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+clc;
+// Given data
+F=500; // N , Vertical force applied to end of lever
+theta=60;// degree, angle made by lever with +ve X axis
+theta=theta*%pi/180;// Conversion of angle into radian
+l=600; // mm , length of lever
+
+// a ) Momemt about O
+d=l*cos(theta);// mm ,perpendicular distance from o to the line of action
+d=d/1000; // m, conversion into meter
+Mo=F*d;// N.m, Magnitude of moment about O
+printf("Magnitude of moment about O of the 500 N is %.2f N.m and it is in clockwise direction as force tends to rotate lever clockwise\n",Mo);
+
+// b) Horizontal force
+
+d=l*sin(theta);//mm, perpendicular distance from o to the line of action
+d=d/1000; // m, conversion into meter
+F=Mo/d;// N, Horizontal Force at A required to produce same Moment about O
+printf("Magnitude of Horizontal Force at A required to produce same Moment about O is %.2f N\n",F);
+
+// c)Smallest force
+
+// F is smaller when d is maximum in expression Mo=F*d, so we choose force perpendicular to OA
+d=0.6;// m ,perpendicular distance from o to the line of action
+F=Mo/d;// N, Smallest Force at A required to produce same Moment about O
+printf("Magnitude of smallest Force at A required to produce same Moment about O is %.2f N\n",F);
+
+//d) 1200 N vertical force
+F=1200;// N, verical force producing same movement on lever acting at pt B
+d=Mo/F;// m, perpendicular distance from o to the line of action of force
+OB=d/cos(theta);//m, distance of point B from O
+OB=OB*1000;//mm, conversion into millimeter
+printf("Verical force of 1200 N must act at %.2f mm far from the shaft to create same moment about O\n",OB);
+
+