(a) Determine the magnitude, direction, and line of action of the horizontal and vertical components of the hydrostatic force acting on curved surface AB shown in Fig. P2.70, which has a radius of 2 m and a width of 4 m.

(b) Assume that the water exists on the opposite side of the vertical barrier (the surface AB remains as shown with point A 8 m below the surface level). Determine the information requested in part (a).

Question

(a) Determine the magnitude, direction, and line of action of the horizontal and vertical components of the hydrostatic force acting on curved surface AB shown in Fig. P2.70, which has a radius of 2 m and a width of 4 m.

(b) Assume that the water exists on the opposite side of the vertical barrier (the surface AB remains as shown with point A 8 m below the surface level). Determine the information requested in part (a).

Accepted Answer

A free-body-diagram of the volume of water in the vicinity of the surface is shown.
Force balances in the horizontal and vertical directions give:

[latex]F_H= F_2[/latex]

[latex]F_V= W +F_1[/latex]

where [latex]F_H[/latex] and [latex]F_V[/latex] Fare the horizontal and vertical components of the force acting on the water by the surface AB. Hence,

[latex]F_H=F_2=(9.81 kN/m^3) (8 +1) (2× 4) = 706.3 kN[/latex]

The line of action of [latex]F_H[/latex] is the same as that of [latex]F_2[/latex]. Its distance from the surface is

[latex]y_p=\overline{y}+\frac{\overline{I} }{\overline{y}A }=9+\frac{4(2^3)/12}{9 imes 8}= 9.037 [/latex] m

To find [latex]F_V[/latex] we find W and [latex]F_1[/latex]:

W=γV = [latex](9.81 kN/m^3 ) \left[2 imes 2-\frac{\pi }{4}(2^2) ight] imes 4=33.7 [/latex] KN

[latex]F_1 = 9.81 kN/m^3 (8×2×4) = 628 [/latex] KN

∴ [latex]F_V= F_1+W = 33.7 + 628 = 662 [/latex] KN

To find the line of action of [latex]F_V[/latex], we take moments at point A:

[latex]F_V imes x_V=F_1 imes d_1+W imes d_2 [/latex]

where [latex]d_1 [/latex] = 1 m, and [latex] d_2 =\frac{2R}{3(4-\pi )} =\frac{2 imes 2}{3(4-\pi )}=1.553 [/latex] m:

∴ [latex] x_V=\frac{F_1 imes d_1+W imes d_2}{F_V}=\frac{628 imes 1+ 33.7 imes 1.553}{662} =1.028 [/latex] m

Finally, the forces [latex]F_H[/latex] and [latex]F_V[/latex] that act on the surface AB are equal and opposite to those calculated above. So, on the surface, [latex]F_H[/latex] acts to the right and [latex]F_V[/latex] acts downward.

Question 2.70: (a) Determine the magnitude, direction, and line of action o...

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