The press plungers of a Bramah (hydraulic) press is $40 c m^{2}$ in cross-section and is used to lift a load of mass $800 k g$ . What minimum force is required to be applied to the pump plunger if its cross-sectional area is $0.02 m^{2}$ ?

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Solution

According to pascal's law, pressure acting on press plunger $=$ pressure acting on pump plunger.
$= \frac{800 \times g}{40 \times 10^{- 4} m^{2}}$
$\Rightarrow∴ F = \frac{800 \times 9.8 \times 0.02}{40 \times 10^{- 4}}$
$= 39200 N$

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The press plungers of a Bramah (hydraulic) press is 40cm2 in cross-section and is used to lift a load of mass 800kg. What minimum force is required to be applied to the pump plunger if its cross-sectional area is 0.02m2?

According to pascal's law, pressure acting on press plunger = pressure acting on pump plunger.=800×g40×10−4m2⇒∴F=800×9.8×0.0240×10−4=39200N

The press plungers of a Bramah (hydraulic) press is $40 c m^{2}$ in cross-section and is used to lift a load of mass $800 k g$ . What minimum force is required to be applied to the pump plunger if its cross-sectional area is $0.02 m^{2}$ ?

According to pascal's law, pressure acting on press plunger $=$ pressure acting on pump plunger.
$= \frac{800 \times g}{40 \times 10^{- 4} m^{2}}$
$\Rightarrow∴ F = \frac{800 \times 9.8 \times 0.02}{40 \times 10^{- 4}}$
$= 39200 N$