Question 3.SP.8: Find the horizontal and vertical components of the force exe......
Find the horizontal and vertical components of the force exerted by the fluids on the horizontal cylinder in Fig. S3.8 if (a) the fluid to the left of the cylinder is a gas confined in a closed tank at a pressure of 35.0 \mathrm{kPa} ;(b) the fluid to the left of the cylinder is water with a free surface at an elevation coincident with the uppermost part of the cylinder. Assume in both cases that atmospheric pressure occurs to the right of the cylinder.
The net projection on a vertical plane of the portion of the cylindrical surface under consideration (see left-hand diagram) is, from the right-hand diagram, ef =2+2 \cos 30^{\circ}=3.73 \mathrm{~m} .
(a) For the gas,
F_{x}=p A_{2}=35.0 \mathrm{kN} / \mathrm{m}^{2}(3.73 \mathrm{~m})=130.5 \mathrm{kN} / \mathrm{m}
The vertical force of the gas on the surface a c is equal and opposite to that on the surface c d . Hence the net projection on a horizontal plane for the gas is a f=2 \sin 30^{\circ}=1 \mathrm{~m} . Thus
F_{z}=p A_{x}=35.0 \mathrm{kN} / \mathrm{m}^{2}(1 \mathrm{~m})=35.0 \mathrm{kN} / \mathrm{m} \text { upward }
(b) For the fluid,
Eq. (3.16): \begin{aligned}F_{x} & =\gamma h_{c} A=9.81 \mathrm{kN} / \mathrm{m}^{2}\left(\frac{1}{2} \times 3.73 \mathrm{~m}\right)(3.73 \mathrm{~m}) \\& =68.3 \mathrm{kN} / \mathrm{m} \end{aligned}
F=\gamma h_c A (3.16)
Net F_{z}= upward force on surface c d e – downward force on surface c a
= weight of volume a b c d e f a – weight of volume a b c a
= weight of cross-hatched volume of liquid
=9.81 \mathrm{kN} / \mathrm{m}^{3}\left[\frac{210}{360} \pi 2^{2}+\frac{1}{2}\left(1 \times 2 \cos 30^{\circ}\right)+(1 \times 2)\right] \mathrm{m}^{2}=100.0 \mathrm{kN} / \mathrm{m} upward .