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A 10.5 N force is exerted on a small piston of a hydraulic system. The cross sectional area of the small piston is .025m2. What is the magnitude of the weight that can be lifted by a large piston which has a surface area of .1m2?
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We must balance the pressures here. For the small piston, \[P_s = {F_s \over A_s}\] For the large piston, \[P_l = {F_l \over A_l}\]There pressure must equal each other, therefore, \[P_s = P_l \rightarrow {F_s \over A_s} = {F_l \over A_l}\]We desire the force of the large piston. So solving for F_l, we get\[F_l = F_s \left( A_l \over A_s \right)\]
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