Question

average mass that must be lifted by a hydraulic press is 80 Kg if the radius of the larger piston is 5 times that of smaller Piston what is the minimum force that must be applied

Answer 1

F2=80*9.8 N

A2=π[5r]^2

F1=?

A1=πr^2

We know,

F2/A2=F1/A1

80*9.8/(π×25r^2)=F1/(πr^2)

F1=31.36 N

Answer 2

Refers to the attachment...

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#hope it helps...