The areas of the effort and load pistons of a hydraulic press are 0.5m² and 5m² respectively. If a force F₁ of 100N is applied on the effort piston, the force F₂ on the load is

Deepthinker581

4 years ago

Effort=0.5
Load=5
Force 1=100
Force 2= ?
Force 2= load/effort times force 1

5/0.5×100
10×100=1000
Force 2=1000

chydirim@gmail.com

10 years ago

isn't the formular- load/area of effort=effort/area of load

L/e=E/l. Pls verify this. it is meant to be option A.