Question

A hydraulic device is used to lift a block of 15 kg mass by 5 cm. (a) If the ratio of areas is A/a = 10, what force is required to do the lifting assuming the weight of the fluid is neglected?

Answer 1

To solve this problem it is necessary to use the concepts related to pressure. The pressure can be defined as

`P = (F)/(A)`

Due to the continuity and pressure conservation relationship, the pressure must be preserved to support the weight (gravity) and to lift it (lift)

In this way,

`P_g = P_l`

`(F_g)/(A) = (F_l)/(a)`

That is equal to

`(F_g)/(F_l)=(A)/(a)`

We have a ratio of A/a = 10, then

`(F_g)/(F_l) = 10`

`(mg)/(F_l) = 10`

`(mg)/(10) = F_l`

`F_l = ((15)(9.81))/(10)`

`F_l = 147.15N`

Therefore the force required to do the lifting is 147.15N