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+//moving weight
+//refer fig.17.6 (a),(b)and(c) 
+//first method
+//For 2000 N block
+W1=2000  //N
+mu=0.2
+N1=W1*cosd(30)  //N
+F1=mu*N1  //N
+//For 1800N block
+W2=1800  //N
+N2=W2*cosd(60)  //N
+F2=mu*N2  //N
+//Let T be the tension in the chord
+u=0
+v=9.81  //m/sec
+//applying impulse momentum equation for the 2000 N block in upward direction parallel to the plane
+//(T-1346.41)*t=2000
+//Applying impulse momentum equation for 1800 N block
+T=1363.48  //N
+//Thus
+t=117.11  //sec
+printf("\nBy first method-\nT=%.2d N\nt=%.2d sec",T,t)
+//second method
+//Writing impulse momentum equation in the direction of motion
+t1=117.11  //sec
+//To find tension in the chord, consider impulse momentum equation of any block
+T1=1363.48  //N
+printf("\nBy second method-\nt=%.3f sec\nT=%.3f N",t1,T1)
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