Solution D2-55 (Figure D2.5 condition 5 SM Targ 1989)

Solution D2-55 (Figure D2.5 condition 5 SM Targ 1989)

Load 1 mass m is fixed on a spring suspension in the elevator (Fig. D2.0 - D2.9, Table D2). The elevator moves vertically according to the law z = 0.5α1t2 + α2sin (ωt) + α3cos (ωt) (the z axis is directed vertically upwards, z is expressed in meters, t is in seconds). The load acts on the weight of the medium´s resistance R = μv, where v - the speed of the cargo in relation to the elevator. Find the law of motion of the cargo in relation to the elevator, ie, x = f (t); place the origin at the point where the end of the spring attached to the load is located, when the spring is not deformed. In order to avoid errors in signs, direct the x-axis towards the extension of the spring, and represent the load in the position at which x> 0, i.e. the spring is stretched. When counting, we can take g = 10 m / s2. Mass of springs and connecting plate 2 are neglected. In the table, c1, c2, c3 are the spring stiffness coefficients, λ0 is the spring extension with the equivalent stiffness at the initial time t = 0, v0 is the initial speed of the load with respect to the elevator (directed vertically upwards). The dash in columns c1, c2, c3 means that the corresponding spring is absent and should not be depicted in the drawing. If, at the same time, the end of one of the remaining springs is free, it should be fixed in the appropriate place or to the load or to the ceiling (floor) of the elevator; The same should be done if the ends of the two remaining springs connected by the strap 2 are free. The condition μ = 0 means that there is no resistance R.