Question #242860

in a hydraulic pressure, a force of 20N is applied to a piston of area 0.2m^2, if the area of the other piston is 2m^2. calculate the force excerted on it.

Expert's answer

Using Pascal's law for F1= 20 N, A1 = 0.2 m2 and A2 = 2 m2:


P1=P2F1A1=F2A2  ⟹  F2=F1A2A1P_1=P_2 \\ \cfrac{F_1}{A_1}=\cfrac{F_2}{A_2} \implies F_2=F_1\cfrac{A_2}{A_1}


We substitute and we can find the force exerted on the piston 2 or F2:

F2=(20 N)(2 m20.2 m2)=200 NF_2=(20\,N)\bigg(\cfrac{2\,\cancel{m^2}}{0.2\,\cancel{m^2}}\bigg)=200\,N


In conclusion, the force exerted on the piston (of area A2 = 2 m2) is 200 N.



Reference:

  • Sears, F. W., & Zemansky, M. W. (1973). University physics.
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