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+clear
+//
+
+//A bar can develop a tensile force or a compressive force. Let the force developed be a compressive force S (push on the cylinder).
+
+//variable declaration
+W=10.0 //weight of Roller,KN
+IL=7.0 //inclined loading at angle of 45°,KN
+H=5.0 //Horizontal loading ,KN
+
+theta=45.0*%pi/180 //angle of loading of IL
+thetaS=30.0*%pi/180.0
+
+//Since there are more than three forces in the system, Lami’s equations cannot be applied. Consider the components in horizontal and vertical directions.
+//sum of vertical Fy & sum of horizontal forces Fx is zero
+//Assume direction of Fx is right
+//Assume direction of Fy is up
+
+S=(-H+IL*cos(theta))/cos(thetaS)
+printf("\n S= %0.3f N",S)
+
+printf("\n Since the value of S is negative the force exerted by the bar is not a push, but it is pull (tensile force in bar) of magnitude %0.3f kN.",-S)
+
+R=W+IL*sin(theta)-S*sin(thetaS)
+printf("\n R= %0.3f kN",R)