# Notes - Force $F_{R}=p_{C}A$ acts through the center of pressure, $p_{c}$ is calculated using the centroid - When calculating $p_{C}=\rho gh,$ use $h=H-\bar{y}$ where $\bar{y}$ is centroid height (from bottom presumably) - Reminder that $\rho_{x}=SG_{x}\rho_{H_{2}O}$ - Buoyancy force $F_{B}=\rho gV_{disp}$ where $V_{disp}$ is the fluid displaced by the object ([[ME 308 HW 02 Solutions#Problem 4|problem 4]]) - For curved surfaces, $F_{V}=\rho gV_{eff}$ where $V$ is the imaginary fluid volume above the wetted curved surface up to the free surface ([[ME 308 HW 02 Solutions#Problem 5|problem 5]]) - Horizontal force on curved surface $F_{H}=p_{C}A_{proj}$ where $A_{proj}$ the projected area of the curved surface onto a flat plane perpendicular to the force direction ([[ME 308 HW 02 Solutions#Problem 5|problem 5]]) # [[ME 308 HW 02.pdf#page=1|Problem 1]] $ F_{R}=p_{C}A $ $ \bar{y}=\frac{4R}{3\pi}=1.27m\to y_{C}=H-\bar{y}=6.73m $ $ p_{C}= \rho gy_{C}=65954 \frac{N}{m^2}\to \underline{ F_{R}=p_{C}A=932kN} $ $ y'=y_{C}+\frac{I_{xc}}{Ay_{C}}=y_{C}+ \frac{.1098R^4}{Ay_{C}}=6.82m $ $ \sum M_{B}=-F_{R}(H-y')+F_{A}(R)=0\to \boxed{ F_{A}=\frac{F_{R}h'}{R}=367kN } $ # [[ME 308 HW 02.pdf#page=2|Problem 2]] $ F_{R}=p_{C}A=(p_{G}+\rho_{oil}gy_{C})A\to y_{C}=d+\frac{H}{2}=\underline{2.3m} $ $ =\left[ p_{G}+SG_{oil}\rho_{H_{2}O}gy_{C} \right](LH)=\boxed{ 25.31kN } $ $ y'=y_{C}+\frac{\rho_{oil}gI_{xc}}{F_{R}}=y_{C}+\frac{SG_{oil}\rho_{H_{2}O}g}{F_{R}}\left( \frac{1}{12}LH^3 \right)=\boxed{ 2.304m\text{ below surface} } $ # [[ME 308 HW 02.pdf#page=3|Problem 3]] $ F_{V}=\rho gV\to \frac{F_{V}}{b}=\rho gA $ $ x=R\cos\theta\quad y=R\sin\theta $ $ h(\theta)=d-R\sin\theta\quad dA=h(\theta)dx=h(\theta)(-R\sin\theta)d\theta $ $ |A|=\Biggr|\int_{- \pi /9} ^{19\pi/9}(d-R\sin\theta)(-R\sin\theta)d\theta\Biggr|=10.02m^2 $ $ F_{V}=\rho g|A|=\boxed{ -103.21kN } $ # [[ME 308 HW 02.pdf#page=4|Problem 4]] Single balloon has 3 forces acting on it: buoyancy, weight of helium, weight of balloon -> equivalent to net upward force, $F_{L}$ $ V=\frac{4}{3}\pi r^3 $ $ F_{L}=g(\rho_{air}V-\rho_{He}V-m_{b})=3.28N $ $ w_{m}=mg=784.6N $ $ n=\frac{w_{m}}{F_{L}}=\boxed{ 240\text{ balloons} } $ # [[ME 308 HW 02.pdf#page=5|Problem 5]] $ F_{b}=\rho gV_{disp}=\rho gA_{disp}L,\quad A_{disp}=\frac{3\pi}{4}R^2+R^2 $ $ \sum F_{y}=F_{b}-mg=0\to m=\frac{F_{b}}{g}=\boxed{ \rho\left( \frac{3\pi}{4}+1 \right)R^2L } $ $ F_{H}=p_{C}A_{proj}=\rho gh_{C}A_{proj} $ $ A_{proj}=2RL,\quad h_{C}=R $ $ \sum F_{x}=F_{H}-F_{N}=0\to F_{N}=F_{H}=\boxed{ 2\rho gR^2L } $