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0593_C11_fm  Page 374  Monday, May 6, 2002  2:59 PM





                       374                                                 Dynamics of Mechanical Systems







                                                      O          + x  1   2   3  + x   Reference Level
                                                                     + x  2
                                                                            3
                                                                  P  1
                                                           θ
                                                                         P
                                                                          2
                                                                                P
                                                                                 3
                       FIGURE 11.7.6
                       Position of tube particles relative to                         T
                       the reference level.

                       Similarly, for the mass center G of T the elevation H relative to O is:

                                                        H =−( ) 2 cosθ                         (11.7.20)
                                                             L

                       Then, from Eq. (11.7.18) we have:
                                                    ˆ
                                                             ˆ
                                                         ˆ
                                                    F =  F = F =  mg cosθ                      (11.7.21)
                                                     x 1  x 2  x 3
                       and
                                                     (
                                                                     (
                                                                                     ( )
                               ˆ
                                    mg
                                          x
                                                          x
                                                                          x
                               F =− (l   + ) sinθ − mg l2  + ) sinθ − mg l3  + ) sinθ − Mg L 2  sinθ
                                θ
                                           1
                                                           2
                                                                           3
                       or
                                                   (
                                                                          ( )
                                           ˆ
                                                              x
                                          F =−mg 6l   + x 1  + x 2 + ) sinθ − Mg L 2  sinθ     (11.7.22)
                                           θ
                                                               3
                       Comparing Eqs. (11.7.21) and (11.7.22) with Eqs. (11.7.14) to (11.7.17), we see that the results
                       are consistent.
                                                                                                 ˆ
                        Next, for the springs, we see from Eq. (11.6.7) that the spring force contributions  F   are
                                                                                                  r
                       given by:
                                                        ˆ
                                                               ∂
                                                                  ∂
                                                        F =− kx x q                            (11.7.23)
                                                         r          r
                       where x is the spring elongation. From Figure (11.7.3), we see that the spring elongations
                                                             ˆ
                       are x , x  – x , x  – x , and –x . Hence, the   are:
                                                             F
                           1  2   1  3   2      3             r
                                            ˆ
                                            F =− kx + (    x     2 kx +  kx
                                                      k x − ) =−
                                             x 1   1     2  1       1   2
                                                 k x − ) + (
                                            F =− (     x   k x − ) = kx +  kx −  2 kx
                                            ˆ
                                                                 x
                                             x 2    2   1     3   2    1   3     2
                                                                                               (11.7.24)
                                            F =− (     x   kx = − 2 kx +  kx
                                            ˆ
                                                 kx − ) −
                                             x 3    3   2    3      3    2
                                            ˆ
                                            F =  0
                                             θ
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