The solution of control K3 Fig 4 srv 9 (option 49) Targ

Solution of K3 Option 49. Target: Flat mechanism consists of rods 1, 2, 3, 4 and slider B or E (Fig. KZ.0 - K3.7) or bars 1, 2, 3 and slider B and E ( Fig. K3.8, K3.9), connected with each other and with fixed bearings O1, O2 hinges; the point D is in the middle of the rod AB. The lengths of the bars are, respectively, l1 = 0,4 m, l1 = 1,2 m, l3 = 1,4 m, l4 = 0,6 m. The position of the mechanism is determined by the angles α, β, γ, φ, θ. The values \u200b\u200bof these angles and other predetermined values \u200b\u200bare listed in Table. ACA (for Figs. 0-4) or in the Table. HSC (for Fig. 5-9); while Table. ACA ω1 and ω4 - constant. Determine the values \u200b\u200bshown in the tables in the columns of "Find." Arc arrows in the figures show how the construction drawing of the mechanism should be postponed corresponding angles: in the course of, or counter-clockwise (for example, the angle γ in Fig. 8 should be deferred from the DB in a clockwise direction, and Fig. 9 - counter-clockwise arrows, and so on. d.). Construction to begin drawing the rod, the direction of which is determined by the angle α; slide with runners for greater clarity to portray as in the example of RS (see. Fig. HSC). Given angular velocity and angular acceleration considered a counterclockwise direction, and set the speed and acceleration vB ab - from B to b (Fig. 5-9).

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