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% This is a comment; to write a comment, you have to start the line with a '%'.
% I will painstakingly do this question on MATLAB in detail for you.
% We start with the RHS, and denote the forces A and B respectively.
F_A = 30; %The magnitude of the force in Newtons
F_B = 120;
A_x = -7;
A_y = -1; %The directions of F_A. Negative because it is pointing left and down.
B_x = 1;
B_y = 5;
F_A_x = F_A * A_x / sqrt(A_x^2 + A_y^2); %The x-direction component of F_A.
% Found by scaling the triangle given in the x-direction. If still confusing ask me.
F_A_y = F_A * A_y / sqrt(A_x^2 + A_y^2);
F_B_x = F_B * B_x / sqrt(B_x^2 + B_y^2);
F_B_y = F_B * B_y / sqrt(B_x^2 + B_y^2);
% Notice the similarity in the expressions written above ^.
% I have written these out in full, but there is a quicker way to do all these repetitive functions on MATLAB!
rhsF_x = F_A_x + F_B_x;
rhsF_y = F_A_y + F_B_y;
rhsF = sqrt(rhsF_x^2 + rhsF_y^2)
% I have written this command without a ';' so the result is printed out
% The part below is beyond the scope of what you should be able to understand on MATLAB
% But I repeat the process of getting x- and y- components for the LHS force, then compare with RHS force to find F
syms F;
expr_x = F * -cosd(17) + 2*F*cosd(27);
expr_y = F * sind(17) + 2*F*sind(27);
expr_F = sqrt(expr_x^2 + expr_y^2);
assume(F > 0);
% The below is the final answer, so I have chosen not to suppress the result.
solF = double(solve(expr_F == rhsF, F))