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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+clear all; clc;
+disp("Ex 6_14")
+disp("Figure 6-27b shows the free body diagram")
+//Applying equations of equilibrium
+//summing moments about C and equating to zero
+a1=60
+a=a1*%pi/180
+F_AB=(2000*2)/(sin(a)*4)
+printf('\n F_AB = %0.1f N',F_AB)
+//summinf forces in X direction
+Cx=F_AB*cos(a)
+printf('\n Cx = %0.0f N',Cx)
+//summing forces in Y direction
+Cy=2000-(F_AB*sin(a))
+printf('\n Cy = %0.0f N\n',Cy)
+
+//Alternate method ignoring AB is not a two force member
+disp("Alternate method")
+A=[3*sin(a),-3*cos(a),0,0;0,-4,0,0;1,0,-1,0;0,1,0,1]
+B=[0;-4000;0;2000]
+C=inv(A)
+D=C*B
+Bx=D(1)
+By=D(2)
+Cx=D(3)
+Cy=D(4)
+
+
+//Result
+printf('\n The values are')
+printf('\n Bx=%0.0f \n By=%0.0f \n Cx=%0.0f \n Cy=%0.0f \n All values are in N',Bx,By,Cx,Cy)