Question

The system in the figure holds the weight of an object. The spring AB has an unstressed length Io = 1.5m, with a constant KAB = 400 N/m. The AC spring has an unstressed length of Io = 1.5m, with a constant Kac = 600 N/m. There is concern that the AC spring allows, when stretched, reach the floor of the structure. To investigate this, determine the weight that the armature W would hold and the length that the AC spring would reach when stretched.

35° DIo 2.0 m

PLEASE HURRY

Answer 1

AD = 2m BA = 2/sin 35 degree = 3.49m stretched length of AB = 3.49m Stretching = 3.49 - 1.5 = 1.98m Tension in spring AB F = kx F = 400 times 1.98 F = 794.76N Now, for equilibrium torque about D must be zero so, tau = F times 2 times cos 55 degree - F_x times 2 = 0 implies F_x = 794.76 times cos 55 degree implies F_x = 455.85N \r\n So, W = 455.85N Now, for stretching in spring AC W = KX implies 455.85 = 600 times x implies x = 0.76m So length = 1.5 + 0.76 = 2.26m