Question

a hydraulic lift requires a minimum effort force of 14.4N to lift a patient of a mass 82kg .How much is the effort piston area if the resistance piston has an area of 1.2m²

Answer 1

The output force of a hydraulic lift is given by:

`F_2=(A_2)/(A_1)F_1`

In this case we need to lift a patient of mass 82 kg, that means that the lifting force has to be equal to:

`F_2=82\cdot9.8=803.6`

Now that we know that we plug the values given, in this case we have:

`\begin{gathered} 803.6=(1.2)/(A_1)(14.4) \\ A_1=(1.2)/(803.6)(14.4) \\ A_1=0.022 \end{gathered}`

Therefore the area of the effort piston is 0.022 squared meters (rounded to three decimal).