The steel cylinder with a cylindrical hole is connected to the copper cone. Find the center of gravity of the assembly. The weight densities of steel and copper are 489 lb/ft3 and 556 lb/ft3, respectively.

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ELEMENT 1:- CYLINDER\text { Volume } \begin{aligned}V_{1} & =\frac{\pi}{4} \times 2.5^{2} \times 8 \\V_{1} & =39.25 \text { in }^{3}\end{aligned}CENTER OF GRAVITY CO-ORDINATES W.R-T. GIVEN REFERENCE\begin{array}{l}y_{1}=\frac{8}{2}=4 \text { in } \\y_{1}=0 \text { in } \\x_{1}=0 \text { in }\end{array}ELEMENT2:- CYLINDER HOLE\text { Vocume } \begin{aligned}v_{2} & =\frac{\pi}{4} \times 6 \times 1.5^{2} \\v_{2} & =10.5975 \text { in }^{3}\end{aligned}\begin{array}{r}z_{2}=\frac{6}{2}=\sin \\y_{2}=0 ; x_{2}=0\end{array}ELEMENT 3:- RIGHT CIRCLUAR CONE\begin{array}{l}v_{3}=\frac{\pi \times 2.5^{2} \times 6}{12}=9.8125 \mathrm{in}^{3} \\z_{3}=8+\frac{6}{4}=9.5 \mathrm{in} \\y_{3}=0 ; x_{3}=0\end{array}GIVENDENSITY of\begin{array}{l}\text { STEL } C_{S}=489 \mathrm{lb} / \mathrm{ft}^{3} \\\text { CoppER } P_{C}=556 \mathrm{lb} / \mathrm{ft} \mathrm{t}^{3}\end{array}THERE FORE CENTRE OF GRAVITY of ASSEMBLY\begin{array}{l}\bar{z}=\frac{\rho_{s} v_{1} z_{1}-\rho_{s} v_{2} z_{2}+e_{c} v_{3} z_{3}}{\rho_{s} v_{1}-\rho_{s} v_{2}+\rho_{c} v_{3}} \\\bar{z}=\frac{(76773-15546.5325+51829.625)}{5455.75)} \\=(19193.25-5182.1775+ \\\bar{z}=5.808 \text { in } \\\bar{x}=0 \\\bar{y}=0\end{array} ...