Reshebnik Targ S.M. task D9 B36 (Fig. 3 conv 6) 1989

Solution of Problem D9 Option 36 - by the thermome of Targ S.M. 1989 year. Condition for the task Dynamics 9 (Full condition on page 86): The mechanism located in the horizontal plane is under the influence of the applied forces in equilibrium; the equilibrium position is determined by the angles a, B, γ, phi, tetta (Fig. D9.0 - D9.9, Tables D9a and D9b). The lengths of the mechanism rods (cranks) are equal: l1 = 0.4 m, l4 = 0.6 m (dimensions l2 and l3 are arbitrary); point E is in the middle of the corresponding rod. The spring force F acts on the slider B of the mechanism; numerically F = c * l, where c is the spring stiffness coefficient, l is its deformation. In addition, in fig. 0 and 1, the force Q acts on the slider D, and the pair of forces with the moment M acts on the crank O1A; in fig. 2–9, pairs of forces with moments M1 and M2 act on the O1A and O2D cranks. Determine what the equilibrium deformation l of the spring is equal to, and indicate whether the spring is stretched or compressed. The values ​​of all given values ​​are given in table. D9a for fig. 0-4 and tab. D9b for fig. 5–9, where Q is expressed in Newtons, and M, M1, M2 in Newtonometers. The construction of the drawing begins with a rod, the direction of which is determined by the angle α; for greater clarity, depict the slider with guides and the spring as in example D9 (see Fig. D9, as well as Fig. D9.10, b). If in the drawing of the solution to the problem being solved, the rod attached to the slider B will be aligned with the spring (as in Fig. D9.10, a), then the spring should be considered attached to the slider on the other side (as in Fig. D9.10, b, where at the same time otherwise guides are shown).