**Content:** d9-86.zip (22.09 KB)

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После оплаты просто скачайте решение задачи Д9 Вариант 86(рисунок 8 условие 6) по теоретической механике из решебника Тарга С.М. 1989 года для студентов-заочников машиностроительных, строительных, транспортных, приборостроительных специальностей высших учебных заведений. Задача оформлена в формате word, добавлена в архив zip.

Solution of the D9 Option 86- termehu of Targa SM 1989. Conditions to task Dynamics 9 (Full condition on page 86): The mechanism located in a horizontal plane under the influence of applied forces in equilibrium; the position of equilibrium is determined angles a, B, γ, phi, theta (Fig. D9.0 - D9.9, Table. D9a and D9b). Length of the rod mechanism (crank) are: l1 = 0,4 m, l4 = 0,6 m (the size of l2 and l3 arbitrary); the point E is in the middle of the corresponding rod. On the slide mechanism of action of the spring elastic force F; numerically with F = * l, where c - the spring constant, l - its deformation. Furthermore, in Fig. 0 and D 1 on the slider the force Q, and to crank O1A - couple with moment M; Fig. 2-9 on cranks and O1A O2D are a pair of forces with moments M1 and M2. Identify what is at equilibrium deformation of the spring l and specify the spring is stretched or compressed. The values \u200b\u200bof all defined variables are shown in Table. D9a to Fig. 0-4 and Table. D9b to Fig. 5-9, where Q is expressed in Newtons, and M, M1, M2 - in newton. Construction to begin drawing the rod, the direction of which is determined by the angle α; for clarity, the slide with the guides and the spring depict as in Example D9 (see. Fig. D9 and Fig. D9.10, b). If the drawing solvable problem embodiment attached to the slider rod would be in aligned with the spring (as in Fig. D9.10 a), the spring must be considered attached to a slider on the other hand (as in Fig. D9.10 b, where both otherwise depicted guides).

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