A mass m is suspended by three cables attached at the three points B, C, and D, as shown in Figure 6.2-2; Let TI’ T2, and T3 be the tensions in the three cables AB, AC, and AD. respectively. If the mass m is stationary, the sum of the tension components in the x, in the y, and in the z directions must each be zero. This requirement gives the following three

A mass suspended by three cables

We can use the following MATLAB script file to solve this system for x and then multiply the result by mg to obtain the desired result.
% File cable.m
% Computes the tensions in three cables.
A1 [l{sqrt(35), -3/sqrt(34) , 1/sqrt(42»);
A2 = [3/sqrt(35), 0, -4/sq~t(~2»);
A3= [5/sqrt(35), 5/sqrt(34) , 5/sqrt(42»);
A = [AI; A2; A3);
B = [0; 0; 1);
x = A\b;
disp (‘The tension T_l is:’)
disp (x(l) )
disp (‘The tension T_2 is:’)
disp (x(2»
disp (‘The tension T_3 is:’)
disp (x(3»
When this file is executed by typing cable, the result is stored in the array x, which gives the values TI = 0.5071, T2 = 0.2915, and T3 = 0.4166. Because MATLAB does not generate an error message when the file is executed, the solution is unique. Using the linearity property, we multiply these results by mg and obtain the following solution to the set AY = bmg: TI = 0.5071 mg, T2 = 0.2915 mg, and T3 = 0.4166 mg.