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+//Least and the greatest value of W for equilibrium
+//refer fig.
+//case (a) for least value
+refer fig. (b)
+//considering equilibrium of 1000 N block
+N1=1000*cosd(30) //N
+//law of friction gives
+F1=0.28*N1 //N
+T=500-242.49 //N
+//consider equilibrium of W
+//F2=0.1W
+Wmin=257.51/(0.1+sind(60)) //N
+//case (b) for greatest value
+//refer fig. 5.11 (c)
+//consider equilibrium of 1000N block
+T=1000*sind(30)+242.49 //N
+//consider equilibrium of W
+//F2=0.2*0.5*W
+Wmax=742.49/(0.1+sind(60)) //N
+printf("\nThe greatest and least values of W are:-\nWmax=%0.2d N\nWmin=%0.2d N",Wmax,Wmin)
+