Question

a hydraulic press consists of 2 cylinders of cross section 1m^2 and 10 m^2. a force of 200N is applied on the smaller cylinder. what force is applied on the larger piston?​

Answer 1

$\LARGE{ \underline{\underline{ \green{ \bf{Required \: answer:}}}}}$

The Hydraulic press or the hydraulic lift works on the same principle that pressure applied to any part of an enclosed fluid is transmitted to all directions.

Given:

• Cross-section of smaller = 1 m²
• Cross-section of larger = 10 m²
• A force of 200 N is applied on the smaller cylinder.

To FinD:

• Force applied on larger cylinder?

Step-wise-Step Explanation:

The pressure remains constant hence,

$\therefore{\boxed{\rm{\dfrac{F_s}{A_s} = \dfrac{F_l}{A_l}}}}$

• Fs = 200 N
• As = 1 m²
• Al = 10 m²

Plugging the given values:

⇒ $\rm{ \dfrac{200}{1} = \dfrac{F_l}{10}}$

⇒ $\rm{F_l= \dfrac{200 \times 10}{1} N}$

⇒ $\rm{F_l = 2000 \: N}$

Hence the force applied on larger piston is 2000 N . And we are done :)

Answer 2

$\sf \underline{Given} : \begin{cases} \: \sf \: F_{s} = 200 N \\ </p><p> \sf \: \: A _{s} = 1 {m}^{2} \\ \sf \: A _{l}= 10 {m}^{2} \end{cases} \: \\ \\ </p><p>$

$\sf \underline{To \: find} :$

• $\text{ \sf \underline{what force is applied on the larger piston \: }}$

$\sf \underline{Solution} : \:$

We have :

$\boxed{ \sf \: \frac{ f_{s}}{a_{s}} = \frac{ f_{1}}{a_{1}}}$

$\bigstar \: \sf \underline{ substitute \: all \: values} :$

$\sf : \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{200}{1} = \frac{ f_{1} }{10} \\ \\ \\ \sf : \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: f_{1} = 200 \times 10 \\ \\ \\ \sf : \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: f_{1} = \underline{ 2000 \: }$

$\mathfrak{ \text{ \sf 2000 N is the force applied on the larger piston?}}$