The average mass that must be lifted by a hydraulic press is 80 Kg .If the radius of the larger piston is 5 times the smaller piston,what is the minimum force that must be applied?

Asked by Abhimanyu Charan | 8th Dec, 2013, 07:46: PM

Expert Answer:

To solve this problem we need to use Pascal's law.

The  mass to be lifted = m = 80 kg

Force Required to lift up = F = mg

= 80 x 9.8 

=784  N

Let  Radius of the larger piston = R

     Radius of the smaller piston = R2

Given that,

R= 5 R2

Area of larger piston = A1 = π R12

Area of smaller piston = A2 = π R22

A1 = 25 A2

Pressure = F/A

According to pascal's law,

Pressure applied will be the same

P1 = P2

F1/A1 = F2/A2  

F1/25A= F2/A2  

F1/25 = F2  = 784 N

Area of larger piston is larger therefore water will exert more force on the larger piston.

Since area of larger piston is 25 times larger,  therefore minimum force we need to apply on the smaller piston is 1/25 times the force required,

minimum force required= 1/25 x 784

       = 31.36 N

 Answer =31.36 N

Answered by Komal Parmar | 9th Dec, 2013, 08:00: AM

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