Question

Consider a hydraulic system with two pistons, piston1 and piston2. A force $F_1$ is applied to piston1 of area $A_1$ and piston2 applies a force $F_2$ over area $A_2$. If we double the force applied by piston1 and reduce the area of piston2 by a factor of three, then the new force applied by piston2 is

• A. $\frac{2}{3}\times F_2$
• B. $\frac{3}{2}\times F_2$
• C. $\frac{2}{5}\times F_2$
• D. $\frac{5}{2}\times F_2$

Answer 1

The correct option is A

$\frac{2}{3}\times F_2$

Given, force applied to piston1 $F_1$, Area of piston1 $A_1$, force applied by piston2 $F_2$ and Area of piston2 $A_2$.

We know we doubled the force applied by piston1 $F_1^{\prime }=2F_1$ and area of piston2 is reduced by factor three $A_2^{\prime }=\frac{A_2}{3}$ and area of piston1 remain same so, $A_1^{\prime }=A_1$. Let the new force applied by piston2 be $F_2^{\prime }$

According to Pascal law,

Initially $\frac{F_1}{A_1}=\frac{F_2}{A_2}$...(1)

From equation (1) we get $F_2=\frac{F_1}{A_1}\times A_2$

After changes in force and area,

$\frac{F_1^{\prime }}{A_1^{\prime }}=\frac{F_2^{\prime }}{A_2^{\prime }}$...(2)

Now substituting values $F_1^{\prime }=2F_1$, $A_2^{\prime }=\frac{A_2}{3}$ and $A_1^{\prime }=A_1$ in equation (2) as

$\frac{2F_1}{A_1}=\frac{F_2^{\prime }}{\frac{A_2}{3}}$

On solving we get $F_2^{\prime }=\frac{2}{3}\frac{F_1}{A_1}\times A_2=\frac{2}{3}\times F_2$