Referring to Fig. 3-12, determine the horizontal and vertical forces due to the water acting on the 6-ft-diameter cylinder per foot of its length.

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(a) Net [latex]F_{H} [/latex] = force on C D A — force on AB. Using the vertical projection C D A and of A B,

[latex]F_{H} [/latex] (C D A) = (62.4)(4 + 2.56)(5.12 x 1) = 2090 Ib to the right

[latex]F_{H} [/latex] (A B) = (62.4)(4 + 4.68)(0.88 x 1) = 477 Ib to the left

Net [latex]F_{H} [/latex] = 2090 - 477 = 1613 Ib to the right.

(b) Net [latex]F_{V} [/latex] = upward force on D A B — downward force on D C

= weight of (volume D A B F E D - volume D C G E D ).

The hatched area (volume) is contained in each of the above volumes, one force being upward and the other downward. Thus they cancel, and

net [latex]F_{V} [/latex] = weight of volume D A B F G C D

Dividing this volume into convenient geometric shapes,

net [latex]F_{V} [/latex] = weight of (rectangle G F J C + triangle C J B + semicircle C D A B)

[latex]=62.4\left[(4 imes 4.24)+\left(\frac{1}{2} imes 4.24 imes 4.24 ight)+\left(\frac{1}{2}\pi 3^2 ight) ight](1)=2500 [/latex] Ib upward

If it is desired to locate this resultant vertical component, the principle of moments is employed. Each part of the 2500-lb resultant acts through the center of gravity of the volume it represents. By static mechanics, the centers of gravity are found and the moment equation is written (see Problems 3.7 and 3.9).

Question 3.13: Referring to Fig. 3-12, determine the horizontal and vertica...

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