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(A) The pressure at depth $h = 3 \ m$ on the water side is $P_w = P_0 + \rho_w gh$,and on the liquid side is $P_{\ell} = P_0 + \rho_{\ell} gh$.The for

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(A) The pressure at depth $h = 3 \ m$ on the water side is $P_w = P_0 + ho_w gh$,and on the liquid side is $P_{\ell} = P_0 + ho_{\ell} gh$.The force exerted by water on the door is $F_w = P_w A = (P_0 + ho_w gh) A$.The force exerted by the liquid on the door is $F_{\ell} = P_{\ell} A = (P_0 + ho_{\ell} gh) A$.For the door to remain closed,the external force $F_{ext}$ applied on the water side must satisfy $F_{ext} + F_w = F_{\ell}$.Therefore,$F_{ext} = F_{\ell} - F_w = (P_0 + ho_{\ell} gh) A - (P_0 + ho_w gh) A = (ho_{\ell} - ho_w) ghA$.Given: $ho_{\ell} = 1500 \ kg/m^3$,$ho_w = 1000 \ kg/m^3$,$g = 10 \ m/s^2$,$h = 3 \ m$,and $A = 100 \ cm^2 = 100 	imes 10^{-4} \ m^2 = 0.01 \ m^2$.Substituting the values: $F_{ext} = (1500 - 1000) 	imes 10 	imes 3 	imes 0.01 = 500 	imes 30 	imes 0.01 = 150 \ N$.

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