40 10 N The 10 N weight is being hung from the cables as shown above. Using Newt

Question

40 10 N The 10 N weight is being hung from the cables as shown above. Using Newton's Second Law v-0 and ,-0), we can form the system of equations CBCOS CACOs30° = 0 MATLAB has two ways to solve equations: 1.) using solve and symbolic math, and 2.) using matrices! You are to utilize the first method. For more information on how to use the second method, see Chapter 14.5.2. This is also the method that you will use when you take Linear Algebra (EK 102) For tips on using solve, refer to the Symbolic Mathematics section of the textbook (Chapter 14.6.4). Here is the line of code that uses solve to solve the system of equations, for the two forces: You are to write a script ek125homeworkla that solves the system of equations for the forces FcA and FCB. Create the symbolic variables FAR and Fea, and write the two equations for Ex and using these symbolic variables. Use the above function call to solve, that receives the two inputs Ex 0 and Ey - 0, and the variables to solve for (Ecb and Eca), and it returns the two solutions Eobsal and Ecasal (this is a glimpse of more complex function calls that you will cover in the future!) Finally, you are to print the solutions. Here is an example of executing the script: Note: These values are NOT the actual solutions >ek125homeworkla The force Ka- 79.97 N, and force Ece = 90.41 N

Explanation / Answer

%%% Matlab program %%

%%% let x=Fcb , y=Fca
syms x y

s=solve('x*cos(40*pi/180)-y*cos(30*pi/180)=0','x*sin(40*pi/180)+y*sin(30*pi/180)=10*9.81');
Fcb=subs(s.x);
Fca=subs(s.y);
fprintf('Fcb= %2.3f N ',Fcb);
fprintf('Fca= %2.3f N ',Fca);

OUTPUT:

Fcb= 90.409 N
Fca= 79.972 N

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