What force is applied on a piston of area of cross section 2 sq. cm to obtain a force of 150 N on the piston of area of cross section 12 sq cm in a hydraulic machine

Dear student,

The pressure experienced in both cases would be same. So, as

 

P=F/AF1/A1=F2/A2or the force on the second piston will be F2=(F1/A1).A2by substituting appropriate values we getF2=(F1/A1).A2F2=(15012x10-4×2x10-4)or the force F2= 25 N



Regards

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