A 1000-lb weight is supported by three cables, as shown in the figure. Note that cable 3 is attached along the y-axis, at a distance of d in feet from the origin O. Using particle equilibrium (usually taught in statics), three simultaneous equations can be formulated. The equations for the tensile forces (in pounds) in the three cables supporting a 1000-lb weight are shown here: [0.348 -0.557 0.870 -0.557 -0.239 0.796 0 d/L 10/L] [T_1 T_2 T_3] = [0 0 1000] Where d = the distance from the origin to the point of attachment of cable 3. and L = the length of cable 3 = Squareroot 10^2 + d^2 (feet) If each cable can support 1500 lb, then the factor of safety FS = 1500/(maximum of the three tensile forces). Write a MATLAB function that calculates the FS for a given value of d. As a check, calculate the FS for d = 2 feet. (Answers: T_1 = 259 lb, T_2 = 162 lb, T_3 = 658 lb, FS = 2.28) Write a MATLAB file to calculate and store the FS for values of d ranging from 0 to 20 feet, in 0.1-foot increments. Plot FS versus d. What value of d maximizes the factor of safety, and what is the maximum factor of safety? Comment on the shape of the curve