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The small and large pistons of a hydraulic press
have areas of 2 m2 and 4 m2. If the load on the large piston in 3200 N, what is the input force (effort) that must be applied on the small piston

Respuesta :

stapmoshun

Answer:

1600 N

Explanation:

pascals principle says..

[tex]\frac{A_{out}}{A_{in}} = \frac{F_{out}}{F_{in}} \\[/tex]

so lets do

[tex]\frac{2}{4} = \frac{x}{3200}[/tex]

so multiply one half (2/4 simplified) by 3200

1600 is equal to x

1600 N

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