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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
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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

User RedKing
by
4.9k points

1 Answer

2 votes

Answer:

1600 N

Step-by-step explanation:

pascals principle says..


(A_(out))/(A_(in)) = (F_(out))/(F_(in)) \\

so lets do


(2)/(4) = (x)/(3200)

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

1600 is equal to x

1600 N

YAAAAAAAAAAAAAAY

User Smentek
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4.4k points
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