182 Dynamics of Mechanical Systems By eliminating S A and S B between these three equations, we have: (6.7.13) Now, suppose that A is represented by a single force, say F A , passing through some common point C of A and B (or A and B extended) together with a couple with torque T A . Similarly, let B be represented by a single force F B passing through C together with a couple with torque T B . Then, because A and B taken together form a zero system (Eq. (6.7.3)), the resultant of A and B and the moment of A and B about C must be zero. That is, (6.7.14) and (6.7.15) Equations (6.7.13), (6.7.14), and (6.7.15), or the equivalent wording, represent the law of action and reaction. 6.8 First Moments Consider a particle P with mass m (or, alternatively, a point P with associated mass m) as depicted in Figure 6.8.1. Let O be an arbitrary reference point, and let p be a position vector locating P relative to O. The first moment of P relative to O, φ P/O , is defined as: (6.8.1) Consider next a set S of N particles P i (i = 1,…, N) having masses as in Figure 6.8.2, where O is an arbitrary reference point. The first moment of S for O, φ S/O , is defined as the sum of the first moments of the individual particles of S for O. That is, (6.8.2) Observe that, in general, φ S/O is not zero. However, if a point G can be found such that the first moment of S relative to G, φ S/G , is zero, then G is defined as the mass center of S. Using this definition, the existence and location of G can be determined from Eq. (6.8.2). Specifically, if G is the mass center and if r i locates p i relative to G, as in Figure 6.8.3, then the first moment of S relative to G may be expressed as: (6.8.3) ˆˆ SS AB +=0 ˆ S ˆ S ˆ S ˆ S ˆ S ˆ S ˆ S ˆ S FF F F AB A B += =−0 or TT T T AB A B += =−0 or φ PO D m= p φ SO D NN ii i N =+++= = ∑ mp mp m p mp 11 22 1 K φ SG ii i N m== = ∑ r 1 0 0593_C06_fm Page 182 Monday, May 6, 2002 2:28 PM Forces and Force Systems 183 From Figure 6.8.3, we have: (6.8.4) Hence, by substituting into Eq. (6.8.3), we obtain: (6.8.5) By solving the last equation for p G , we obtain: (6.8.6) where M is the total mass of S. That is, (6.8.7) FIGURE 6.8.1 A particle P and reference point O. FIGURE 6.8.2 A set S of particles and a reference point O. FIGURE 6.8.3 A set S of particles with mass center G. p O P(m) p P (m ) P (m ) P (m ) P (m ) O 1 1 2 2 i i N N i S p P (m ) P (m ) P (m ) O 1 1 2 2 i i i S G r i p G pp r rpp iGi iiG =+ =− or mm mm mm ii i N ii G i N ii iG i N i N ii i i N i N G rpp pp pp == == == ∑∑ ∑∑ ∑∑ =− () =− =−       = 11 11 11 0 pp Gii i N Mm= () = ∑ 1 1 Mm i i N = = ∑ 1 0593_C06_fm Page 183 Monday, May 6, 2002 2:28 PM 184 Dynamics of Mechanical Systems Equation (6.8.6) demonstrates the existence of G by providing an algorithm for its location. We may think of a body as though it were composed of particles, just as a sandstone is composed of particles of sand. Then, the sums in Eqs. (6.8.2) through (6.8.7) become very large, and in the limit they may be replaced by integrals. For homogeneous bodies, the mass is uniformly distributed throughout the region, or geometric figure, occupied by the body. The mass center location is then solely determined by the shape of the figure of the body. The mass center position is then said to be at the centroid of the geometric figure of the body. The centroid location for common and simple geometric figures may be determined by routine integration. The results of such integra- tions are listed in figurative form in Appendix I. As the name implies, a centroid is at the intuitive center or middle of a figure. As an illustration of these concepts, consider the gravitational forces acting on a body B with an arbitrary shape. Let B be composed of N particles P i having masses m i (i = 1,…, N), and let G be the mass center of B, as depicted in Figure 6.8.4. Let the set of all the gravitational forces acting on B be replaced by a single force W passing through G together with a couple with torque T. Then, from the definition of equivalent force systems, W and T are: (6.8.8) and (6.8.9) where k is a downward-directed unit vector as in Figure 6.8.4 and M is the total mass of particles of B. The last equality of Eq. (6.8.9) follows from the definition of the mass center in Eq. (6.8.3). Equations (6.8.7) and (6.8.8) show how dramatic the reduction in forces can be through the use of equivalent force systems. 6.9 Physical Forces: Inertia (Passive) Forces Inertia forces arise due to acceleration of particles and their masses. Specifically, the inertia force on a particle is proportional to both the mass of the particle and the acceleration of FIGURE 6.8.4 Gravitational forces on the particles of a body. 1 i S P P G k m g m g m g i 1 r i 2 P 2 Wk kk==       = == ∑∑ mg m g Mg i i N i i N 11 Trk rk=×=       ×= == ∑∑ ii i N ii i N mg m g 11 0 0593_C06_fm Page 184 Monday, May 6, 2002 2:28 PM Forces and Force Systems 185 the particle. However, the inertia force is directed opposite to the acceleration. Thus, if P is a particle with mass m and with acceleration a in an inertial reference frame R, then the inertia force F * exerted on P is: (6.9.1) An inertial reference frame is defined as a reference frame in which Newton’s laws of motion are valid. From elementary physics, we recall that from Newton’s laws we have the expression: (6.9.2) where F represents the resultant of all applied forces on a particle P having mass m and acceleration a. By comparing Eqs. (6.9.1) and (6.9.2), we have: (6.9.3) Equation (6.9.3) is often referred to as d’Alembert’s principle. That is, the sum of the applied and inertia forces on a particle is zero. Equation (6.9.3) thus also presents an algorithm or procedure for the analysis of dynamic systems as though they were static systems. For rigid bodies, considered as sets of particles, the inertia force system is somewhat more involved than for that of a single particle due to the large number of particles making up a rigid body; however, we can accommodate the resulting large number of inertia forces by using equivalent force systems, as discussed in Section 6.5. To do this, consider the representation of a rigid body as a set of particles as depicted in Figure 6.9.1. As B moves in an inertial frame R, the particles of B will be accelerated and thus experience inertia forces; hence, the inertia force system exerted on B will be made up of the inertia forces on the particles of B. The inertia force exerted on a typical particle P i of B is: (6.9.4) where m i is the mass of P i and A i is the acceleration of P i in R. Using the procedures of Section 6.5, we can replace this system of many forces by a single force F * passing through an arbitrary point, together with a couple having a torque FIGURE 6.9.1 Representation of a rigid body as a set of particles. Fa * =−m Fa= m FF+= * 0 Fa iii m * =− P (m ) P (m ) P (m ) 1 1 2 2 N N i B r P (m ) i i R 0593_C06_fm Page 185 Monday, May 6, 2002 2:28 PM 186 Dynamics of Mechanical Systems T * . It is generally convenient to let F * pass through the mass center G of the body. Then, F * and T * are: (6.9.5) and: (6.9.6) where r i locates P i relative to G. Using Eq. (4.9.6), we see that because both P i and G are fixed on B, a i may be expressed as: (6.9.7) where αα αα and ωω ωω are the angular acceleration and angular velocity of B in R. Hence, by substituting into Eq. (6.9.5), F* becomes: (6.9.8) or (6.9.9) where M is the total mass of B and where the last two terms of Eq. (6.9.8) are zero because G is the mass center of B (see Eq. (6.8.3)). Similarly, by substituting for a i in Eq. (6.9.6), T * becomes: (6.9.10) Fa * =− () = ∑ m ii i N 1 Tra * =×− () = ∑ iii i N m 1 aa r r iG i i =+×+×× () ααωωωω Farr ar r a * =− +×+×× () [] =−       −× () −× × ()         =− − × − × × () = == = ∑ ∑∑ ∑ m mm m M iG i i i N i i N Gii i N ii i N G ααωωωω ααωωωω ααωωωω 1 11 1 00 Fa=−M G Tr ar r ra r r r r arr r * =×− () +×+× × () [] =−       ×− ×× () −××× () [] =− × − × × () −× = == = ∑ ∑∑ ∑ ii i N Gi i ii i N Giii i N ii i i N Giii ii m mm m mm 1 11 1 0 ααωωωω ααωωωω ααωω××× () == ∑∑ ωω r i i N i N 11 0593_C06_fm Page 186 Monday, May 6, 2002 2:28 PM Forces and Force Systems 187 or (6.9.11) where the first term of Eq. (6.9.10) is zero because G is the mass center, and the last term is obtained from its counterpart in the previous line by using the identity: (6.9.12) To see this, simply expand the triple products of Eq. (6.9.10) using the identity. Specifically, (6.9.13) and (6.9.14) where the first terms are zero because r i and ωω ωω are perpendicular to ωω ωω × r i . Comparing Eqs. (6.9.13) and (6.9.14), we see the results are the same. That is, (6.9.15) Neither of the terms of Eq. (6.9.11) is in a form convenient for computation or analysis; however, the terms have similar forms. Moreover, we can express these forms in terms of the inertia dyadic of the body as discussed in the next chapter. This, in turn, will enable us to express T * in terms of the moments and products of inertia of B for its mass center G. References 6.1. Kane, T. R., Analytical Elements of Mechanics, Vol. 1, Academic Press, New York, 1961, p. 150. 6.2. Kane, T. R., Dynamics, Holt, Rinehart & Winston, New York, 1968, pp. 92–115. 6.3. Huston, R. L., Multibody Dynamics, Butterworth-Heinemann, Stoneham, MA, 1990, pp. 153–212. Problems Section 6.2 Forces and Moments P6.2.1: A force F with magnitude 12 lb acts along a line L passing through points P and Q as in Figure P6.2.1. Let the coordinates of P and Q relative to a Cartesian reference frame be as shown, measured in feet. Express F in terms of unit vectors n x , n y , and n z , which are parallel to the X-, Y-, and Z-axes. Trr rr * =×× ()         −× × × ()         ∑∑ mm ii i i N ii i i N ααωωωω ====11 abc acbabc×× () ≡⋅ () −⋅ () r rrrrrrr i iiiiiii ××× () [] =⋅ × () −⋅ () ×=−⋅ () ×ωωωωωωωωωωωωωωωω ωωωωωωωωωωωωωωωω××× () [] =⋅ × () −⋅ () ×=− ⋅ () ×r r rr rr rr i i ii ii ii rrrr iiii ××× () [] =×× × () [] ωωωωωωωω 0593_C06_fm Page 187 Monday, May 6, 2002 2:28 PM 188 Dynamics of Mechanical Systems P6.2.2: See Problem P6.2.1. Find the moment of F about points O, A, and B of Figure P6.2.1. Express the results in terms of n x , n y , and n z . P6.2.3: A force F with magnitude 52 N acts along diagonal OA of a box with dimensions as represented in Figure P6.2.3. Express F in terms of the unit vectors n 1 , n 2 , and n 3 , which are parallel to the edges of the box as shown. P6.2.4: See Problem P6.2.3. Find the moment of F about the corners A, B, C, D, E, G, and H of the box. Express the results in terms of n 1 , n 2 , and n 3 . Section 6.3 Systems of Forces P6.3.1: Consider the force system exerted on the box as represented in Figure P6.3.1. The force system consists of ten forces with lines of action and magnitudes as listed in Table P6.3.1. Let the dimensions of the box be 12 m, 4 m, and 3 m, as shown. a. Express the forces F 1 ,…, F 10 in terms of the unit vectors n 1 , n 2 , and n 3 shown in Figure P6.3.1. b. Find the resultant of the force system expressed in terms of n 1 , n 2 , and n 3 . c. Find the moment of the force system about O (expressed in terms of n 1 , n 2 , and n 3 ). FIGURE P6.2.1 A force F acting along a line L. FIGURE P6.2.3 A force F acting along a box diagonal. O Z Y X n n n z y x P(1,2,4) Q(2,5,3) B(0,3,0) A(2,0,0) F L n n n 1 2 3 F C D H G E B O A 12m 4m 3m 0593_C06_fm Page 188 Monday, May 6, 2002 2:28 PM Forces and Force Systems 189 P6.3.2: See Problem 6.3.1, and (a) find the moment of the force system about points C and H, and (b) verify Eq. (6.3.6) for these results. That is, show that: where R is the resultant of the force system. P6.3.3: A cube with 2-ft sides has forces exerted upon it as shown in Figure P6.3.3. The magnitudes and lines of action of these forces are listed in Table P6.3.3. FIGURE P6.3.1 A force system exerted on a box. TABLE P6.3.1 Forces, Their Magnitudes, and Lines of Action for the Force System of Figure P6.3.1 Force Magnitude (N) Line of Action F 1 18 BA F 2 25 BO F 3 30 AO F 4 10 CB F 5 12 OC F 6 26 OG F 7 20 DH F 8 25 EO F 9 16 DG F 10 24 CD FIGURE P6.3.3 Forces exerted on a cube. n n n A H F F F F F F F F F F B O C D E 3m 4m 12m G 8 10 9 4 3 5 1 2 2 3 1 7 6 M M CH R CH =+× A 2 ft D E H G B C O F F F F F F n n n 2 6 5 4 3 1 1 2 3 0593_C06_fm Page 189 Monday, May 6, 2002 2:28 PM 190 Dynamics of Mechanical Systems a. Determine the resultant R of this system of forces. b. Find the moments of the force system about points C, H, E, and G. Express the results in terms of the unit vectors n 1 , n 2 , and n 3 shown in Figure P6.3.3. 6.4 Special Force Systems: Zero Force Systems and Couples P6.4.1: In Figure P6.4.1, the box is subjected to forces as shown. The magnitudes and directions of the forces are listed in Table P6.4.1. Determine the magnitudes of forces F 1 , F 3 , F 4 , F 5 , F 6 , and F 10 so that the system of forces on the box is a zero system. TABLE P6.3.3 Forces, Their Magnitudes, and Lines of Action for the Force System of Figure P6.3.3 Force Magnitude (lb) Line of Action F 1 10 AB F 2 10 GH F 3 8 CB F 4 8 GD F 5 12 CD F 6 12 EO FIGURE P6.4.1 Force system exerted on a box. TABLE P6.4.1 Forces, Their Magnitudes, and Lines of Action for the Forces of Figure P6.4.1 Force Magnitude (N) Line of Action F 1 10 AB F 2 8 CB F 3 8 AO F 4 6 HG F 5 7 GO F 6 5 AE F 7 10 GD F 8 16 DB F 9 12 HD F 10 9 CO F 11 26 OD n n A H F F F F F F F F B O C D E 3m 4m 12m G 8 9 4 3 5 1 2 2 3 6 n 1 F 10 F F 11 7 0593_C06_fm Page 190 Monday, May 6, 2002 2:28 PM Forces and Force Systems 191 P6.4.2: Three cables support a 1500-lb load as depicted in Figure P6.4.2. By considering the connecting joint O of the cables to the load to be in static equilibrium and by recalling that forces in cables are directed along the cable, determine the forces in each of the cables. The location of the fixed ends of the cables (A, B, and C) are determined by their x, y, z coordinates (measured in feet) as shown in Figure P6.4.2. P6.4.3: Repeat Problem P6.4.2 if the load is 2000 lb. P6.4.4: Repeat Problem P6.4.3 if the coordinates of the cable supports (A, B, and C) are given in meters instead of feet. P6.4.5: Show that the force system exerted on the box shown in Figure P6.4.5 is a couple. The magnitudes and directions of the forces are listed in Table P6.4.5. FIGURE P6.4.2 Cables OA, OB, and OC supporting a vertical load. FIGURE P6.4.5 Forces system exerted on a box. TABLE P6.4.5 Forces, Their Magnitudes, and Lines of Action for the Forces of Figure P6.4.5 Force Magnitude (N) Line of Action F 1 8 OC F 2 8 BA F 3 15 BO F 4 12 CB F 5 10 GA F 6 10 CD F 7 15 HE F 8 12 AO Z Y X O(0,0,0) C(3,7,5) B(-2,2,10) A(6,-3,8) 1500 lb A H F F F F O C D E G 8 4 3 5 2 6 1 F F 7 B F F 60 cm 40 cm 30 cm 0593_C06_fm Page 191 Monday, May 6, 2002 2:28 PM [...]... j (7 .5. 4) Then, by forming the projection of I P O onto nb we obtain the product of inertia I P O, which a ab in view of Eq (7 .5. 3) can be expressed as: I P O = n b ⋅ I P O = ai bj n j ⋅ IiP O a ab or P I P O = ai bjIij O ab (7 .5. 5) 059 3_C07_fm Page 206 Monday, May 6, 2002 2:42 PM 206 Dynamics of Mechanical Systems Observe that the scalar components ai and bi of na and nb of Eqs (7 .5. 2) and (7 .5. 3)... terms of the unit vectors n1, n2, and n3 of Figure P6 .5. 1.) FIGURE P6 .5. 1 A force system exerted on a box TABLE P6 .5. 1 Forces, Their Magnitudes, and Lines of Action for the Force System of Figure P6 .5. 1 Force F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 Magnitude (N) Line of Action 18 25 30 10 12 26 20 25 16 24 BA BO AO CB OC OG DH EO DG CD 059 3_C06_fm Page 193 Monday, May 6, 2002 2:28 PM Forces and Force Systems. .. Dyadics 2 05 Finally, for systems of particles or for rigid bodies, the inertia dyadic is developed from the contributions of the individual particles That is, for a system S of N particles we have: N IS O = ∑I Pi O (7.4.13) i =1 7 .5 Transformation Rules Consider again the definition of the second moment vector of Eq (7.2.1): I P O = mp × (n a × p) a (7 .5. 1) Observe again the direct dependency of I P O... 059 3_C06_fm Page 192 Monday, May 6, 2002 2:28 PM 192 Dynamics of Mechanical Systems P6.4.6: See Problem P6.4 .5 Find the torque of the couple P6.4.7: See Problems P6.4 .5 and P6.4.6 Find the moment of the force system about points O, A, D, and G P6.4.8: Show that the magnitude of the torque of a simple couple (two equal-magnitude but oppositely directed forces) is simply the product of the magnitude of. .. the Geometric Parameters and Weights of Figures P6.7.1A and B i θi (°) ri (in.) ᐉi (in.) Wi (lb) 1 2 3 45 15 30 4. 45 6 .5 2 .5 11.7 14 .5 6.0 5. 0 3.0 1. 15 P6.7.3: See Problems P6.7.1 and P6.7.2 Suppose the equivalent force system is to be a wrench, where the couple torque M is a minimum Locate a point G on the line of action of the equivalent force Find the magnitudes of the equivalent wrench force and... 45 15 30 4. 45 6 .5 2 .5 11.7 14 .5 6.0 5. 0 3.0 1. 15 059 3_C06_fm Page 198 Monday, May 6, 2002 2:28 PM 059 3_C07_fm Page 199 Monday, May 6, 2002 2:42 PM 7 Inertia, Second Moment Vectors, Moments and Products of Inertia, Inertia Dyadics 7.1 Introduction In this chapter we review various topics and concepts about inertia Many readers will be familiar with a majority of these topics; however, some topics, particularly... and the xi, yi, zi in feet? TABLE P6.8.1 Masses and Coordinates of a Set of Particles Pi mi xi yi zi P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 6 4 3 8 1 9 4 5 5 2 –1 3 4 8 –2 1 7 –3 1 6 0 –7 5 0 –3 –2 –8 4 –8 5 2 4 5 7 –9 0 3 –2 9 1 P6.8.2: See Problem P6.8.1 From the definition of mass center as expressed in Eq (6.8.3) show that the coordinates of the mass center may be obtained by the simple expressions: N... distance of P from O and it also depends upon the direction of the unit vector na The form of the definition of Eq (7.2.1) is motivated by the form of the terms of the inertia torque of Eq (7.1.1) Indeed, for a set S of particles, representing a rigid body B (Figure 7.2.2), the second moment is defined as the sum of the second moments of the individual particles That is, N D IS O = a ∑ N I Pi a O = i =1 ∑... I13 − I 31I13 I 22 (7.7. 15) These coefficients are seen to be directly related to the elements Iij of the inertia matrix Indeed, they may be identified as: II = sum of diagonal elements of Iij III = sum of diagonal elements of the cofactor matrix of Iij IIII = determinant of Iij 059 3_C07_fm Page 211 Monday, May 6, 2002 2:42 PM Inertia, Second Moment Vectors, Moments and Products of Inertia, Inertia Dyadics... (7.6 .5) where IG O is defined as: a D where M is the total mass,     N ∑ m  , i i =1 of the particles of S; and IG O is the second moment of a particle located at G with a mass a equal to the total mass M of S Eq (7.6.4) is often called a parallel axis theorem The reason for this name can be seen from the analogous equation for the moments of inertia: that is, by examining the projection of the . Weights of Figures P6.8.7A and B i θθ θθ i (%) r i (in.) ᐉ i (in.) W i (lb) 1 45 4. 45 11.7 5. 0 2 15 6 .5 14 .5 3.0 3 30 2 .5 6.0 1. 15 C A D B O 3.7 in. 31 in. 17 in. OB = 13 in. E 10. 05 in. C. P6.8.1 Masses and Coordinates of a Set of Particles P i m i x i y i z i P 1 6 –102 P 2 43–74 P 3 34 5 5 P 4 8807 P 5 1 –2 –3 –9 P 6 91–20 P 7 47–83 P 8 5 –34–2 P 9 51 –89 P 10 26 51 FIGURE P6.8.4 A thin,. Geometric Parameters and Weights of Figures P6.7.1A and B i θθ θθ i (°) r i (in.) ᐉ i (in.) W i (lb) 1 45 4. 45 11.7 5. 0 2 15 6 .5 14 .5 3.0 3 30 2 .5 6.0 1. 15 FIGURE P6.7.4 Three springs in