Question

write a MATLAB SCRIPT that will output A and B.

Problem 7.4 - Weight Supported by Two Cables PROBLEM DESCRIPTION This example is illustrated in the fure below. A weight W is supported by two cables anchored a distance D apart. The cable length Lan is given, but the length, Lac is to be determined. Each cable can support a maximum tension force equal to W. For the weight to remain stationary, the total horizontal force and total vertical force much each be zero. This principle leads to the following equations: -TASCOS + Tac cos 0 = 0 and Tas sin + Tac sin $ = W and . We can solve these equations for the tension forces Tas and Tac if we know the angles From the Law of Cosines: e = cos(- D2+ LB-LC) 2DLAB From the Law of Sines: $ = sin 2 LAB Sine Lac Assume the following information: D = 6', LAB = 3', and W = 2000 lbs. For these given values use a loop in your script to find LACmin, the shortest length Lac we can use without Tabor Tac exceeding 2000 lb. Note that the largest Lac can be is 6.7' (which corresponds to 0 = 90°.

Answer 1

clc
clear all

syms TAB TAC % sybolic variable for T_AB and T_AC
%% Given values
D = 6;
L_AB = 3;
W = 2000;
%% Computations
L_AC = 3.01:0.005:6.7; % array of length L_AC
% define empty arrays
T_AB = [];
T_AC = [];
for i =1:length(L_AC)
% calculate theta and phi
theta = acosd((D^2+L_AB^2-L_AC(i)^2)/2/D/L_AB);
phi = asind(L_AB*sind(theta)/L_AC(i));
% system of equations has to be solved
eq1 = -TAB*cosd(theta) +TAC*cosd(phi)==0;
eq2 = TAB*sind(theta)+TAC*sind(phi)==W;
% solving system of equations
k = solve([eq1 eq2],TAB, TAC);
T_AB = [T_AB, double(k.TAB) ];
T_AC = [T_AC, double(k.TAC) ];

end
%% Find the element in array L_AC for which L_AC is minumum and neither
% T_AC nor T_AB greater than 2000 lb
flag = 0;
i = 0;
while flag ==0
i = i +1;
if(T_AB(i)<=2000 && T_AC(i)<=2000)
flag = 1;
break
end
end
% displaying and plotting results
fprintf('Answer a): The shortest length L_AC can be without\n');
fprintf(' T_AB nor T_AC exceeding 2000 lb is %0.3f''\n',L_AC(i));
plot(L_AC(i:end),T_AC(i:end),'-k',L_AC(i:end),T_AB(i:end),'--k');
xlabel('Length L_{AC} (ft)');
ylabel('T_{AB} and T_{AC} (lb)');
legend('T_{AB}','T_{AC}');
title('Answer b) The plot of the tension forces T_{AB} and T_{AC} vs L_{AC}');

Command Window Answer a): The shortest length L AC can be without T_AB nor T_AC exceeding 2000 lb is 4.025' fx >>|

Answer b) The plot of the tension forces Tae and Tac Vs LAC 2000 - - 1800 - - - - - - - - - - -- - - - - - - - - -- - - - - - -- - - - - - - - - 1600 1400 1200 TAR and Tac (lb) 600 400 4.5 6.5 5.5 Length Lac (ft)