Question 2.4: A circular viewing port is to be located 1.5 ft belowthe sur...
A circular viewing port is to be located 1.5 ft belowthe surface of a tank as shown in Figure 2.6. the magnitude and location of the force acting Find on the window.
The force on the window is
F=\rho g \sin \alpha A \bar{\eta}where
a = π/2 and \bar{\eta}=1.5 \mathrm{ft};
the force is
F=\rho g A \bar{\eta}=\frac{\left(62.4 \mathrm{lb}_{\mathrm{m}} / \mathrm{ft}^{3}\right)\left(32.2 \mathrm{ft} / \mathrm{s}^{2}\right)\left(\pi / 4 \mathrm{ft}^{2}\right)(1.5 \mathrm{ft})}{32.2 \mathrm{lb}_{\mathrm{m}} \mathrm{ft} / \mathrm{s}^{2} \mathrm{lb}_{\mathrm{f}}}=73.5 \mathrm{lb}_{\mathrm{f}}(327 \mathrm{~N})
The force F acts at \bar{\eta}+\frac{I_{\text {centroid }}}{A \bar{\eta}}. For a circular area, I_{\text {centroid }}=\pi R^{4} / 4, so we obtain
\eta_{\text {c.p. }}=1.5+\frac{\pi R^{4}}{4 \pi R^{2} 1.5} \mathrm{ft}=1.542 \mathrm{ft}Related Answered Questions
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