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+//minimum weight W to prevent downward motion of 1000N block
+mu1=0.2
+mu2=0.3
+//Refer fig. 5.20
+alpha=atand(3/4)  //degree
+//considering equilibrium of block W
+//N1=W*cosd(alpha)
+//F1=mu2*N1
+//T1=0.84*W
+theta=180  //degree
+//Friction equation of rope gives
+//T2=T1*%e^(mu2*theta)
+//solving
+//T2=2.156*W
+//Consider equilibrium of 1000 N block
+//N2-N1=800
+//N2=0.8*W+800
+//F2=0.3*N2
+//F1+F2+T2-1000*sind(alpha)=0
+//solving we get
+W=(1000*sind(alpha)-240)/(0.24+0.24+2.156)  //N
+printf("\nRequired force is W=%0.2f N",W)
+
+
+