Question

what force is applied on a piston of area of cross section 2cm^2 to obtain a force 150N on the piston of area of cross section 12cm^2 in a hydraulic machine ?​

Answer 1

ANSWER:

• The force on first piston = 25 N.

GIVEN:

• Area of first piston = 2 cm².

• Force on second piston = 150 N.

• Area of second piston = 12 cm².

• It is observed in a hydraulic machine.

TO FIND:

• The force on first piston.

EXPLANATION:

$\boxed{ \gray{\sf For \ a \ hydraulic \ machine \ P_1 = P_2}}$

$\boxed{ \gray{ \large{ \bold{Pressure = \dfrac{Force}{Area}}}}}$

$\sf \mapsto P_1 = \dfrac{F_1}{A_1}$

$\sf \mapsto P_2= \dfrac{F_2}{A_2}$

$\sf We\ know\ that \ P_1 = P_2$

$\sf \mapsto \dfrac{F_1}{A_1}= \dfrac{F_2}{A_2}$

$\sf F_1 = F_1 \ | \ F_2 = 150\ N$

$\sf A_1 = 2 \ cm^2 \ | \ A_2 = 12\ cm ^{2}$

$\sf \mapsto \dfrac{F_1}{2}= \dfrac{150}{12}$

$\sf \mapsto F_1= \dfrac{150}{6}$

$\sf \mapsto F_1= \dfrac{50}{2}$

$\sf \mapsto F_1= 25 \ N$

Hence force applied on the first piston will be 25 N.

NOTE: Here higher force is obtained (150 N) by applying smaller force (25 N) and this is one of the advantages of hydraulic machines.

Answer 2

By the principle of hydraulic machine,

Pressure on smaller piston = pressure on wider piston.

.•. P1 = P2

$\frac{F1}{A1} = \frac{F2}{A2} \\ \\ \implies\frac{F1}{2 \times {10}^{ - 4} } = \frac{150}{12 \times {10}^{ - 4} } \\ \\ \implies \: F1 \: = \frac{150}{12 \times {10}^{ - 4} } \times 2 \times {10}^{ - 4} \\ \\ \implies \: F1 \: = \frac{ \cancel{150} \: \: ²⁵ }{ \cancel{12 }\times \cancel{ {10}^{ - 4} }} \times \cancel{2 }\times \cancel{{10}^{ - 4} } \\ \\ \ = 25 \: N....... \: \: \: (ans)$