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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+clc;
+W=-1200;//N,j Weight
+BD=[-2.4,1.2,-2.4];//m, Vector BD
+EC=[-1.8,0.9,0.6];//m, Vector EC
+//T_BD=norm(T_BD)*BD/norm(BD);// m, vector of tension in BD
+//T_EC=norm(T_EC)*EC/norm(EC);// m, vector of tension in EC
+// Applying equillibrium conditions we get
+// Sum_F=0, and Sum(M_A)=0 and setting co-efficient equal to zero
+A=[0.8,0.771;1.6,-0.514];//MAtrix of co-efficient
+b=[-1440;0];//matrix b
+x=linsolve(A,b);// solution matrix
+T_BD=x(1);// N,Tension in BD
+T_EC=x(2);//N, Tension in EC
+printf("T_BD= (%.0f N) and T_EC= (%.0f N) \n",x(1),x(2));
+
+Ax=2/3*T_BD+6/7*T_EC;//N, x component of reaction at A
+Ay=-(1/3*T_BD+3/7*T_EC+W);//N, Y component of rection at A
+Az=2/3*T_BD-2/7*T_EC;//N, z component of reaction at A
+
+printf("Reaction at A is A=(%.0f N)i +(%.0f N)j +(%.1f N)k \n",Ax,Ay,Az);
+//Answe in Newton instead of lbs
+//1lbs=4.44N