Question

if you apply a Force of F1 to area A1 on one side of a hydraulic jack, and the second side of the jack has an area that is twice of A1, what will F2 be in comparison to F1

Answer 1

Answer:

$2F_{1}$

Explanation:

F₁ = Force on one side of the jack

A₁ = Area of cross-section of one side of the jack

F₂ = Force on second side of the jack

A₂ = Area of cross-section of second side of the jack = 2 A₁

Using pascal's law

$\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{A_{_{2}}}$

$\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{2A_{_{1}}}$

$F_{1}= \frac{F_{2}}{2}$

$F_{2}= 2F_{1}$