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

`2F_(1)`

Step-by-step 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

`(F_(1))/(A_(1))= \frac{F_{_(2)}}{A_{_(2)}}`

`(F_(1))/(A_(1))= \frac{F_{_(2)}}{2A_{_(1)}}`

`F_(1)= (F_(2))/(2)\\`

`F_(2)= 2F_(1)`