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Previous Problem List Next (1 point) A flexible cable suspended between two vertical supports is hanging under its own weight. The weight of a horizontal roadbed is distributed evenly along the x-axis with the density ρ 0.7 lb/ft. The coordinate system is chosen so that they-axis passes through the lowest point Pi and the x axis is a -1.5 ft below Pi, as shown in the picture. If the absolute value of tension force in the cable at point Pi is notation dyldx for the first derivative 1, then the initial value problem that governs the shape of the cable is (Use the x (0) The solution of this problem is y y(x), where

Answer 1

Given p = 0.7 lb/ft: a = 1.5 ft: T = 1 As the cable is in equilibrium So ef_x = 0 = ef_y. from the above diagram we have ef_x = T_2 cos Theta - T = 0 implies T_2 cos Theta = T. ef_y = T_2 cos Theta - w = 0 implies T_2 sin Theta = w. T_2 sin Theta/T_2 cos Theta = W/T implies Tan Theta = W/T. We know tan Theta is slope dy/dx at rho_2. dy/dx = W/Y. The Wright of the cabel is w = rho_x arc length l = pl) (lim arc length) dy/dx = rho l/T. are length (l) = integral^P_2_P_1 squareroot 1 + (dy/dx)^2 middot dx. \r\n dy/dx = rho/T integral^P_2_P_1 squareroot 1 + (dy/dx)^2 middot dx. d^2 y/d^2 x = rho/T squareroot 1 + (dy/dx)^2 middot dx. d^2 y/d^2 x = 0.7/1 squareroot 1 + [dy/dx]^2. E initial conditionery are y (0) = a = 1.5 ft. dy/dx (0) = 0 \r\n