# Solution D8-42 (Figure D8.4 condition 2 SM Targ 1989)

If the total amount of your purchases from the seller Михаил_Перович more than:

- 200 $ the discount is 15%

- 100 $ the discount is 10%

- 50 $ the discount is 7%

- 20 $ the discount is 5%

- 10 $ the discount is 3%

- 5 $ the discount is 2%

- 1 $ the discount is 1%

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Solution D8-42 (Figure D8.4 condition 2 SM Targ 1989)

Vertical shaft AK (Fig. D8.0 - D8.9), rotating with a constant angular velocity ω = 10 s-1, is fixed by a thrust at point A and a cylindrical bearing at the point indicated in Table. D8 in column 2 (AB = BD = DE = EK = a). A thin, uniform broken rod of mass m = 10 kg, consisting of parts 1 and 2, is rigidly attached to the shaft (the dimensions of the parts of the rod are shown in the figures, where b = 0.1 m, and their masses m1 and m2 are proportional to the lengths), and a weightless length bar l = 4b with point mass m3 = 3 kg at the end; Both bars lie in one plane. The fixing points of the rods are indicated in the table in columns 3 and 4, and the angles α, β, γ, φ are given in columns 5-8. Disregarding the weight of the shaft, determine the reaction of the thrust bearing and the bearing. When counting, take a = 0.6 m.

Vertical shaft AK (Fig. D8.0 - D8.9), rotating with a constant angular velocity ω = 10 s-1, is fixed by a thrust at point A and a cylindrical bearing at the point indicated in Table. D8 in column 2 (AB = BD = DE = EK = a). A thin, uniform broken rod of mass m = 10 kg, consisting of parts 1 and 2, is rigidly attached to the shaft (the dimensions of the parts of the rod are shown in the figures, where b = 0.1 m, and their masses m1 and m2 are proportional to the lengths), and a weightless length bar l = 4b with point mass m3 = 3 kg at the end; Both bars lie in one plane. The fixing points of the rods are indicated in the table in columns 3 and 4, and the angles α, β, γ, φ are given in columns 5-8. Disregarding the weight of the shaft, determine the reaction of the thrust bearing and the bearing. When counting, take a = 0.6 m.