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authorprashantsinalkar2018-02-03 11:01:52 +0530
committerprashantsinalkar2018-02-03 11:01:52 +0530
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+//maximum mechanical advantage and maximum efficiency
+//Effort
+P=150 //N
+W=7700 //N
+MA=W/P //mechanical advantage
+//If efficiency=60%
+eff=0.6
+VR=(MA)/(eff)
+//When an effort of 250 N raised a load of 13200 N
+P1=250 //N
+W1=13200 //N
+MA1=(W1)/(P1)
+eff1=MA1*100/VR //percent
+//assume law of machine as P=m*W+C
+//from first case 150=7700*m+C
+//from second case 250=13200*m+C
+//Solving we get
+m=100/5500
+//maximum mechanical advantage
+MAmax=1/(m)
+//maximum efficiency
+Effmax=100/(m*VR) //percent
+printf("\nMechanical advantage=%0.2f\nVelocity ratio=%0.2f\nEfficiency=%0.2f percent\nMaximum mechanical advantage=%0.2f\nMaxumum efficiency=%0.2f percent",MA,VR,eff1,MAmax,Effmax)