Hydrostatic Force on a Curved Surface A submarine, whose hull has a circular cross section of diameter D, is just submerged in water of density ρ, as shown in Fig. E1.6. Derive an equation that gives the total horizontal force Fx on the left half of the hull, for a distance W normal to the plane of

Question

Hydrostatic Force on a Curved Surface

A submarine, whose hull has a circular cross section of diameter [latex]D[/latex], is just submerged in water of density [latex]\rho[/latex], as shown in Fig. E1.6. Derive an equation that gives the total horizontal force [latex]F_{x}[/latex] on the left half of the hull, for a distance [latex]W[/latex] normal to the plane of the diagram. If [latex]D=8 \mathrm{~m}[/latex], the circular cross section continues essentially for the total length [latex]W=50 \mathrm{~m}[/latex] of the submarine, and the density of sea water is [latex]\rho=1,026 \mathrm{~kg} / \mathrm{m}^{3}[/latex], determine the total horizontal force on the left-hand half of the hull.

Accepted Answer

The force is obtained by evaluating the integral of Eqn. (1.41), which is identical to that for the rectangle in Fig. 1.15:

[latex]\begin{array}{c}F=\int_{A} p \sin heta d A=\int_{A^{*}} p d A^{*}, \ \end{array}[/latex] (1.41)

[latex]\begin{array}{c}F_{x}=\int_{A^{*}} p d A=\int_{z=0}^{z=D} ho g W(D-z) d z=\frac{1}{2} ho g W D^{2}\end{array}.[/latex] (E1.6.1)

Insertion of the numerical values gives:

[latex]F_{x}=\frac{1}{2} imes 1,026 imes 9.81 imes 50 imes 8.0^{2}=1.61 imes 10^{7} \mathrm{~N} .[/latex] (E1.6.2)

Thus, the total force is considerable---about [latex]3.62 imes 10^{6} \mathrm{lb}_{\mathrm{f}}[/latex].

Question 1.6: Hydrostatic Force on a Curved Surface A submarine, whose hul...

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