JAMB - Physics (2004 - No. 26)
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
4m2
8m2
2m2
1m2
Selitys
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
∴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