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- 1. Mechanical Translations Systems Solution First, notice that this is the same system as used in Example 2.17 of Reference [1], however here we have no friction between the masses and the ground. Secondly, we will use the common symbols for the mechanical components and not the fancy ones introduced by Nise. (a) First draw the FBD's of the two masses, see the figure below. u k2$ x1 K x2 k2$ x2 K x1 .. .. m2 k3$x2 m1 m1$x1 m2$x2 k1$x1 . . . . b$ x1 K x2 b$ x2 K 1 x n Using FDB's of the two masses and Newton's second law ( >F = 0 ) we can obtain the following k= 1 k equations of motion: FBD mass 1 : Keep m2 still , move m1 in positive direction and evaluate the forces acting on m1 .. . . m1$x1 Cb$ x1 Kx2 Ck1$x1 Ck2$ x1 Kx2 = u FBD mass 2 : Keep m1 still , move m2 in positive direction and evaluate the forces acting on m2 .. . . m2$x2 Cb$ x2 Kx1 Ck3$x2 Ck2$ x2 Kx1 = 0 Simplifying and rearranging we obtain 2 d x1 d x1 d x2 m1$ Cb$ C k1 Ck2 $x1 = b$ Ck2$x2 Cu d t2 dt dt 2 d x2 d x2 d x1 m2$ 2 Cb$ C k2 Ck3 $x2 = b$ Ck2$x1 dt dt dt Control Systems Engineering Engineering Mathematics 1 2 of 5
- 2. Mechanical Translations Systems Taking the laplace transform of these two equations assuming zero initial conditions, we have m1$s2 Cb$s Ck1 Ck2 $X1 = b$s Ck2 $X2 CU m2$s2 Cb$s Ck2 Ck3 $X2 = b$s Ck2 $X1 We solve these two equations for X1 , X2 with Maple's solve and we find X1 s m2 s2 Cb s Ck2 Ck3 G1 s = = m1$s2 Cb$s Ck1 Ck2 $ m2$s2 Cb$s Ck2 Ck3 K b$s Ck2 U s 2 X2 s b s Ck2 G2 s = = m1$s Cb$s Ck1 Ck2 $ m2$s2 Cb$s Ck2 Ck3 K b$s Ck2 U s 2 2 Look behind the button, to see how it is done in Maple. We also could have used Cramer's rule, as done in Example 2.17, but having Maple at hand this is only a waste of time! (b) The two not simplified equations of motion were: .. . . m1$x1 Cb$ x1 Kx2 Ck1$x1 Ck2$ x1 Kx2 = u .. . . m2$x2 Cb$ x2 Kx1 Ck3$x2 Ck2$ x2 Kx1 = 0 Taking the Laplace transform and rearranging for matrix-vector form we have m1$s2 Cb$s Ck1 Ck2 $X1 C K Kk2 $X2 = U s$b K Kk2 $X1 C m2$s2 Cb$s Ck2 Ck3 $X2 = 0 s$b The matrix-vector form A.X = B in the s-domain now is Control Systems Engineering Engineering Mathematics 1 3 of 5
- 3. Mechanical Translations Systems m1$s2 Cb$s Ck1 Ck2 K Kk2 s$b X1 u . = K Kk2 s$b m2$s2 Cb$s Ck2 Ck3 X2 0 Solutions can be found using matrix inverse and is found to be the same as before: X = AK $B 1 Look behind the button to see how this result is achieved. Example 2 (Skill-Assessment Exercise 2.8) Problem Find the transfer function G s = X2 s /F s for the translational mechanical system shown in Fig. 2.21 of Reference [1]. Solution First, rename the component symbols as follows: M1 = m1 , M2 = m2 = 1 kg fv = bi , i = 1 ..4 = 1 Ns$mK1 i K = k = 1 N$mK1 f t =F t N To see if the force F is in the t-domain or in the s-domain use F t respectively F s . Next, draw the FDB's and write the following equations of motion .. . . . . . . Mass 1 : m1$x1 C b1$ x1 K 2 C b2$ x1 K 2 C b3$ x1 K 2 Ck$ x1 Kx2 = F s x x x .. . . . . . . . Mass 2 : m2$x2 C b1$ x2 K 1 C b2$ x2 K 1 C b3$ x2 K 1 Cb4$x2 Ck$ x2 Kx1 = 0 x x x Filling in component values we have .. . . x1 C3$ x1 K 2 C x1 Kx2 = F s x .. . . . x2 C 3$ x2 K 1 Cx2 C x2 Kx1 = 0 x Then, take the Laplace transform of these equations and rearrange Control Systems Engineering Engineering Mathematics 1 4 of 5
- 4. Mechanical Translations Systems s2 C3$s C1 $X1 K 3$s C1 $X2 = F s K 3$s C1 $X1 C s2 C4$s C1 $X2 = 0 Finally, we could find the required transfer function by solving these equations with Cramer's rule ( see solution on CD-ROM) or Maple's solve command. Here, however, we will use the matrix-vector notation and Maple. O restart : O with LinearAlgebra : s2 C3$s C1 K 3$s C1 O Ad : K 3$s C1 s2 C4$s C1 F O Bd : 0 O X d AK .B 1 2 s C4 s C1 F s s3 C7 s2 C5 s C1 X := 3 s C1 F s s C7 s2 C5 s C1 3 O Θ1 d X1 : Θ2 d X2 : Required transfer function Θ2 O Gd F 3 s C1 G := s s C7 s2 C5 s C1 3 End of Mechanical Translation System Control Systems Engineering Engineering Mathematics 1 5 of 5

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