Problem Statement
The 0.9m diameter cylinder in the figure is 7m long and rests in static equilibrium against a frictionless wall at point B. Compute the specific gravity of the cylinder.
Solution
1. Vertical Force \( F_{V} \)
Vertical force is the weight of the water volume vertically above ADBECA:
\( F_{V} = \gamma \cdot \text{Volume}_{AOEC} + \text{Volume}_{BOE} + \text{Volume}_{ADBO} \)
Substitute the values:
\( F_{V} = 9810 \cdot \left( \frac{1}{2} \pi \cdot (0.45)^2 \cdot 7 + \frac{1}{4} \pi \cdot (0.45)^2 \cdot 7 + 0.45 \cdot 0.45 \cdot 7 \right) \)
Calculation:
\( F_{V} = 9810 \cdot (2.23 + 1.11 + 1.42) = 46670 \, \text{N (up)} \)
2. Weight of Cylinder (\( W \))
The weight of the cylinder is equal to the vertical force:
\( W = F_{V} = 46670 \, \text{N} \)
3. Specific Gravity of the Cylinder
Using the equation:
\( \gamma_{\text{cyl}} \cdot \text{Volume}_{\text{cyl}} = 46670 \)
Substitute \( \text{Volume}_{\text{cyl}} = \pi \cdot (0.45)^2 \cdot 7 \):
\( \gamma_{\text{cyl}} = \frac{46670}{\pi \cdot (0.45)^2 \cdot 7} = 10480 \, \text{N/m}^3 \)
Finally, compute the specific gravity:
\( \text{Sp Gr}_{\text{cyl}} = \frac{\gamma_{\text{cyl}}}{\gamma} = \frac{10480}{9810} = 1.07 \)
Result:
- Specific Gravity of the Cylinder: \( 1.07 \)
