A hydraulic system has small and large pistons with cross-sectional area of 3x10^-4 m^2 and 6x10^-3m^2 respectively. When a firce of 30N is applied to small piston, it pushes down the piston by 30cm. Determine pressure transmitted in the fluid?

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A hydraulic system has small and large pistons with cross-sectional area of 3x10^-4 m^2 and 6x10^-3m^2 respectively. When a firce of 30N is applied to small piston, it pushes down the piston by 30cm. Determine pressure transmitted in the fluid?

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Answer 1 · 2015-08-10T23:37:33

Have a look, if it makes sense:

Explanation:

Considering Pascal's principle, pressure is transmitted entirely to the whole fluid (see the toothpaste tube effect! :-) so that:
$P_{1} = P_{2}$ $P_1=P_2$
but $P_{1} = \frac{F_{1}}{A_{1}} = \frac{30}{3 \times 10^{- 4}} = 100, 000 P a$ $P_1=F_1/A_1=30/(3xx10^-4)=100,000Pa$
BUT energy must be conserved!
so ${Work}_{1} = {Work}_{2}$ $"Work"_1="Work"_2$ (force times distance)
$F_{1} \cdot h_{1} = F_{2} \cdot h_{2}$ $F_1*h_1=F_2*h_2$
where $F_{2} = P_{1} \cdot A_{2}$ $F_2=P_1*A_2$ (the pressure is the same on both pistons)
So:
$F_{2} = 100, 000 \cdot 6 \times 10^{- 3} = 600 N$ $F_2=100,000*6xx10^-3=600N$
So:
$h_{2} = F_{1} \cdot \frac{h_{1}}{F_{2}} = 30 \cdot \frac{30}{600} = 1.5 c m$ $h_2=F_1*h_1/F_2=30*30/600=1.5cm$

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A hydraulic system has small and large pistons with cross-sectional area of 3x10^-4 m^2 and 6x10^-3m^2 respectively. When a firce of 30N is applied to small piston, it pushes down the piston by 30cm. Determine pressure transmitted in the fluid?

1 Answer

Explanation:

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