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+// SAMPLE PROBLEM 6/6
+clc;funcprot(0);
+// Given data
+alpha_0=3;// rad/s^2
+m=70;// kg
+k=0.250;// The radius of gyration in m
+mu_s=0.25;// The coefficient of static friction
+g=9.81;// The acceleration due to gravity in m/s^2
+DCbar=0.30;// m
+r_A=0.250;// m
+r_Bi=0.150;// m
+r_Bo=0.450;// m
+
+// Calculation
+a_t=r_A*alpha_0;// m/s^2
+alpha=a_t/DCbar;// rad/s^2
+abar=r_Bo*alpha;// m/s^2
+function[X]=force(y)
+    // SigmaF_x=m*abar_x
+    X(1)=(y(1)-y(2))-(m*-abar);
+    N=(m*g);// N
+    // SigmaM_G=Ibar*alpha
+    X(2)=((r_Bo*y(1))-(r_Bi*y(2)))-(m*k^2*alpha);
+endfunction
+y=[10 100];
+z=fsolve(y,force);
+F=z(1);// N
+T=z(2);// N
+printf("\nThe tension in the cable,T=%3.1f N \nThe friction force exerted by the horizontal surface on the spool,F=%2.1f N",T,F);
+N=(m*g);// N
+F_max=mu_s*N;// N
+// If the coefficient of static friction had been 0.1
+mu_s=0.1;// The coefficient of static friction
+F=mu_s*(m*g);// N
+// SigmaM_C=Ibar*alpha + m*abar*r
+T=((m*(r_A^2)*alpha)+(m*abar*r_Bo))/DCbar;// N
+printf("\nThe tension in the cable,T=%3.1f N",T);