The solution of control K3 Fig 5 srv 6 (option 56) Targ

The solution of control K3 Fig 5 srv 6 (option 56) Targ

Right after the payment you receive a link to the archive with the solution of the problem in theoretical mechanics K3 V56 (figure 5 the condition 6) of manuals Targ SM 1989 for part-time students of engineering, construction, transport, instrument-specialties of higher education. The task is made in the format word, packed to the archive zip.

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Solution of the K3 task Option 56. Task: The flat mechanism consists of rods 1, 2, 3, 4 and the slider B or E (Fig. KZ.0 - K3.7) or of rods 1, 2, 3 and the sliders B and E ( Fig. K3.8, K3.9) connected to each other and to the fixed supports O1, O2 by hinges; point D is in the middle of rod AB. The lengths of the rods 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 ​​of these angles and other specified values ​​are shown in table. KZa (for Fig. 0-4) or in table. KZB (for Fig. 5-9); while in the table. KZa ω1 and ω4 are constant values. Determine the values ​​indicated in the tables in the “Find” columns. The arc arrows in the figures show how, when constructing a drawing of the mechanism, the corresponding angles should be laid down: clockwise or counterclockwise (for example, the angle γ in Fig. 8 should be postponed from DB clockwise, and in Fig. 9, counterclockwise arrows, etc.). The construction of the drawing begins with a rod whose direction is determined by the angle α; for greater clarity, depict the slider with guides as in the KZ example (see Fig. KZb). Given angular velocity and angular acceleration, consider counter-clockwise, and given velocity vB and acceleration aB from point B to b (Fig. 5-9).
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