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+// SAMPLE PROBLEM 6/2
+clc;clear;funcprot(0);
+// Given data
+m=150;// kg
+M=5;// kN
+theta=30;// degree
+ACbar=1.5;// m
+BDbar=1.5;// m
+ABbar=1.8;// m
+g=9.81;// The acceleration due to gravity in m/s^2
+
+// Calculation
+// SigmaM_C=0
+A_t=M/ACbar;// kN
+// SigmaF_t=m*abar_t
+// alpha=14.81-6.54*cos(theta);
+wsquare_30=(29.6*theta*%pi/180)-(13.08*sind(theta));// (rad/s)^2
+alpha_30=14.81-(6.54*cosd(theta));// rad/s^2
+A_n=(m/1000)*ACbar*wsquare_30;// kN
+A_t=(m/1000)*BDbar*alpha_30;// kN
+// SigmaM_A=m*abar*d
+B=((A_n*(ABbar-0.6)*cosd(theta))+(A_t*0.6))/(ABbar*cosd(theta));// kN
+printf("\nThe force in the link DB,B=%1.2f kN",B);