Question

The areas of pistons in a hydraulic machine are 6 cm^2 and 576 cm^2.what force on the smaller piston will support a load of 1152 N on the larger piston? State the assumption made in the above calculation.

Answer 1

Answer:

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Answer 2

The force on smaller piston is 12 N.

Explanation:

Given that,

Area of larger piston = 576 cm²

Area of smaller piston = 6 cm²

Load = 1152 N

Pressure remains same all over

We need to calculate the force on smaller piston

Using formula of pressure

$P=\dfrac{F}{A}$

Where, F = load

P= pressure

A = area

For large piston,

$P=\dfrac{1152}{576}$...(I)

For smaller piston,

$P=\dfrac{F}{6}$...(II)

Divided equation (I) by equation (II)

$\dfrac{1152}{576}=\dfrac{F}{6}$

$F=\dfrac{6}{576}\times1152$

$F=12\ N$

Hence, The force on smaller piston is 12 N.

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Topic : Pressure

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