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In a simple hydraulic press, the cross sectional area of the two cylinders is `5xx10^(-1)m^(2)` and `10^(-2)m^(2)` respectively. A force of `20 N` is applied at the small plunger. a. What is the pressure produced in the cylinders? b. What is the thrust exerted on the large plunger? c. How much work is done by the operator, if the smaller plunger moves down `0.1m`? |
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Answer» In a hydraulic press a force `F_(1)` applied to the smaller plunger creats a pressure `(F_(1))/(A_(1))` in the liquid and this pessure is transmitted equally throughout the liquid and acts on the larger plunger. The thrust acting on te larger plunger upwards due to tis pressure is `F_(2)=A_(2)((F_(1))/(A_(1)))` Hence, the thrust on `A_(2)` is magnified by `(A_(2))/(A_(1))` times `a. (F_(1))/(A_(1))=20/(5xx10^(-4))=40000N//m^(2)` b. `F_(2)=F_(1)xx((A_(2))/(A_(1)))=20xx(10^(-2))/(5xx10^(-4))=400N` work done by force `F_(1)=F_(1)d_(1)=20xx0.1=2J` |
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