In a hydraulic press , a force of 40N is applied on the effo...
In a hydraulic press , a force of 40N is applied on the effort piston of area 0.4m2. If the force exerted on the load piston is 400N, the area of the large piston is
Pressure developed in the effort piston = F/a
∴F/a = 40/0.4 = 400/4 = 100Nm2
this pressure is transmitted to the larger piston of area A; => P = F/A
∴ 100 = 400/A
=> A = 400/100 = 4m2