Problem Statement
The 1.8 m diameter cylinder in the figure weighs 100000 N and is 1.5 m long. Determine the reactions at A and B, neglecting friction.
Solution
1. Horizontal Force
The horizontal force is calculated as:
\( F_H = \gamma \cdot A \cdot \bar{y} \)
\( F_H = 0.8 \times 9810 \times (1.8 \times 1.5) \times \frac{1.8}{2} \)
\( F_H = 19071 \, \text{N} = 19.071 \, \text{kN} \) (right)
2. Vertical Force
The vertical force is equal to the weight of the volume of water vertically above BDC:
\( F_V = \gamma \cdot \text{Volume}_{BCD} \)
Calculating the volume:
\( F_V = 0.8 \times 9810 \times \left( \frac{\pi \times 0.9^2}{2} \times 1.5 \right) \)
\( F_V = 14978 \, \text{N} = 14.978 \, \text{kN} \) (upward)
3. Reactions at A and B
The reaction at A is equal to the horizontal force:
\( R_A = F_H = 19071 \, \text{N} \) (left)
The reaction at B is calculated as:
\( R_B = \text{Weight of cylinder} – F_V \)
\( R_B = 100000 – 14978 = 85022 \, \text{N} = 85.022 \, \text{kN} \) (upward)
Results:
- Reaction at A: \( R_A = 19.071 \, \text{kN} \) (left)
- Reaction at B: \( R_B = 85.022 \, \text{kN} \) (upward)
Explanation
- Horizontal Force: The horizontal reaction at A balances the hydrostatic force acting on the curved surface of the cylinder.
- Vertical Force: The vertical reaction at B accounts for the weight of the cylinder minus the buoyant force due to the displaced water volume.
Physical Meaning
This problem demonstrates the principles of hydrostatics and buoyancy, showing how forces and reactions are distributed in systems involving submerged or partially submerged bodies.
