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+//Two incloned planes
+//refer fig.15.10 (a),(b) and (c)
+//Let the assembly move down the 60 degree plane by an acceleration a m/sec^2
+//Consider the block weighing 100 N
+//Applying equilibrium conditions
+N1=50  //N
+mu=1/3
+//From law of friction
+F1=mu*N1  //N
+//T+((100*a)/(9.81))=69.93
+//Now consider 50 N block
+N2=50*cosd(30)  //N
+//From the law of friction
+F2=mu*N2
+//((50*a)/(9.81))-T=-39.43
+//Solving we get
+a=(69.93-39.43)*9.81/(100+50)  //m/sec^2
+T=69.93-(100*1.9947/9.81) //N
+printf("\na=%.4f m/sec^2\nT=%.2f N",a,T)
+
+
+
+