi.sOL is defined as the scalar Fig. 3.2r Por,:Pcos0 (3.33) We note that the projection Pe1 is equal in absolute value to the length of the segmentOA; it will be positive if OA has the same senseas the axis OL, that is, if 0 is acute, and negative otherwise. If P and OL are at a right angle, the projection of P on OL is zero. Consider now a vector Q directed along OL and of the same senseas OL (Fig. 3.22). The scalarproduct of P and Q can be expressed as P .Q: P Q c o s0 : P o r Q

tip of Sl. The mixed triple product will be zero if S, P, and Q are coplanar. Since the parallelepipeddefined in the preceding paragraphis independent of the order in which the three vectors are taken, the six mixed triple products which can be formed with S, P, and Q *ill all have the same absolutevalue, although not the same sign. It is easily shown that s.(P x Q): p.(Q x s): Q.(s x P)

3.1 I Momentof o Forceobout o Given Axis

97

: - S . ( Q x P )- - P . ( Sx e ) : - e . ( p x S ) (3.3e)

Arranging in a circle and in counterclockwiseorder the letters representingthe three vectors(Fig. 3.26),we observethat the sign of the mixed triple product remains unchangedif the vectors are permuted in such a way that they are still read in counterclockwise order. Such a permutation is said to be a ci,rcular permutation. It also folloys from Eq. (3.39)and from the commutaiiveproperty of scalarproducts that the mlred triple product of S, P, and Q can be defined equallywell as S . (P x Q) or (S x P) . Q. The mixed triple product of the vectors S, P, and Q can be expressed terms of the rectangular in components thesevectors. of DenotingP x Q by V and using formula (3.30)to express scalar the product of S and V, we write S . ( P x Q ) : S . V : S " V ,+ S u V y S = V , + Substitutingfrom the relations(3.9) for the cornponents V, we of obtain s . (p x Q) : s,(pyQ.- p,Q) + sy(p.p, _ p,Q^) + s.(P.gv - PrQ") (3.40) This expression be written in a more compactform if we observe can that it representsthe expansionof a determinant:

a'\/ v'Fig.3.26

lsS . ( P x Q )=

lp, Qv

l;" P.,J

),,

p;l

P-,1

s,l

(3.41)

By applying the rules goveming the permutation of rows in a determinant, we could easilyverify the relations (3.39)whlch were derived earlier from geometricalconsiderations.

3.I I

MOMENT A FORCE OF ABOUT GIVEN AXIS A

Now that we have further increasedour knowledge of vector algebra, we can introduce a new concept,the concept of moment of a force about an axis. Consider again a force F acting on a rigid body and the moment M6 of that force about O (Fie. 3.27). Let OL be

z

Fig.3.27

tip of Sl. The mixed triple product will be zero if S, P, and Q are coplanar. Since the parallelepipeddefined in the preceding paragraphis independent of the order in which the three vectors are taken, the six mixed triple products which can be formed with S, P, and Q *ill all have the same absolutevalue, although not the same sign. It is easily shown that s.(P x Q): p.(Q x s): Q.(s x P)

3.1 I Momentof o Forceobout o Given Axis

97

: - S . ( Q x P )- - P . ( Sx e ) : - e . ( p x S ) (3.3e)

Arranging in a circle and in counterclockwiseorder the letters representingthe three vectors(Fig. 3.26),we observethat the sign of the mixed triple product remains unchangedif the vectors are permuted in such a way that they are still read in counterclockwise order. Such a permutation is said to be a ci,rcular permutation. It also folloys from Eq. (3.39)and from the commutaiiveproperty of scalarproducts that the mlred triple product of S, P, and Q can be defined equallywell as S . (P x Q) or (S x P) . Q. The mixed triple product of the vectors S, P, and Q can be expressed terms of the rectangular in components thesevectors. of DenotingP x Q by V and using formula (3.30)to express scalar the product of S and V, we write S . ( P x Q ) : S . V : S " V ,+ S u V y S = V , + Substitutingfrom the relations(3.9) for the cornponents V, we of obtain s . (p x Q) : s,(pyQ.- p,Q) + sy(p.p, _ p,Q^) + s.(P.gv - PrQ") (3.40) This expression be written in a more compactform if we observe can that it representsthe expansionof a determinant:

a'\/ v'Fig.3.26

lsS . ( P x Q )=

lp, Qv

l;" P.,J

),,

p;l

P-,1

s,l

(3.41)

By applying the rules goveming the permutation of rows in a determinant, we could easilyverify the relations (3.39)whlch were derived earlier from geometricalconsiderations.

3.I I

MOMENT A FORCE OF ABOUT GIVEN AXIS A

Now that we have further increasedour knowledge of vector algebra, we can introduce a new concept,the concept of moment of a force about an axis. Consider again a force F acting on a rigid body and the moment M6 of that force about O (Fie. 3.27). Let OL be

z

Fig.3.27

of the unit vectors i, j, and k for r\ in (3.42), we obselve that the expressions thus obtained for the moments of F about the coordtinate ax.es are respectively equal to the expressions obtained in Sec. 3.8 for the components of the moment M6 of F about O:

3.1 I Momentof o Forceobout o Given Axis

99

M,: AF^- zFu Mo:zF*-xF^ M^: xF, - !lF,

(3.18)

We observethat-just_as the componentsF,, F* and F^ of a force F acting on a rigid body measure,respectively, ihe tendency of F to move the rigid body in the x, y, and z directions, the moments M", Mu, and M^ of F about the coor&nate axesmeasurethe tendencv of F io impart to the rigid body a motion of rotation about the r, y, and z axes,respectively. More generally,the moment of a force F applied at A about an axis whlch does not pass through the origin is obtained by choosingan arbitrary point B on the axis (Fig. S.Zg)and determining the projection on the axisBL of the moment M6 of F about B. We write Mnr: i . Mr: ,[ . (ra4 x F)

(3.45)

where tNB : tA - rn represents the vector drawn from B to A. Expressing Me; in the form of a determinant, we have

Msr:

| ,t. l

l,I

i,

lxe73 Aen F,, I F,I

zdn F^

(3.46)

z

Fis. 3.29

where \", \y, \. : direction cosines of axis BL XA/B: Xa - Xn rjdn: At - ln F,, Fy, F. : components of force F

Zdtn: Za - Zn

It should be noted that the result obtained is independent of the choice of the point B on the given a,nis.Indeed, denoting by M67 the result obtained with a different point C, we have

Mcr:,\.[(ra - r") x tr']:.[.[(ra-rr) x F ] + z \ . [ ( " " - 1 6 , )x F ] But, since the vectors,\ and tB - tc lie in the sameline, the volume of the parallelepipedhaving the vectors i, rr - 16, and F for sidesis zero, as is the mixed triple product of these three vectors (Sec.3.f0). The expression obtained for M61 thus reducesto its first term, which is the expression used earlier to define Mpy. In addition, it follows from Sec. 3.6 that, when computing the moment of F about the given axis,A can be any point on the line of action of F.

of the unit vectors i, j, and k for r\ in (3.42), we obselve that the expressions thus obtained for the moments of F about the coordtinate ax.es are respectively equal to the expressions obtained in Sec. 3.8 for the components of the moment M6 of F about O:

3.1 I Momentof o Forceobout o Given Axis

99

M,: AF^- zFu Mo:zF*-xF^ M^: xF, - !lF,

(3.18)

We observethat-just_as the componentsF,, F* and F^ of a force F acting on a rigid body measure,respectively, ihe tendency of F to move the rigid body in the x, y, and z directions, the moments M", Mu, and M^ of F about the coor&nate axesmeasurethe tendencv of F io impart to the rigid body a motion of rotation about the r, y, and z axes,respectively. More generally,the moment of a force F applied at A about an axis whlch does not pass through the origin is obtained by choosingan arbitrary point B on the axis (Fig. S.Zg)and determining the projection on the axisBL of the moment M6 of F about B. We write Mnr: i . Mr: ,[ . (ra4 x F)

(3.45)

where tNB : tA - rn represents the vector drawn from B to A. Expressing Me; in the form of a determinant, we have

Msr:

| ,t. l

l,I

i,

lxe73 Aen F,, I F,I

zdn F^

(3.46)

z

Fis. 3.29

where \", \y, \. : direction cosines of axis BL XA/B: Xa - Xn rjdn: At - ln F,, Fy, F. : components of force F

Zdtn: Za - Zn

It should be noted that the result obtained is independent of the choice of the point B on the given a,nis.Indeed, denoting by M67 the result obtained with a different point C, we have

Mcr:,\.[(ra - r") x tr']:.[.[(ra-rr) x F ] + z \ . [ ( " " - 1 6 , )x F ] But, since the vectors,\ and tB - tc lie in the sameline, the volume of the parallelepipedhaving the vectors i, rr - 16, and F for sidesis zero, as is the mixed triple product of these three vectors (Sec.3.f0). The expression obtained for M61 thus reducesto its first term, which is the expression used earlier to define Mpy. In addition, it follows from Sec. 3.6 that, when computing the moment of F about the given axis,A can be any point on the line of action of F.

{n the problems for this lessonyou will apply the scalar product or dot prodact Iof trvo vectors to determine the angleformed by tuo giaen oectors and the projection of a force on a gioen axis.Yot will also use the mixed triple product of three vectorsto find the mnment of a force about rLgi.aen axis and the perpendicular distancebetueen two lines,

l. clculctlng the ongle formed by fwo given vectars. First ex?ress vectors the in terms of their componentsand determine the magnitudesof the two vectors. The cosine of the desired angle is then obtained by dividing the scalarproduct of the two vectorsby the product of their magnitudes[Eq. (3.32)J.

2. fomputing the projeclion of s vecfor P on a given cxis Ol"" tn general, begin by expressing and the unit vector .tr,that defines the direction of the axis, P in component form. Take care that i has the correct sense(that is, .f, is directed from O to L). The required projection is then equal to the scalarproduct P . ,1,. However,if you know the angle 0 formed by P and r\, the projection is also given by P cos0.

3. Determining fhe mcmenf lll6;.al c force sbout a given uxis OL We defined Ms; as

Mor:i.Mo:.tr.(rxF)

(3,42)

where r\ is the unit vector along OL and r is a position vectorfrom any point on the line OL to any point on the line of action of F. As was the casefor the moment of a force about a point, choosingthe most convenientposition vector will simplify your calculations. Also, recall the waming of the previouslesson: vectorsr and F the must have the correct sense,and they must be placed in the proper order. The procedure you should follow when computing the moment of a force about an axis is illustrated in part c of Sample Prob. 3.5. The two essentialstepsin this procedure are to first express,\, r, and F in terms of their rectangularcomponentsand to then evaluatethe mixed triple product ,\ . (r x F) to determine the moment about the axis. In most three-dimensionalproblems the most convenient wav to compule the mixed triple product is by using a determinant

z -.-.. -.,..:..:....

:

As noted in the text, when ,,\ is directed along one of the coordinate axes,Ms1 is equal to the scalarcomponent of Me along that axis. (contirrued)

roI

{n the problems for this lessonyou will apply the scalar product or dot prodact Iof trvo vectors to determine the angleformed by tuo giaen oectors and the projection of a force on a gioen axis.Yot will also use the mixed triple product of three vectorsto find the mnment of a force about rLgi.aen axis and the perpendicular distancebetueen two lines,

l. clculctlng the ongle formed by fwo given vectars. First ex?ress vectors the in terms of their componentsand determine the magnitudesof the two vectors. The cosine of the desired angle is then obtained by dividing the scalarproduct of the two vectorsby the product of their magnitudes[Eq. (3.32)J.

2. fomputing the projeclion of s vecfor P on a given cxis Ol"" tn general, begin by expressing and the unit vector .tr,that defines the direction of the axis, P in component form. Take care that i has the correct sense(that is, .f, is directed from O to L). The required projection is then equal to the scalarproduct P . ,1,. However,if you know the angle 0 formed by P and r\, the projection is also given by P cos0.

3. Determining fhe mcmenf lll6;.al c force sbout a given uxis OL We defined Ms; as

Mor:i.Mo:.tr.(rxF)

(3,42)

where r\ is the unit vector along OL and r is a position vectorfrom any point on the line OL to any point on the line of action of F. As was the casefor the moment of a force about a point, choosingthe most convenientposition vector will simplify your calculations. Also, recall the waming of the previouslesson: vectorsr and F the must have the correct sense,and they must be placed in the proper order. The procedure you should follow when computing the moment of a force about an axis is illustrated in part c of Sample Prob. 3.5. The two essentialstepsin this procedure are to first express,\, r, and F in terms of their rectangularcomponentsand to then evaluatethe mixed triple product ,\ . (r x F) to determine the moment about the axis. In most three-dimensionalproblems the most convenient wav to compule the mixed triple product is by using a determinant

z -.-.. -.,..:..:....

:

As noted in the text, when ,,\ is directed along one of the coordinate axes,Ms1 is equal to the scalarcomponent of Me along that axis. (contirrued)

roI

S"35 Given the vectors P : 3i - j + 2k, e - 4i + 5j - 3k, ancl S : -2i + 3j - k, compute the scalarproductsp . e, p . S, andQ.S. 3.$& Form the scalarproductsB . C and B' . C, where B : B', and use the results obtained to prove the identity c o s a c o s B : ] c o s ( a + 9 1 + j c o s( a - F ) . *.3? Section AB of a pipeline lies in the yz plane and forms an angle r-rf37" with the a aiis. Branch lines Cb and EF ioin AB as sho.Jn. Determine the angle formed by pipes AB arxl CD. Section Ats of a pipeline lies in the yz pl,aneand forrns an angle of 37" with the ziiis. BranchIines iD anclEF ioin AB as shorin. Determine the angle formed by pipes AB and EF.

S-Sf

3.SS Considerthe volleyball net shown. Determine the angle formed by guy wires AB and AC.

Fig. P3.39 ond P3.4O 3"1*S Considerthe volleyball net shown. Determine the angle formed by guy wires AC and AD. S"4l Knorvingthat the tensionin cableAC is 1260 N, determine(a)the angle between cable AC and the boom AB, (b) the projection on AB of the force exertedby cableAC at point A.

S.42 Knowing that the tension in cable AD is 405 N, determine (a) the angle between cable AD and the boom AB, (b) the projection on AB of the force exertedbv cableAD at ooint A.

a

Fig. P3.41 ond P3.42

r03

S"35 Given the vectors P : 3i - j + 2k, e - 4i + 5j - 3k, ancl S : -2i + 3j - k, compute the scalarproductsp . e, p . S, andQ.S. 3.$& Form the scalarproductsB . C and B' . C, where B : B', and use the results obtained to prove the identity c o s a c o s B : ] c o s ( a + 9 1 + j c o s( a - F ) . *.3? Section AB of a pipeline lies in the yz plane and forms an angle r-rf37" with the a aiis. Branch lines Cb and EF ioin AB as sho.Jn. Determine the angle formed by pipes AB arxl CD. Section Ats of a pipeline lies in the yz pl,aneand forrns an angle of 37" with the ziiis. BranchIines iD anclEF ioin AB as shorin. Determine the angle formed by pipes AB and EF.

S-Sf

3.SS Considerthe volleyball net shown. Determine the angle formed by guy wires AB and AC.

Fig. P3.39 ond P3.4O 3"1*S Considerthe volleyball net shown. Determine the angle formed by guy wires AC and AD. S"4l Knorvingthat the tensionin cableAC is 1260 N, determine(a)the angle between cable AC and the boom AB, (b) the projection on AB of the force exertedby cableAC at point A.

S.42 Knowing that the tension in cable AD is 405 N, determine (a) the angle between cable AD and the boom AB, (b) the projection on AB of the force exertedbv cableAD at ooint A.

a

Fig. P3.41 ond P3.42

r03

3.49

To lift a heavy crate, a man usesa block and tackle attachedto the bottom of an I-beam at hook B. Knowing that the moments about the y and the z axesof the force exertedat B by portion AB of the rope are, respectiveln 120 N . m and -460 N . m, determine the distancea.

Problems | 05

3.5O To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the man applies a 195-N force to end A of the rope and that the moment & ttrut force about the y axis is 132 N . m, determine the distance a.

3.5t

A small boat hangs from two davits, one of which is shown in the figure. It is known that the moment about the z axis of the resultant force Ra exerted on the davit at A must not exceed 279 lb . ft in absolutevalue. Determine the largest allowabletension in line ABAD when r : 6 ft.

9.52 For the davit of Prob. 3.5I, determine the larsest allowable distance r when the tension in line ABAD is 60 lb.

Fig. P3.49 ond P3.50

3.53

To loosena frozen valve, a force F of magnitude70 lb is applied to the handle of the valve. Knowing that g : 25, M, : -6f lb . ft, and M, : -43 lb . ft, determine @ and d.

3.54 When a force F is applied to the handle of the valve shown, itsmomentsabout the x and z axesare, respectively, : -77 lb . {t M, atd M^: -81 lb . ft. For d : 27 in., determinethe momentM, of F about the y axis.

3.49

To lift a heavy crate, a man usesa block and tackle attachedto the bottom of an I-beam at hook B. Knowing that the moments about the y and the z axesof the force exertedat B by portion AB of the rope are, respectiveln 120 N . m and -460 N . m, determine the distancea.

Problems | 05

3.5O To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the man applies a 195-N force to end A of the rope and that the moment & ttrut force about the y axis is 132 N . m, determine the distance a.

3.5t

A small boat hangs from two davits, one of which is shown in the figure. It is known that the moment about the z axis of the resultant force Ra exerted on the davit at A must not exceed 279 lb . ft in absolutevalue. Determine the largest allowabletension in line ABAD when r : 6 ft.

9.52 For the davit of Prob. 3.5I, determine the larsest allowable distance r when the tension in line ABAD is 60 lb.

Fig. P3.49 ond P3.50

3.53

To loosena frozen valve, a force F of magnitude70 lb is applied to the handle of the valve. Knowing that g : 25, M, : -6f lb . ft, and M, : -43 lb . ft, determine @ and d.

3.54 When a force F is applied to the handle of the valve shown, itsmomentsabout the x and z axesare, respectively, : -77 lb . {t M, atd M^: -81 lb . ft. For d : 27 in., determinethe momentM, of F about the y axis.

3.61 A sign erected on uneven ground is guyed by cablesEF and EG. If the force exertedby cable EF at E is 46 lb, determine the moment of that force about the line joining points A and D.

Problems 107

Fig. P3.6f ond P3.62 3"62 A sign erected on unevenground is guyed by cablesEF atdBc.If the force exertedby cable EC at E is 54 lb, determine the moment of that force about the line joining points A and D. 3,63 Two forces F1 and F2 in spacehave the same magnitude F. Prove that the moment of F1 about the line of action of F2 is equal to the moment of Fe about the line of action of F1.

"3"64 In Prob. 3.55, determine the perpendicular distance between portion BII of the cable and the diagonal AD. *3.65 In Prob. 3.56, determine the perpendicular distancebetween portion BG of the cable and the diagonalAD.

*3.66 In Prob. 3.57, determine the perpendicular distance between cable AE and the line joining points D and B. "3.67 In Prob. 3.58, determine the perpendicular distance between cable CF and the line joining points D and B. *3.68 In Prob. 3.61, determine the perpendicular distance between cable EF and the line joining points A and D. *3.69 In Prob. 3.62, determine the perpendicular distance between cableEC and the line joining points A and D.

3.61 A sign erected on uneven ground is guyed by cablesEF and EG. If the force exertedby cable EF at E is 46 lb, determine the moment of that force about the line joining points A and D.

Problems 107

Fig. P3.6f ond P3.62 3"62 A sign erected on unevenground is guyed by cablesEF atdBc.If the force exertedby cable EC at E is 54 lb, determine the moment of that force about the line joining points A and D. 3,63 Two forces F1 and F2 in spacehave the same magnitude F. Prove that the moment of F1 about the line of action of F2 is equal to the moment of Fe about the line of action of F1.

"3"64 In Prob. 3.55, determine the perpendicular distance between portion BII of the cable and the diagonal AD. *3.65 In Prob. 3.56, determine the perpendicular distancebetween portion BG of the cable and the diagonalAD.

*3.66 In Prob. 3.57, determine the perpendicular distance between cable AE and the line joining points D and B. "3.67 In Prob. 3.58, determine the perpendicular distance between cable CF and the line joining points D and B. *3.68 In Prob. 3.61, determine the perpendicular distance between cable EF and the line joining points A and D. *3.69 In Prob. 3.62, determine the perpendicular distance between cableEC and the line joining points A and D.

3.I3 EQUIVATENT COUPLESFigure 3.34 showsthree coupleswhich act successively the same on rectangularbox. As seenin the preceding section,the only motion a couple can impart to a rigid body is a rotation. Since each of the three couples shown has the same moment M (same direction and samemagnitudeM : L20lb . in.), we can expectthe three couples to have the same effect on the box.

3.13 Equivolent Couples | 09

30 lbz

Fig. 3.34

(a)

(b)

(c)

As reasonableas this conclusionappears,we should not accept it hastily.While intuitive feeling is of great help in the study of mechanics, it should not be accepted as a substitute for logical reasoning. Before stating that two systems(or groups) of forces have the same effect on a rigid body, we should prove that fact on the basisof the experimental evidence introduced so far. This evidence consistsof the parallelogramlaw for the addition of two forces (Sec. 2,2) and the principle of transmissibility (Sec. 3.3). Therefore, we will state that tu;o systemsof forces are equioalent (i.e., they have the same effect on a rigid body) if u;e can transform one of them i.ntothe other by meansof one or seoeralof the following operations:(1) replacing two forcesacting on the sameparticle by their resultant;(2) resolving a force into two components;(3) canceling two equal and opposite forces acting on the sameparticle; (4) attachingto the sameparticle two equal and opposite forces; (5) moving a force along its line of action. Each of these operationsis easilyjustified on the basisof the parallelogramlaw or the principle of transmissibility. Let us now prove thattuoo coupleshaoingthe sampmoment M are equi,oalenf.First consider two couples contained in the same plane, and assumethat this plane coincides with the plane of the figure (Fig, 3,35),The first coupleconsists the forcesF1 and -F1 of of magnitude F,, which are located at a distanced1 from each other (Fig. 3.35a),and the secondcoupleconsists the forcesF2 and -F2 of of magnitude F2, which are located at a distanced,2from each other (Fig. 3.35d).Sincethe two couples havethe samemoment M, which is perpendicular to the plane of the figure, they must have the same sense(assumedhere to be counterclockwise), and the relationF1d1: F2d2

(3.4e)

must be satisfted.To prove that they are equivalent, we shall show that the first couple can be transformed into the second by means of the operationslisted above.

3.I3 EQUIVATENT COUPLESFigure 3.34 showsthree coupleswhich act successively the same on rectangularbox. As seenin the preceding section,the only motion a couple can impart to a rigid body is a rotation. Since each of the three couples shown has the same moment M (same direction and samemagnitudeM : L20lb . in.), we can expectthe three couples to have the same effect on the box.

3.13 Equivolent Couples | 09

30 lbz

Fig. 3.34

(a)

(b)

(c)

As reasonableas this conclusionappears,we should not accept it hastily.While intuitive feeling is of great help in the study of mechanics, it should not be accepted as a substitute for logical reasoning. Before stating that two systems(or groups) of forces have the same effect on a rigid body, we should prove that fact on the basisof the experimental evidence introduced so far. This evidence consistsof the parallelogramlaw for the addition of two forces (Sec. 2,2) and the principle of transmissibility (Sec. 3.3). Therefore, we will state that tu;o systemsof forces are equioalent (i.e., they have the same effect on a rigid body) if u;e can transform one of them i.ntothe other by meansof one or seoeralof the following operations:(1) replacing two forcesacting on the sameparticle by their resultant;(2) resolving a force into two components;(3) canceling two equal and opposite forces acting on the sameparticle; (4) attachingto the sameparticle two equal and opposite forces; (5) moving a force along its line of action. Each of these operationsis easilyjustified on the basisof the parallelogramlaw or the principle of transmissibility. Let us now prove thattuoo coupleshaoingthe sampmoment M are equi,oalenf.First consider two couples contained in the same plane, and assumethat this plane coincides with the plane of the figure (Fig, 3,35),The first coupleconsists the forcesF1 and -F1 of of magnitude F,, which are located at a distanced1 from each other (Fig. 3.35a),and the secondcoupleconsists the forcesF2 and -F2 of of magnitude F2, which are located at a distanced,2from each other (Fig. 3.35d).Sincethe two couples havethe samemoment M, which is perpendicular to the plane of the figure, they must have the same sense(assumedhere to be counterclockwise), and the relationF1d1: F2d2

(3.4e)

must be satisfted.To prove that they are equivalent, we shall show that the first couple can be transformed into the second by means of the operationslisted above.

in whetherthey are contained the samemomentM are equivalentplaneor in parallelplanes. the sarne is The property *" hu,r" juit established very important for the of correct .,ni"trtandlng of thb mechanics rigid bodies.It indicates that when a couple acts on a rigid body, it does not matter where the two forces foiming the couple act or what magnitude and direction they have. The only thing which corrnts is the moment of the couple (magnitudeand direction).Coupleswith the samemoment will^harethi sameeffect on the rigid body.

by 3.15 CouplesCon Be Represented Vectors I I I

Ot 3.| 4 ADDITION COUPIESConsider two intersecting planes P1 and P2 and tr,vocouples acting ^We can, without any loss of generality, respectively in P1 and pr. -F1 perof urri-" that the couplein P1consists two forcesF1 and and-acting -Fr pendicular to the line of intersection of the two planes '::l that the iespectively at A and B (Fig. 3.37a). Similarly, we assume. to of in i2 consists twoTorcesF2 and -F2 perpendicular AB "o.bl" the resultant R of and'acting respectivelyat A and B. It is clear thit -F1 and -F2 form a couple' F1 and f", a"i the resultant -R of Denoting by r the vector joining B to A and recalling the definition of the mlment of a coupie (Sec.3.12),we er?ressthe moment M of the resulting couple as follows: M:rxR:rx(F1 and, by Varignon'stheorem, M : r x F 1 + r X F 2But the first term in the expressionobtained represents the moment M1 of the couple in P1.and'the second term representsthe moment M2 of the couple in P2. We have M : Ml * M2 (3.50) Fig.3.37

tF2)

and we conclude that the sum of two couples of moments M1 and M2 is a couple of moment M equal to the vector sum of M1 and M2

(Fig.3.37b).

BYVECTORS CAN O.I5 COUPLES BEREPRESENTEDAs we saw in Sec. 3.13, couples which have the same moment, whether they act in the sameplane or in parallel planes,are equivalent. There is therefore no need to draw the actual fbrces forming a givencouplein order to define its effect on a rigid body (Fig. 3'384)' it is sufficient to draw an arrow equal in magnitude and direction to the moment M of the couple (Fi-g.3.38b).bn the other hand, we saw in Sec.3.14 that the sitm of two couplesis itself a couple and that the moment M of the resultantcouplecan be obtainedby fbrmins the vector sum of the momentsMr and Me of the givencouples' obeythe law of additionofvectors,and the arrowused TFus,couples in Fig. S.igb to representthe couple defined in Fig. 3'38a can truly a be considered vector.

in whetherthey are contained the samemomentM are equivalentplaneor in parallelplanes. the sarne is The property *" hu,r" juit established very important for the of correct .,ni"trtandlng of thb mechanics rigid bodies.It indicates that when a couple acts on a rigid body, it does not matter where the two forces foiming the couple act or what magnitude and direction they have. The only thing which corrnts is the moment of the couple (magnitudeand direction).Coupleswith the samemoment will^harethi sameeffect on the rigid body.

by 3.15 CouplesCon Be Represented Vectors I I I

Ot 3.| 4 ADDITION COUPIESConsider two intersecting planes P1 and P2 and tr,vocouples acting ^We can, without any loss of generality, respectively in P1 and pr. -F1 perof urri-" that the couplein P1consists two forcesF1 and and-acting -Fr pendicular to the line of intersection of the two planes '::l that the iespectively at A and B (Fig. 3.37a). Similarly, we assume. to of in i2 consists twoTorcesF2 and -F2 perpendicular AB "o.bl" the resultant R of and'acting respectivelyat A and B. It is clear thit -F1 and -F2 form a couple' F1 and f", a"i the resultant -R of Denoting by r the vector joining B to A and recalling the definition of the mlment of a coupie (Sec.3.12),we er?ressthe moment M of the resulting couple as follows: M:rxR:rx(F1 and, by Varignon'stheorem, M : r x F 1 + r X F 2But the first term in the expressionobtained represents the moment M1 of the couple in P1.and'the second term representsthe moment M2 of the couple in P2. We have M : Ml * M2 (3.50) Fig.3.37

tF2)

and we conclude that the sum of two couples of moments M1 and M2 is a couple of moment M equal to the vector sum of M1 and M2

(Fig.3.37b).

BYVECTORS CAN O.I5 COUPLES BEREPRESENTEDAs we saw in Sec. 3.13, couples which have the same moment, whether they act in the sameplane or in parallel planes,are equivalent. There is therefore no need to draw the actual fbrces forming a givencouplein order to define its effect on a rigid body (Fig. 3'384)' it is sufficient to draw an arrow equal in magnitude and direction to the moment M of the couple (Fi-g.3.38b).bn the other hand, we saw in Sec.3.14 that the sitm of two couplesis itself a couple and that the moment M of the resultantcouplecan be obtainedby fbrmins the vector sum of the momentsMr and Me of the givencouples' obeythe law of additionofvectors,and the arrowused TFus,couples in Fig. S.igb to representthe couple defined in Fig. 3'38a can truly a be considered vector.

couple tends to impart to the rigid body the samerotational motion O that the foice F tended-toprod.t"" before it was transferred "boirtThe couple is represented a couplevector Me perpendicuiar by to O. plane containing r and F. Since Me is a free vector, it may to the however,the couple vector is be applied anywhere;fir convenience, ,suuliu attachLdat O, together with F, and the combinationobtained (Fig' 3.39c)' is referred to as aforcelcouplesVstem If the force F had been moved from A to a different point O' (Fig. 3.404and c), the moment Mo, : r' x F of F about O' should hu# b""" computed, and a new force-couple system,corrsistingof F and of the cbuple vector Me,, would have been attachedal O'' The relation existingbetween the moments of F about O and O' is obtained by writing Mo, : r' X F = (r * s) x F = r X F * s x F Mo,:Mo*sXF

of 3.16 Resolutiono GivenForceinto " l-orce ol I ond o Louple

I l3

(3.51)

where s is the vector joining O' to O, Thus, the moment Mp' of F about O' is obtained by adding to the moment Me of F about O the vector product s X F representingthe moment about O' of the force F applied at O.

\a)

(b)

lc)

Fis.3.40

This result could also have been establishedby observingthat, in order to transfer to O' the force-couple system attached at O (Fig. 3.40band c), the couplevector M6 can be freely movedto O'; to to irove the force F from-O to O', however,it is necessary add to F a couple vector whose moment is equal to the moment about O' of the force F applied at O. Thus, the couple vector Me' must be the sum of Ms and the vector s x F' As noted above,the force-couplesystemobtainedby transferring of a force F from a point A to a point O consists F and a couplevector *y to F. Conversely, force-couplesystemcon-sistilg Me perpendicular of-"tot"" F and a couplevector Mo -hi"h aremutualiy perpendiatlar can be replacedby a iingle equivalentforce. This is done by mwing the force F in the-planeperpendicularto Me until its moment about O is equal to the momenl of the couple to be eliminated.

Pholo 3.2 The force exe*ed by eoch hond on the wrench could be replocedwith on equivolent system octing on the nut. force-couple

couple tends to impart to the rigid body the samerotational motion O that the foice F tended-toprod.t"" before it was transferred "boirtThe couple is represented a couplevector Me perpendicuiar by to O. plane containing r and F. Since Me is a free vector, it may to the however,the couple vector is be applied anywhere;fir convenience, ,suuliu attachLdat O, together with F, and the combinationobtained (Fig' 3.39c)' is referred to as aforcelcouplesVstem If the force F had been moved from A to a different point O' (Fig. 3.404and c), the moment Mo, : r' x F of F about O' should hu# b""" computed, and a new force-couple system,corrsistingof F and of the cbuple vector Me,, would have been attachedal O'' The relation existingbetween the moments of F about O and O' is obtained by writing Mo, : r' X F = (r * s) x F = r X F * s x F Mo,:Mo*sXF

of 3.16 Resolutiono GivenForceinto " l-orce ol I ond o Louple

I l3

(3.51)

where s is the vector joining O' to O, Thus, the moment Mp' of F about O' is obtained by adding to the moment Me of F about O the vector product s X F representingthe moment about O' of the force F applied at O.

\a)

(b)

lc)

Fis.3.40

This result could also have been establishedby observingthat, in order to transfer to O' the force-couple system attached at O (Fig. 3.40band c), the couplevector M6 can be freely movedto O'; to to irove the force F from-O to O', however,it is necessary add to F a couple vector whose moment is equal to the moment about O' of the force F applied at O. Thus, the couple vector Me' must be the sum of Ms and the vector s x F' As noted above,the force-couplesystemobtainedby transferring of a force F from a point A to a point O consists F and a couplevector *y to F. Conversely, force-couplesystemcon-sistilg Me perpendicular of-"tot"" F and a couplevector Mo -hi"h aremutualiy perpendiatlar can be replacedby a iingle equivalentforce. This is done by mwing the force F in the-planeperpendicularto Me until its moment about O is equal to the momenl of the couple to be eliminated.

Pholo 3.2 The force exe*ed by eoch hond on the wrench could be replocedwith on equivolent system octing on the nut. force-couple

SAMPLE PROBLEM3.7Replace the couple and lorce shown by an equivalent single force applied to the lever. Determine the distancefrom the shaft to the point of application of this equivalent force.

60 mm

200

'/ '260mm

a// / /

/i gOtlJTlOtil i SoLuTloN

t

I

4 N'nJk

First the given force and couple are replacedby an equivalentforce-couple systemat O. We move the force F : -(400 N)j to O and at tle sametime : add a couple of moment M6 equal to the moment about O of the force in # k' its original position. {W} _ l + r . r , , F l 4 . g { r\ . r r ,k + 150mm ] M o - O B x F : [ ( 0 . 1 5 0 m )+ ( 0 . 2 6 0 m ) j x ( - 4 0 0 N ) j i $ ,*nnr,1 =-(60N.m)k This couple is added to the couple of moment -(24 N ' m)k formed by the two 200-N forces, and a couple of moment -(84 N . m)k is obtained. This Iast couple can be eliminated by applying F at a point C chosenin such a way that

F = - r4(x) N\j

-(84N'm)k: od , r' : t(OC)cos + 60oi (OC)sin60Jl x (-400N)j(81N'rrr)k

: - (OC)cos 0"(400N)k 6

We concludethat_ (400 N)i

(OC) cos 60" : 0.210 m : 210 mm- (2'l N'rn)k

BC : 12-t) nm

ts

&tlr*rnutiv* Sslufio$, Since the effect of a couple does not depend on its location, the couple of moment -(24 N ' m)k can be moved to B; we thus obtain a force-couple system at B. The couple can now be eliminated by applpng F at a point C chosenin such a way that

-(24N'm)k:nd*r: - ( B C ) c o s6 0 " ( 4 0 0 N ) k150mm

We conclude that (BC) cos 60' : 0.060 m : 60 mm BC : 120 mm : 300 mm * 120 mm OC : 42t) x*tz OC : OB + BC

- (100 N)j r4(r0 \)j

il5

SAMPLE PROBLEM3.7Replace the couple and lorce shown by an equivalent single force applied to the lever. Determine the distancefrom the shaft to the point of application of this equivalent force.

60 mm

200

'/ '260mm

a// / /

/i gOtlJTlOtil i SoLuTloN

t

I

4 N'nJk

First the given force and couple are replacedby an equivalentforce-couple systemat O. We move the force F : -(400 N)j to O and at tle sametime : add a couple of moment M6 equal to the moment about O of the force in # k' its original position. {W} _ l + r . r , , F l 4 . g { r\ . r r ,k + 150mm ] M o - O B x F : [ ( 0 . 1 5 0 m )+ ( 0 . 2 6 0 m ) j x ( - 4 0 0 N ) j i $ ,*nnr,1 =-(60N.m)k This couple is added to the couple of moment -(24 N ' m)k formed by the two 200-N forces, and a couple of moment -(84 N . m)k is obtained. This Iast couple can be eliminated by applying F at a point C chosenin such a way that

F = - r4(x) N\j

-(84N'm)k: od , r' : t(OC)cos + 60oi (OC)sin60Jl x (-400N)j(81N'rrr)k

: - (OC)cos 0"(400N)k 6

We concludethat_ (400 N)i

(OC) cos 60" : 0.210 m : 210 mm- (2'l N'rn)k

BC : 12-t) nm

ts

&tlr*rnutiv* Sslufio$, Since the effect of a couple does not depend on its location, the couple of moment -(24 N ' m)k can be moved to B; we thus obtain a force-couple system at B. The couple can now be eliminated by applpng F at a point C chosenin such a way that

-(24N'm)k:nd*r: - ( B C ) c o s6 0 " ( 4 0 0 N ) k150mm

We conclude that (BC) cos 60' : 0.060 m : 60 mm BC : 120 mm : 300 mm * 120 mm OC : 42t) x*tz OC : OB + BC

- (100 N)j r4(r0 \)j

il5

3.7S

Two parallel 60-N forces are applied to a lever as shown' Determine the moment of the couple formed by the two forces (a) by resolving each force into horlontal and vertical components and adding tie moments of the two resulting couples. (b) by using the perp.ndi"ular distancebetween the two forces.tt'r by surnming the moments of the two forces about point A.

Fig. P3.7O 3.71 A plate in the shapeof a parallelogramis acted upon,bl' tYo coupl&. D"tettoine (a) the tnoment ol the couple formed by,Jhg two 'zt-lb fo.""r, (b) the perpendiculardistancebetweenthe 12-lb forces if the resultant of tire i"o couplesis zero, (c) the value of a if the resultant couple is 72 lb ' in. clockwise and d is 42 in. g.72 A couple M of magnitude 18 N ' m is applied to the handle of u ,"r"i"dti r"r to tfuhten a screw into a block of wood Determine the magnitudLs of the two smallest horizontal forces that are equivalenl to M if they are applied (a) at corners A and D, (b) at corners B and C, (c) anywhere on the block.21lb C 12th

-Tl lb l6,inI

I

B

Fis. P3.71

roon.AD/

\

160 rnr.n

160mm

Fis. P3.72

r17

3.7S

Two parallel 60-N forces are applied to a lever as shown' Determine the moment of the couple formed by the two forces (a) by resolving each force into horlontal and vertical components and adding tie moments of the two resulting couples. (b) by using the perp.ndi"ular distancebetween the two forces.tt'r by surnming the moments of the two forces about point A.

Fig. P3.7O 3.71 A plate in the shapeof a parallelogramis acted upon,bl' tYo coupl&. D"tettoine (a) the tnoment ol the couple formed by,Jhg two 'zt-lb fo.""r, (b) the perpendiculardistancebetweenthe 12-lb forces if the resultant of tire i"o couplesis zero, (c) the value of a if the resultant couple is 72 lb ' in. clockwise and d is 42 in. g.72 A couple M of magnitude 18 N ' m is applied to the handle of u ,"r"i"dti r"r to tfuhten a screw into a block of wood Determine the magnitudLs of the two smallest horizontal forces that are equivalenl to M if they are applied (a) at corners A and D, (b) at corners B and C, (c) anywhere on the block.21lb C 12th

-Tl lb l6,inI

I

B

Fis. P3.71

roon.AD/

\

160 rnr.n

160mm

Fis. P3.72

r17

of 3.8O ShaftsA and B connect the gear box to the wheel assemblies ShaftsA and.B a tractor, and shaft C conneits it to the engine' lie in the vertical qz plane,while shaft C is directed along the r axis. Replacethe couples applied to the shafts with a single eq'uiualenfcouple, speciiying iti magnitude and the direction of its axis. 3.81 The tension in the cable attached to the end C of an adjustable boom ABC is 560 Ib. Replacethe force exerted by the cable at C with an equivalent force-couple system (a) at A, (b) at B' Fig. P3.80

Problems | 9 I

Fig. P3.81

3.82 A 1601b force P is applied at point A of a structural member, Replace P with (a) an equivalen[ force-couple system at C, (b) ar' f-r o*f.eqriivalent systemconsisiing of a vertical force at B and a second Fig. P3.82 force at D. 3.83 The 80-N horizontal force P acts on a bell crank as shown' (a) ReplaceP with an equivalentforce-couplesystemat B' (b) Find it" two verticalforcesai C and D that areiquivalent to the couple found in part a.

4ft----]

t_Fis. P3.83

loomm

,,t--l

g.84 A dirigible is tethered by a cable attached to its cabin at B. If the tensioi in the cable is i040 N, replace the force exerted by the cable at B with an equivalentsystemformed by two parallel forces applied at A and C.

Fig. P3,84

of 3.8O ShaftsA and B connect the gear box to the wheel assemblies ShaftsA and.B a tractor, and shaft C conneits it to the engine' lie in the vertical qz plane,while shaft C is directed along the r axis. Replacethe couples applied to the shafts with a single eq'uiualenfcouple, speciiying iti magnitude and the direction of its axis. 3.81 The tension in the cable attached to the end C of an adjustable boom ABC is 560 Ib. Replacethe force exerted by the cable at C with an equivalent force-couple system (a) at A, (b) at B' Fig. P3.80

Problems | 9 I

Fig. P3.81

3.82 A 1601b force P is applied at point A of a structural member, Replace P with (a) an equivalen[ force-couple system at C, (b) ar' f-r o*f.eqriivalent systemconsisiing of a vertical force at B and a second Fig. P3.82 force at D. 3.83 The 80-N horizontal force P acts on a bell crank as shown' (a) ReplaceP with an equivalentforce-couplesystemat B' (b) Find it" two verticalforcesai C and D that areiquivalent to the couple found in part a.

4ft----]

t_Fis. P3.83

loomm

,,t--l

g.84 A dirigible is tethered by a cab