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+//To find the inertia torque on the crankshaft
+clc
+//given
+CP=60//in
+l=CP/12
+a=41
+cg=19
+g=32.2//ft/s^2
+m1=580//lb
+Mr=500//lb
+n=5//from example 12.3
+x=40*%pi/180
+N=120
+r=1//ft
+k=25
+w=N*%pi/30
+Rm=m1+(cg/CP)*Mr
+fp=w^2*r*(cos(x)+cos(2*x)/n)
+Fp=-Rm*fp/g
+OM=0.7413//ft -from example 12.3
+Tp=Fp*OM//from 12.6
+L=a+k^2/a//length for simple equivalent pendulum
+L1=L/12
+Tc=-Mr*(a/12)*(l-L1)*w^2*sin(2*x)/(g*2*n^2)//from 12.10
+Tw=-Mr*a*cos(x)/(n*12)
+T=Tp+Tc+Tw
+printf("\nTp= %.f lbft\nTc = %.1f lbft\nTw = %.1f lbft\nTotal torque exerted on the crankshaft due to the inertia of the moving parts = Tp+Tc+tw = %.1f lbft",Tp,Tc,Tw,T)