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Solution K2-50 (Figure K2.5 condition 0 SM Targ 1989) Solution K2-50 (Figure K2.5 condition 0 SM Targ 1989)
Solution K2-50 (Figure K2.5 condition 0 SM Targ 1989)
Solution K2-50 (Figure K2.5 condition 0 SM Targ 1989) Solution K2-50 (Figure K2.5 condition 0 SM Targ 1989)

Solution K2-50 (Figure K2.5 condition 0 SM Targ 1989)

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Solution K2-50 (Figure K2.5 condition 0 SM Targ 1989)

The mechanism consists of stepped wheels 1-3, engaged or connected by belt drive, rack 4 and load 5, tied to the end of a thread wound on one of the wheels (Fig. K2.0 - K2.9, Table K2). The radii of the wheel stages are equal: wheel 1 has r1 = 2 cm, R1 = 4 cm, wheel 2 has r2 = 6 cm, R2 = 8 cm, wheel 3 has r3 - 12 cm, R3 = 16 cm. the points A, B and C are located in the "Given" column of the table. The law of motion or the law of change in the speed of the leading link of the mechanism, where φ1 (t) is the law of rotation of the wheel 1, s2 (f) is the law of motion of the rack 4, ω2 (t ) Is the law of variation of the angular velocity of the wheel 2, v5 (t) is the law of the change in the speed of the load 5, etc. (everywhere φ is expressed in radians, s is in centimeters, t is in seconds). Positive direction for φ and ω against the clockwise direction, for s4, s5 and v4, v5 - down. Determine the velocities (v - linear, ω - angular) and the accelerations (a - linear, ε - angular) of the corresponding points or bodies (v5 - the speed of the load 5, etc., at the time t1 = 2) etc.).

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  • Added to the site 21.11.2017

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Cumulative discount

200 $ 15% the discount is
100 $ 10% the discount is
50 $ 7% the discount is
20 $ 5% the discount is
10 $ 3% the discount is
5 $ 2% the discount is
1 $ 1% the discount is

Cumulative discount

200 $ 15% the discount is
100 $ 10% the discount is
50 $ 7% the discount is
20 $ 5% the discount is
10 $ 3% the discount is
5 $ 2% the discount is
1 $ 1% the discount is

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