Moment of a force about a point and about an axis

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1 3. STATICS O RIGID BODIES In the precedng chapter t was assumed that each of the bodes consdered could be treated as a sngle partcle. Such a vew, however, s not always possble, and a body, n general, should be treated as a combnaton of a large number of partcles. The sze of the body wll have to be taken nto consderaton, as well as the fact that forces wll act on dfferent partcles and thus wll have dfferent ponts of applcaton. 3.1 Moment of a force about a pont and about an axs. Let us consder a force actng on a rgd body (RB) (g.3.1). We shall defne the moment of about 0 as the vector product of r and : M 0 = r (3.1) Accordng to the defnton of the cross product (see ch.a.7) the moment M 0 must be perpendcular to the plane contanng the pont 0 and the force, the sense of M 0 moment M 0 s s furnshed by the rght-hand rule, and the magntude of the M 0 = rsnθ = d (3.2) The magntude of M 0 measures the tendency of the force to make the RB rotate about a fxed axs drected along M 0. Recallng (A.21), the cross-product (3.1) can be convenently expressed by the followng determnant j k M = r = x y z (3.3) x y z Thus, the rectangular components of M 0 are M x = y z - z y M y = z x - x z (3.4) M z = x y - y x StRB-1

2 Let us now ntroduce a new concept, the concept of moment of a force about an axs. Consder agan a force actng on a rgd body and the moment M 0 of that force about 0 (g.3.2). Let 0L be an axs through 0; we defne the moment M 0L of about 0L as the projecton 0C of the moment M 0 on the axs 0L. Denotng by λ the unt vector along 0L, and recallng the expresson (A.14), we wrte M OL =λ M 0 = λ (r ) (3.5) whch shows that the moment M 0L of about the axs 0L s the scalar obtaned by formng the mxed trple product of λ, r and. Expressng M 0L n the form of a determnant, we wrte OL λ x x λ y y λ z M = x y z (3.6) where λ x, λ y, λ z are the drecton cosnes of axs 0L. The physcal sgnfcance of the moment M 0L becomes more apparent f we resolve nto two rectangular components 1 and 2, wth 1 parallel to 0L and 2 lyng n a plane P perpendcular to 0L (g.3.3). Resolvng r smlarly nto two components r 1 and r 2 and substtutng for and r nto (3.5), we get M OL = λ [(r 1 + r 2 ) ( )] = λ (r 1 1 ) + λ (r 1 2 ) + λ (r 2 1 ) + λ (r 2 2 ) (3.7) Notng that all mxed trple products, except the last one, are equal to zero, snce they nvolve vectors whch are coplanar when drawn from a common orgn, we have M OL = λ (r 2 2 ) (3.8) It s easly to check that the scalar M 0L measures the tendency of 2 to make the RB rotate about the fxed axs 0L. Snce the other component 1 of does not tend to make the body rotate about 0L, we conclude that: z StRB-2

3 The moment M 0L of about 0L measures the tendency of the force to mpart the rgd body a moton of rotaton about the fxed axs 0L. It s easly to check that the moment of about a coordnate axs s equal to the component of M 0 along that axs. It means that the formulae (3.4) can be regarded as moments of about respectve axes. 3.2 A couple. Two forces and, havng the same magntude, parallel lnes of acton, and opposte sense are sad to form a couple (g.3.4). Clearly, the sum of the components of the two forces n any drecton s zero. The sum of the moments of the two forces about a gven pont, however, s not zero. Whle the two forces wll not translate the body on whch they act, they wll tend to make t rotate. Let us calculate now the sum of the moments of the two forces formng a couple about a gven pont 0 (g.3.5). We fnd easly that r A + r B ( ) = (r A r B ) Settng r A r B = r, where r s the vector jonng the ponts of applcaton of the two forces, we conclude that the sum of the moments of and about 0 s represented by the vector M = r (3.9) whch s called the moment of the couple; t s a vector perpendcular to the plane contanng the two forces and ts magntude s M = r snθ (3.10) where d s the perpendcular dstance between the lnes of acton and. Snce the moment M n (3.9) s ndependent of the choce of the orgn 0 of the coordnate axes, we note that the same result would have been obtaned f the moments of and had been computed about a dfferent pont 0. Thus, the moment M of a couple s a free vector whch may be appled at any pont (g.3.6). StRB-3

4 rom the defnton of the moment of a couple, t also follows that two couples, one consstng of the forces 1 and 1, the other of the forces 2 and 2 (g.3.7), wll have equal moments f 1 d 1 = 2 d 2 (3.11) and f the two couples le n parallel planes (or n the same plane) and have the same sense. In fact, t means that two couples are the same f ther moments are equal. nally, snce the only characterstc of a couple s the moment whch s a free vector, we conclude that two couples havng the same moment M are equvalent, whether they are contaned n the same plane or n parallel planes. The property we have just establshed s very mportant for the correct understandng of the mechancs of rgd bodes. It ndcates that, when a couple acts on a RB, t does not matter where the two forces formng the couple act, or what magntude and drecton they have. The only thng whch counts s the moment of the couple. Couples wth the same moment wll have the same effect on the RB. There s therefore no need to draw the actual forces formng a gven couple n order to defne ts effect on a rgd body. It s suffcent to draw an arrow equal n magntude and drecton to the moment M of a couple. Couples obey the law of addton of vectors and therefore the sum of two couples of moments M 1 and M 2 s M = M 1 +M 2 (3.12).e. a couple of the moment M equal to the vector sum of M 1 and M 2 Establshng the equvalency of couples we have referred mostly to the defnton of the moment of the couple (3.9) and to the fact that ths s a free vector. However, proofs of the equvalency of any force systems may obtan more rgorous form f they wll use prncples of statcs (see 1.3) only. Therefore, we shall state that two systems of forces are equvalent (.e. they have the same effect on a RB) f we can transform one of them nto the other by means of one or several of the followng operatons: (1) replacng two forces StRB-4

5 actng on the same partcle by ther resultant; (2) resolvng a force nto two components; (3) cancelng two equal and opposte forces actng on the same partcle; (4) attachng to the same partcle two equal and opposte forces; (5) movng a force along ts lne of acton. Each of these operatons s easly justfed on the bass of the parallelogram law or the prncple of transmssblty. Let us now prove that two couples havng the same moment M are equvalent. Ths tme, however, we shall use only the operatons lsted above. rst, we shall consder two couples contaned n the same plane (g.3.8). Next we shall consder two couples contaned n parallel planes P 1 and P 2 (g.3.9). The extended explanatons of both fgures and proofs wll be gven n the lecture. 3.3 Resultant of parallel forces. Let us fnd the resultant of two parallel forces actng on a RB. Two cases are possble: 1) the forces are of the same sense, and 2) the forces are of opposte sense. Composton of two forces of the same sense Consder a RB on whch act two parallel forces 1 and 2 (g.3.10). Applyng the 1 st and 2 nd prncples of statcs we can replace the gven system of parallel forces wth an equvalent system of concurrent forces Q 1 and Q 2. Summng them accordng to the parallelogram law we easly obtan that Q 1 + Q 2 = Then transferrng the sum to C we get the resultant R of two forces 1 and 2. Consderng the constructon of the resultant R as well as the trangles A0C, Akl, B0C and Bmn we get: 1) the magntude of the resultant R = (3.13) and 2) the poston of pont of applcaton C of the resultant BC/ 1 = AC/ 2 = AB/R (3.14) Thus, the resultant of two parallel forces of the same sense actng on a RB s equal to the sum of ther magntudes, parallel to them, and of the same sense; StRB-5

6 the lne of acton of the resultant les between the ponts of applcaton of the component forces, ts dstances from the ponts beng nversely proportonal to the magntudes of the forces. Composton of two forces of the opposte sense. Consder the concrete case of 1 > 2 (g.3.11). Take a pont C on the extenson of BA and apply two balanced forces P and P parallel to the gven forces 1 and 2. The magntudes P and P are chosen to satsfy the equaton: P = 1 2 (3.15) Compoundng forces 2 and P, we fnd from (3.13) and (3.14) that ther resultant Q s equal n magntude to 2 + P,.e., t s equal to force 1 and appled at A. orces 1 and 2 are replaced by a sngle force P = R, ther resultant. The magntude and pont of applcaton of the resultant s determned by (3.15) and (3.14). Thus, the resultant of two parallel forces of the opposte sense actng on a RB s equal to the dfference between ther magntudes, parallel to them, and has the same sense as the greater force; the lne of acton of the resultant les on an extenson of the lne segment connectng the ponts of applcaton of the component forces, ts dstances from the ponts beng nversely proportonal to the magntudes of the forces. If several parallel forces act on a body, ther resultant, f any, can be found by consecutvely applyng the rule of composton of two forces, or by a method whch wll be examned n chapter Equvalent transformatons of a gven system of forces. Resoluton of a gven force nto a force at 0 and a couple. Consder a force actng on a rgd body at a pont A defned by the poston vector r (g.3.12a). Suppose that for some reason we would rather have the force act at pont 0. We know that we can move along ts lne of acton but we cannot move t to a pont 0 away from the orgnal lne of acton wthout modfyng the acton of on the RB. StRB-6

7 We may, however, attach a balanced system of two forces and at pont 0 (g.3.12b). As a result of ths transformaton, a force s now appled at 0 ; the other two forces form a couple of a moment M 0 = r. Thus, any force actng on a rgd body may be moved to an arbtrary pont 0, provded that a couple s added, of moment equal to the moment of about 0. The combnaton obtaned (g.3.12c) s referred to as force-couple system. Reducton of a system of forces to one force and one couple. Consder a system of forces 1, 2, 3,, actng on a rgd body at ponts A 1, A 2, A 3,, defned by the poston vectors r 1, r 2, r 3,, (g.3.13a). Suppose we need reduce the system of forces to a gven centre 0. As seen just above, each force may be moved from A to the gven centre 0 f a couple of moment M = r s added to the orgnal system of forces. Repeatng ths procedure for all the forces, we obtan the system shown n g.3.13b. Snce the forces are now concurrent, they may be replaced by ther resultant R. Smlarly, the couple vectors M may be added vectorally and replaced by a sngle couple vector M R 0, called moment resultant,.e.: R = and R 0 = M = ( r ) M (3.16) Thus, we may fnally state that any system of forces, however complex, may be reduced to an equvalent force-couple system actng at a gven pont 0 (g.3.13c) Equvalent systems of forces It appears an mportant queston: when two systems of forces can be regarded as equvalent? We have seen just above that any system of forces actng on a RB may be reduced to a force-couple system at a gven pont 0. Ths equvalent force-couple system characterzes completely the effect of the gven system on same force-couple at a gven pont 0. Recallng that the force-couple at 0 s defned by the relatons (3.16), we state: Two systems of forces 1, 2, 3,, and 1, 2, 3,, are equvalent f, and only f, the sums of forces and the StRB-7

8 sums of moments about a gven pont 0 of the forces the two systems are, respectvely, equal,.e. = and M = M (3.17) Note that, to prove that two systems of forces are equvalent, the second of relatons (3.17) needs to be establshed wth respect to only one pont 0. It wll hold, however to any pont f the two systems are equvalent. Reducton of a system of forces to a wrench. In the general case of a system of forces n space, the force-couple system at 0 R conssts of the force R and a couple vector M 0 whch are not perpendcular, and nether of whch s zero (g.3.14a). Thus, the system of forces cannot be reduced to a sngle force or a sngle couple. The couple vector, however, may be replaced by two other vectors obtaned by resolvng M R 0 nto a component M 1 along R and a component M 2 n a plane perpendcular to R (g.3.14b). The couple vector M 2 and the force R may then be replaced by a sngle force R actng along a new lne of acton. The orgnal system of forces thus reduces to R and to the couple vector M 1 (g.3.14c),.e. to R and a couple actng n the plane perpendcular to R. Ths partcular force-couple system s called a wrench because the resultng combnaton of push and twst s the same that would be caused by an actual wrench. The lne of acton of R s known as the axs of wrench, and the rato M 1 /R s called the ptch of the wrench. System of forces equvalent to zero An mportant partcular case of the reducton of a system of forces to a forcecouple system occurs when both the force R and the couple vector M R 0 are equal to zero. The system of forces s sad to be equvalent to zero. Such a system has no effect on the rgd body on whch t acts, and the rgd body s sad to be n equlbrum. Ths case wll be consdered n detal n the next secton. Bblography Beer.P, Johnston E.R., Jr., Vector Mechancs for Engneers, McGraw-Hll StRB-8

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