A piston has an area 10 times as large as an input piston. How large a force must be applied to the inputpisto?
pressure formula p=F/A p is the pressure, F is the normal force, A is the area of the surface area on contact p=F1/A1=F2/A2 F2=F1*A2/A1 where A1=10A2; F1=2,500 lbs. F2=2500/10=250 lbs
If the larger piston supports a load of 5300 N, how large a force must be applied to the input piston?
in a hydraulic lever,the ratio of the area of input and the force of the input is equal to the ratio of the ratio of area and pressure of output. In this case let area of input be ai=y force of input be fi=x area of output be ao=60y force of output be fo=5300 fi/ai=fo/ao fi=fo/ao*ai =5300*y/60y =5300/60 =83.33 therefore the foce which has to be supplied is 83.33 N