Dynamics of Mechanical Systems 2009 Part 4 docx

132 Dynamics of Mechanical Systems where ωω ωω QP and αα αα QP are the angular velocity and angular acceleration, respectively, of the connecting rod QP. Because QP has planar motion, we see from Figure 5.3.4 that ωω ωω QP and αα αα QP may be expressed as: (5.3.17) where n z (= n x × n y ) is perpendicular to the plane of motion. Also, the position vector QP may be expressed as: (5.3.18) By carrying out the indicated operations of Eqs. (5.3.15) and (5.3.16) and by using Eqs. (5.3.13) and (5.3.14), v P and a P become: (5.3.19) and (5.3.20) Observe from Figure 5.3.4 that P moves in translation in the n x direction. Therefore, the velocity and acceleration of P may be expressed simply as: (5.3.21) where x is the distance OP. By comparing Eqs. (5.3.19) and (5.3.20) with (5.3.21), we have: (5.3.22) FIGURE 5.3.4 Geometrical parameters and unit vectors of the piston, connecting rod, and crank arm system. Q ᐉ φ θ r O n n n n y θ r z x P n x ωωαα QP z z =− =− ˙˙˙ φφnnand QP QP n n=−llcos sinφφ xy vnn Pxy rr=− − () +− () ΩΩsin ˙ sin cos ˙ cosθφφ θφll an n Px y r r =− − − () +− − + () Ω Ω 22 22 cos ˙˙ sin ˙ cos sin ˙˙ cos ˙ sin θφφφ φ θφ φφ φ ll ll vn an Px x xx== ˙˙˙ and P ˙ sin ˙ sinxr=− −Ωθφφl 0593_C05_fm Page 132 Monday, May 6, 2002 2:15 PM Planar Motion of Rigid Bodies — Methods of Analysis 133 (5.3.23) (5.3.24) and (5.3.25) Observe further from Figure 5.3.4 that x may be expressed as: (5.3.26) and that from the law of sines we have: (5.3.27) By differentiating in Eqs. (5.3.26) and (5.3.27), we immediately obtain Eqs. (5.3.22) and (5.3.23). Finally, observe that by differentiating in Eqs. (5.3.22) and (5.3.23) we obtain Eqs. (5.3.24) and (5.3.25). 5.4 Instant Center, Points of Zero Velocity If a point O of a body B with planar motion has zero velocity, then O is called a center of zero velocity. If O has zero velocity throughout the motion of B, it is called a permanent center of zero velocity. If O has zero velocity during only a part of the motion of B, or even for only an instant during the motion of B, then O is called an instant center of zero velocity. For example, if a body B undergoes pure rotation, then points on the axis of rotation are permanent centers of zero velocity. If, however, B is in translation, then there are no points of B with zero velocity. If B has general plane motion, there may or may not be points of B with zero velocity. However, as we will see, if B has no centers of zero velocity within itself, it is always possible to mathematically extend B to include such points. In this latter context, bodies in translation are seen to have centers of zero velocity at infinity (that is, infinitely far away). The advantage, or utility, of knowing the location of a center of zero velocity can be seen from Eq. (5.3.10): (5.4.1) where P and Q are points of a body B and where r locates P relative to Q. If Q is a center of zero velocity, then v Q is zero and v P is simply: (5.4.2) 0 =−rΩcos ˙ cosθφ φl ˙˙ cos ˙˙ sin ˙ cosxr=− − −Ω 22 θφφφ φll 0 22 =− − +rΩ sin ˙˙ cos ˙ sinθφ φφ φll xr=+cos cosθφl l sin sinθφ = r vv r PQ =+×ωω vr P =×ωω 0593_C05_fm Page 133 Monday, May 6, 2002 2:15 PM 134 Dynamics of Mechanical Systems This means that during the time that Q is a center of zero velocity, P moves in a circle about Q. Indeed, during the time that Q is a center of zero velocity, B may be considered to be in pure rotation about Q. Finally, observe in Eq. (5.4.2) that ω is normal to the plane of motion. Hence, we have: (5.4.3) where the notation is defined by inspection. If a body B is at rest, then every point of B is a center of zero velocity. If B is not at rest but has planar motion, then at most one point of B, in a given plane of motion, is a center of zero velocity. To prove this last assertion, suppose B has two particles, say O and Q, with zero velocity. Then, from Eq. (5.3.10), we have: (5.4.4) where r locates O relative to Q. If both O and Q have zero velocity then, (5.4.5) If O and Q are distinct particles, then r is not zero. Because ωω ωω is perpendicular to r, Eq. (5.4.5) is then satisfied only if ωω ωω is zero. The body is then in translation and all points have the same velocity. Therefore, because both O and Q are to have zero velocity, all points have zero velocity, and the body is at rest — a contradiction to the assumption of a moving body. That is, the only way that a body can have more than one center of zero velocity, in a plane of motion, is when the body is at rest. We can demonstrate these concepts by graphical construction. That is, we can show that if a body has planar motion, then there exists a particle of the body (or of the mathematical extension of the body) with zero velocity. We can also develop a graphical procedure for finding the point. To this end, consider the body B in general plane motion as depicted in Figure 5.4.1. Let P and Q be two typical distinct particles of B. Then, if B has a center O of zero velocity, P and Q may be considered to be moving in a circle about O. Suppose the velocities of P and Q are represented by vectors, or line segments, as in Figure 5.4.2. Then, if P and Q rotate about a center O of zero velocity, the velocity vectors of P and Q will be perpen- dicular to lines through O and P and Q, as in Figure 5.4.3. Observe that unless the velocities of P and Q are parallel, the lines through P and Q perpendicular to these velocities will always intersect. Hence, with non-parallel velocities, the center O of zero velocity always exists. FIGURE 5.4.1 A body B in general plane motion. FIGURE 5.4.2 Vector representations of the velocities of particles P and Q. vr PPP vr==ωω or ω vv r OQ =+×ωω ωω× =r 0 P Q B P Q B v v P Q 0593_C05_fm Page 134 Monday, May 6, 2002 2:15 PM Planar Motion of Rigid Bodies — Methods of Analysis 135 If the velocities of P and Q are parallel with equal magnitudes, and the same sense, then they are equal. That is, (5.4.6) where the last equality follows from Eq. (5.3.10) with r locating P relative to Q. If P and Q are distinct, r is not zero; hence, ωω ωω is zero. B is then in translation. Lines through P and Q perpendicular to v P and v Q will then be parallel to each other and thus not intersect (except at infinity). That is, the center of zero velocity is infinitely far away (see Figure 5.4.4). If the velocities of P and Q are parallel with non-equal magnitudes, then the center of zero velocity will occur on the line connecting P and Q. To see this, first observe that the relative velocities of P and Q will have zero projection along the line connecting P and Q: That is, from Eq. (5.3.10), we have: (5.4.7) Hence, v P/Q must be perpendicular to r. (This simply means that P and Q cannot approach or depart from each other; otherwise, the rigidity of B would be violated.) Next, observe that if v P and v Q are parallel, their directions may be defined by a common unit vector n. That is, (5.4.8) where v P , v Q , and v P/Q are appropriate scalars. By comparing Eqs. (5.4.7) and (5.4.8) we see that n must be perpendicular to r. Hence, when v P and v Q are parallel but with non-equal magnitudes, their directions must be perpendicular to the line connecting P and Q. There- fore, lines through P and Q and perpendicular to v P and v Q will coincide with each other and with the line connecting P and Q (see Figure 5.4.5). Next, observe from Eqs. (5.3.10) and (5.4.3) that, if O is the center of zero velocity, then the magnitude of v P is proportional to the distance between O and P. Similarly, the magnitude of v Q is proportional to the distance between O and Q. These observations enable us to locate O on the line connecting P and Q. Specifically, from Eq. (5.4.3), the distance from P to O is simply v P /ω. From a graphical perspective, O can be located as in Figure 5.4.6. Similar triangles are formed by O, P, Q and the “arrow ends” of v P and v Q . FIGURE 5.4.3 Location of center O of zero velocity by the inter- section of lines perpendicular to velocity vectors. FIGURE 5.4.4 A body in translation with equal velocity particles and center of zero velocity infinitely far away. P Q B O P Q v v P Q vv r PQ =×and = 0ωω vv r v r PQ =+× =×ωωωωor P/Q vvnv n v n PP QQ PQ vv== =,, / and P/Q 0593_C05_fm Page 135 Monday, May 6, 2002 2:15 PM 136 Dynamics of Mechanical Systems To summarize, we see that if a body has planar motion, there exists a unique point O of the body (or the body extended) that has zero velocity. O may be located at the intersection of lines through two points that are perpendicular to the velocity vectors of the points. Alternatively, O may be located on a line perpendicular to the velocity vector of a single point P of the body at a distance v P /ω from P (see Figure 5.4.7). Finally, when the zero velocity center O is located, the body may be considered to be rotating about O. Then, the velocity of any point P of the body is proportional to the distance from O to P and is directed parallel to the plane of motion of B and perpendicular to the line segment OP. 5.5 Illustrative Example: A Four-Bar Linkage Consider the planar linkage shown in Figure 5.5.1. It consists of three links, or bars (B 1 , B 2 , and B 3 ), and four joints (O, P, Q, and R). Joints O and R are fixed while joints P and Q may move in the plane of the linkage. The ends of each bar are connected to a joint; thus, the bars may be identified (or labeled) by their joint ends. That is, B 1 is OP, B 2 is PQ, B 3 is QR. In this context, we may also imagine a fourth bar B 4 connecting the fixed joints O and R. The system then has four bars and is thus referred to as a four-bar linkage. The four-bar linkage may be used to model many physical systems employed in mech- anisms and machines, particularly cranks and connecting rods. The four-bar linkage is thus an excellent practical example for illustrating the concepts of the foregoing sections. In this context, observe that bars OP(B 1 ) and RQ(B 3 ) undergo pure rotation, while bar PQ(B 2 ) undergoes general plane motion, and bar OR(B 4 ) is fixed, or at rest. FIGURE 5.4.5 A body B with particles P and Q having parallel velocities with non-equal magnitudes. FIGURE 5.4.6 Location of the center of zero velocity for a body having distinct particles with parallel but unequal velocities. FIGURE 5.4.7 Location of the center for zero velocity knowing the velocity of one particle and the angular speed. FIGURE 5.5.1 A four-bar linkage. P Q v v P Q B P Q v v P Q B O B P O v / v | | P P ω P Q R O B B B 1 2 3 0593_C05_fm Page 136 Monday, May 6, 2002 2:15 PM Planar Motion of Rigid Bodies — Methods of Analysis 137 The system of Figure 5.5.1 has one degree of freedom: The rotations of bars OP(B 1 ) and RQ(B 3 ) each require two coordinates, and the general motion of bar PQ(B 2 ) requires an additional three coordinates for a total of five coordinates. Nevertheless, requiring joint P to be connected to both B 1 and B 2 and joint Q to be connected to both B 2 and B 3 produces four position (or coordinate) constraints. Thus, there is but one degree of freedom. A task encountered in the kinematic analyses of linkages is that of describing the orientation of the individual bars. One method is to define the orientations of the bars in terms of angles that the bars make with the horizontal (or X-axis) such as θ 1 , θ 2 , and θ 3 as in Figure 5.5.2. Another method is to define the orientation in terms of angles that the bars make with the vertical (or Y-axis) such as φ 1 , φ 2 , and φ 3 as in Figure 5.5.2. A third method is to define the orientations in terms of angles that the bars make with each other, as in Figure 5.5.3. The latter angles are generally called relative orientation angles whereas the former are called absolute orientation angles. Relative orientation angles are usually more meaningful in describing the configuration of a physical system. Absolute orientation angles are usually easier to work with in the analysis of the problem. In our example, we will use the first set of absolute angles θ 1 , θ 2 , and θ 3 . Because the system has only one degree of freedom, the orientation angles are not independent. They may be related to each other by constraint equations obtained by considering the linkage of four bars as a closed loop: Specifically, consider the position vector equation: (5.5.1) This equation locates O relative to itself through position vectors taken around the loop of the mechanism. It is called the loop closure equation. Let ᐉ 1 , ᐉ 2 , ᐉ 3 , and ᐉ 4 be the lengths of bars B 1 , B 2 , B 3 , and B 4 . Then, Eq. (5.5.1) may be written as: (5.5.2) where λλ λλ 1 , λλ λλ 2 , λλ λλ 3 , and λλ λλ 4 are unit vectors parallel to the rods as shown in Figure 5.5.4. These vectors may be expressed in terms of horizontal and vertical unit vectors n x and n y as: (5.5.3) FIGURE 5.5.2 Absolute orientation angles. FIGURE 5.5.3 Relative orientation angles. Y X φ θ φ θ θ 2 2 3 1 1 3 φ 3 1 β β 2 β OP PQ QR RO+++=0 llll 11 22 33 44 0λλλλλλλλ+++= λλλλ λλλλ 11 1 22 2 33 344 =+ =+ =− =− cos sin , cos sin cos sin , θθ θθ θθ nn nn nn n xy xy xx 0593_C05_fm Page 137 Monday, May 6, 2002 2:15 PM 138 Dynamics of Mechanical Systems Hence, by substituting into Eq. (5.5.2) we have: (5.5.4) This immediately leads to two scalar constraint equations relating θ 1 , θ 2 , and θ 3 : (5.5.5) and (5.5.6) The objective in a kinematic analysis of a four-bar linkage is to determine the velocity and acceleration of the various points of the linkage and to determine the angular velocities and angular accelerations of the bars of the linkage. In such an analysis, the motion of one of the three moving bars, say B 1 , is generally given. The objective is then to determine the motion of bars B 1 and B 2 . In this case, B 1 is the driver, and B 2 and B 3 are followers. The procedures of Section 5.4 may be used to meet these objectives. To illustrate the details, consider the specific linkage shown in Figure 5.5.5. The bar lengths and orientations are given in the figure. Also given in Figure 5.5.5 are the angular velocity and angular acceleration of OP(B 1 ). B 1 is thus a driver bar and PQ(B 2 ) and QR(B 3 ) are follower bars. Our objective, then, is to find the angular velocities and angular accelerations of B 2 and B 3 and the velocity and acceleration of P and Q. To begin the analysis, first observe that, by comparing Figures 5.5.4 and 5.5.5, the angles and lengths of Figure 5.5.5 satisfy Eqs. (5.5.5) and (5.5.6). To see this, observe that ᐉ 1 , ᐉ 2 , ᐉ 3 , ᐉ 4 , θ 1 , θ 2 , θ 3 , and θ 4 have the values: (5.5.7) Then Eqs. (5.5.5) and (5.5.6) become: (5.5.8) FIGURE 5.5.4 Linkage geometry and unit vectors. Y X θ θ θ 2 3 1 3 λ λ λ λ n O 1 B 2 2 n R B 3 B 1 4 x y ll ll ll l 1122334 112233 0 cos cos cos sin sin sin θθθ θθθ ++− () +++ () = n n x y ll l l 1122334 cos cos cosθθθ++= ll l 112233 0sin sin sinθθθ++= ll l l 134 23 20 30 495 6098 90 30 315 == = = =° =° = ° ° () . . ,, m, 2 m , m , m or – 45 1 θθθ 2 0 90 3 0 30 4 95 315 6 098. cos . cos . cos . () + () += 0593_C05_fm Page 138 Monday, May 6, 2002 2:15 PM Planar Motion of Rigid Bodies — Methods of Analysis 139 and (5.5.9) Next, recall that B 1 and B 3 have pure rotation about points O and R, respectively, and that B 2 has general plane motion. Third, let us introduce unit vectors λλ λλ i and νν νν i (i = 1, 2, 3) parallel and perpendicular to the bars as in Figure 5.5.6. Then, in the configuration shown, the λλ λλ i and νν νν i may be expressed in terms of horizontal and vertical unit vectors n x and n y as: (5.5.10) (5.5.11) (5.5.12) Consider the velocity analysis: because B 1 has pure rotation, its angular velocity is: (5.5.13) The velocity of joint P is then: (5.5.14) (Recall that O is a center of zero velocity of B 1 and that P moves in a circle about O.) FIGURE 5.5.5 Example four-bar linkage. FIGURE 5.5.6 Unit vectors for the analysis of the linkage of Figure 5.5.5. 45° 30° B B B Q 4.95 m 6.098 m 3.0 m R O P 2.0 m α = 4 rad/sec ω = 5 rad/sec 2 1 3 2 OP OP 45° 30° B B B Q R O P α = 4 rad/sec ω = 5 rad/sec 2 1 3 2 OP OP n n n λ λ 1 1 λ 2 2 3 3 y z x ν ν 2 0 90 3 0 30 4 95 315 0. sin . sin . sin () + () += λλνν 11 ==−nn yx and λλνν 2 3 2 12 12 3 2= ( ) + () =− () + ( ) //nn n n xy x y and 2 λλνν 3 22 22 22 22= ( ) − ( ) = ( ) + ( ) // //nn nn xy xy and 3 ωωωω OP z = =− D rad sec 1 5n vOPn n P z x =× =− × =− = ωωλλ νν 11 1 520 10 10 . m sec 0593_C05_fm Page 139 Monday, May 6, 2002 2:15 PM 140 Dynamics of Mechanical Systems Because B 2 has general plane motion, the velocity of Q may be expressed as: (5.5.15) where ω 2 is the angular speed of B 2 . Note that Q is fixed in both B 2 and B 3 . Because B 3 has pure rotation with center R, Q moves in a circle about R. Hence, v Q may be expressed as: (5.5.16) where ω 3 is the angular speed of B 3 . Comparing Eqs. (5.5.15) and (5.5.16) we have the scalar equations: (5.5.17) and (5.5.18) Solving for ω 2 and ω 3 we obtain: (5.5.19) Hence, v Q becomes: (5.5.20) Observe that in calculating the angular speeds of B 2 and B 3 we could also use an analysis of the instant centers as discussed in Section 5.4. Because the velocities of P and Q are perpendicular to, respectively, B 1 (OP) and B 3 (QR), we can construct the diagram shown in Figure 5.5.7 to obtain ωω ωω 2 , ωω ωω 3 , and v Q . By extending OP and RQ until they intersect, we obtain the instant center of zero velocity of B 2 . Then, IP and IQ are perpendicular to, respectively, v P and v Q . Triangle IOR forms a 45° right triangle. Hence, the distance between I and P is (6.098 – 2.0) m, or 4.098 m. Because v P  is 10 m/sec, ω 2 is: (5.5.21) v v PQ n n n nnn nn QP yz x xxy xy =+× = + × () =+ =+− + ( () [] =− () [] + () ωωλλνν 22222 2 22 10 3 0 10 3 10 3 1 2 3 2 10 3 2 3 2 ωω ω ωω . // // vRQn nn nn Q z xy xy =× = ×− () =− ( ) + ( ) [] =− − ωωνν 33 33 323 495 495 2 2 2 2 35 35 ωω ωωω . ./ / 10 15 35 23 −=− ωω 332 35 23 ( ) =−/.ωω ωω 23 244 181==− rad sec and rad sec vnn Q xy =+634 634. . m sec ω 2 10 4 098 2 44== =v P IP/ . . rad sec 0593_C05_fm Page 140 Monday, May 6, 2002 2:15 PM Planar Motion of Rigid Bodies — Methods of Analysis 141 Similarly, the distance IQ is 3.67 m, and v Q is, then, (5.5.22) Then ω 3 becomes: (5.5.23) Next, consider an acceleration analysis. Because B 1 has pure rotation, P moves in a circle about O and its acceleration is: (5.5.24) Because P and Q are both fixed on B 2 , the acceleration of Q may be expressed as: (5.5.25) FIGURE 5.5.7 Instant center of zero velocity of B 2 . B B B Q R O P 2 1 3 4.95 m 6.098 m 45° v v 4.098 m 3.67 m I 2.0 m P Q vnn Q xy IQ== ()() ==+ω 23 3 3 367 244 895 634 634νννννν . . . ω 3 895495 181=− =− =−vQR Q / . . . rad sec aOP OPn nn nn P zz xy =× +× × () =×+− () ×− () × [] =− =−− ααωωωωλλλλ ννλλ 111 31 1 11 42 5 5 2 8 50 8 50 m sec 2 a a PQ PQ nnn n n nn nn n QP xyz z xy xy x =+× +× × () =− − + × () + () × () × () [] −− + − =− − + () + ( ) ααωωωω ααλλλλ ννλλ 222 222 2 22 2 2 8 50 3 0 2 44 2 44 3 0 85030 1786 85030 32 . . ./ α α –12 nnnn nn yxy xy [] − ( + () ) [] =− − () +− + () 17 86 3 2 23 467 1 5 58 93 2 6 22 ./ 12 αα 0593_C05_fm Page 141 Monday, May 6, 2002 2:15 PM [...]... (5.8 .42 ) Then, ∆ becomes: ( ) ( ) (5.8 .43 ) α + 4 (5.8 .44 ) 2 2 ∆ = −  − rω 2 − 0 + ( − rα + 2rα )  = − r 2 ω 4 + α 2     Hence, ξ and η are: ξ= η= −1 r 4 + α2 2 ( 0 − (2r − r ) ( − rα + 2rα )] = α )[ −1 r 4 + α2 2 ( 2 [(2r − r) (−rω − 0) − 0] = α ω ω + ) 2 2 These expressions are identical with those of Eq (5.8.18) 2 4 (5.8 .45 ) 0593_C05_fm Page 156 Monday, May 6, 2002 2:15 PM 156 Dynamics of Mechanical. .. on the line of action of F, relative to O Figure 6.2.2 depicts point O, a typical point Q, vectors p and F, and the line of action L of F 163 0593_C06_fm Page 1 64 Monday, May 6, 2002 2:28 PM 1 64 Dynamics of Mechanical Systems FIGURE 6.1.1 A force applied at different ends of a rod FIGURE 6.2.1 A force F, its line of action L , and point of application P FIGURE 6.2.2 A force F with line of action L,... system of forces Systems of forces — particularly systems with many forces — are conveniently characterized by (and thus represented by) two vectors: the resultant and the moment about some point O FIGURE 6.3.1 A mechanical system FIGURE 6.3.2 A set of forces acting on a mechanical system 0593_C06_fm Page 166 Monday, May 6, 2002 2:28 PM 166 Dynamics of Mechanical Systems The resultant R of a system of. ..0593_C05_fm Page 142 Monday, May 6, 2002 2:15 PM 142 Dynamics of Mechanical Systems Because Q also moves in a circle about R, we have: aQ = α 3 × RQ + ω 3 × (ω 3 × RQ) [ ] = α 3n z × ( 4. 95λ 3 ) + ( −1.81n z ) × ( −1.81n z ) × ( 4. 95λ 3 ) = 4. 95α 3 ν3 + 16.21λ 3 = 4. 95α 3 [( 2/2) n x + ( ] [( 2/2) n y + 16.21 2/2) n x − ( 2/2) n y ] (5.5.26) = −3.5α 3n x − 3.5α 3n y + 11 .46 n x − 11 .46 n y = (11 .46 − 3.5α... FIGURE P5 .4. 3 A four-bar linkage P5 .4. 4: Repeat Problem P5 .4. 3 if the angular speed of link AB is 5 rad/sec clockwise P5 .4. 5: Consider the crank, connecting rod, and piston system shown in Figure P5 .4. 5 Let the crank OA have a clockwise angular speed of 300 rpm as shown 0593_C05_fm Page 160 Monday, May 6, 2002 2:15 PM 160 Dynamics of Mechanical Systems a Locate or identify the instant center of zero... O, we find the coordinates of C to be: xC = xQ + ˙˙ ˙˙ ω 2 xQ − αyQ α2 + 4 = 0+ ω 2 ( − rα ) − α(0) rαω 2 =− 2 α2 + 4 α + 4 (5.8. 34) α( − rα ) + ω 2 (0) rω 4 = 2 2 4 α +ω α + 4 (5.8.35) and yC = yQ + ˙˙ ˙˙ αxQ + ω 2 yQ α +ω 2 4 =r+ For positive values of ω and α, the position of C is depicted in Figure 5.8.3 To verify the results, consider calculating the acceleration of C using the expression:... Monday, May 6, 2002 2:15 PM 144 Dynamics of Mechanical Systems NX O X n 11 NY G θ1 1 O2 B1 G2 θ2 n 21 n 13 O3 B2 θ3 n 23 G3 O4 O N-1 B3 N n N1 GN-1 θN-1 ON Z B N-1 θN n N3 GN BN Z FIGURE 5.6 .4 Numbering and labeling of the systems and [( ) ( ) ] ˙˙ ˙2 ˙˙ ˙2 aG1 = (l 2) θ1 cos θ1 − θ1 sin θ1 N X + −θ1 sin θ1 − θ1 cos θ1 N Z (5.6 .4) Similarly, the velocity and acceleration of O2 are: ˙ ˙˙ ˙2 vO2 = lθ1n11... intersection of the line through O and perpendicular to L Finally, observe that if O is a point on L, then MO is zero 6.3 Systems of Forces Consider a mechanical system V consisting of particles and rigid bodies as depicted in Figure 6.3.1 Let S be a set of forces acting on V That is, let S consist of forces Fi (i = 1,…, N) with lines of action passing through points Pi of the particles and bodies of V as... angular speed of 5 rad/sec, as shown a Locate or identify the instant centers of zero velocity for links AB, BC, and CD for the position shown in Figure P5 .4. 3 b Use the results of (a) to determine the velocities of joints B and C and the angular speeds of links BC and CD c Express the velocities of B and C in terms of unit vectors and of Figure P5 .4. 3 B 7.0 m C 5 rad/sec ny 3.5 m 4. 04 m A nx 60° 9.02... Motion of Rigid Bodies — Methods of Analysis 159 Section 5 .4 Instant Center, Points of Zero Velocity P5 .4. 1: A 2-ft-radius wheel rolls to the right so that its center O has a velocity of 3 ft/sec, as presented in Figure P5 .4. 1 a Locate (or identify) the instant center of zero velocity b Determine the angular velocity of the wheel c Determine the velocities of points A, Q, P, and B on the rim of the . ᐉ j1 j3 k1 k3 0593_C05_fm Page 143 Monday, May 6, 2002 2:15 PM 144 Dynamics of Mechanical Systems and (5.6 .4) Similarly, the velocity and acceleration of O 2 are: (5.6.5) and in terms of N X and N Z , they. 3 33 3 3 49 5 181 181 49 5 49 5 1621 49 5 2 2 2 2 1621 2 2 2 2 ./ / ./ / α nn nnnn nn y xyxy xy [] =− − + − =− () +− − () 3 5 3 5 11 46 11 46 11 46 3 5 11 46 3 5 33 33 αααα αααα −−=−23 46 7 1. − ωωνν 33 33 323 49 5 49 5 2 2 2 2 35 35 ωω ωωω . ./ / 10 15 35 23 −=− ωω 332 35 23 ( ) =−/.ωω ωω 23 244 181==− rad sec and rad sec vnn Q xy =+6 34 6 34. . m sec ω 2 10 4 098 2 44 == =v P IP/ . .