A cylinder 3 m in diameter and 4 m long retains water on one side and is supported as shown. Determine the horizontal reaction at A and the vertical reaction at B. The cylinder weighs 196.2 kN. Ignore friction.

Diagram of Supported Cylinder

Solution

1. Horizontal Force from Water (\(F_x\))

The horizontal force is the pressure on the vertically projected area (BOC).

$$ A_{proj} = \text{Diameter} \times \text{Length} $$ $$ = 3 \, \text{m} \times 4 \, \text{m} = 12 \, \text{m}^2 $$

The depth of the centroid of this projected area is:

$$ \bar{h} = \frac{\text{Diameter}}{2} = \frac{3}{2} = 1.5 \, \text{m} $$

The horizontal force is:

$$ F_x = \rho g A_{proj} \bar{h} $$ $$ = 1000 \times 9.81 \times 12 \times 1.5 $$ $$ = 176580 \, \text{N} $$

2. Vertical Force from Water (\(F_y\))

The vertical force is the weight of the water in the volume of the semi-cylinder BDCOB. This creates an upward buoyant force.

$$ \text{Volume (V)} = \text{Area of Semicircle} \times \text{Length} $$ $$ = \left( \frac{\pi R^2}{2} \right) \times L $$ $$ = \left( \frac{\pi (1.5)^2}{2} \right) \times 4 $$ $$ \approx 14.137 \, \text{m}^3 $$

The upward vertical force is the weight of this water volume:

$$ F_y = \rho g V $$ $$ = 1000 \times 9.81 \times 14.137 $$ $$ \approx 138684 \, \text{N} $$

3. Reaction Forces (Equilibrium)

For the cylinder to be in equilibrium, the support reactions must balance the applied forces.

Horizontal Reaction at A (\(R_A\)):

The horizontal reaction at support A must balance the horizontal force from the water.

$$ \sum F_x = 0 $$ $$ R_A - F_x = 0 $$ $$ R_A = F_x = 176580 \, \text{N} $$

Vertical Reaction at B (\(R_B\)):

The vertical reaction at support B, combined with the upward buoyant force from the water, must balance the downward weight of the cylinder.

$$ \sum F_y = 0 $$ $$ R_B + F_y - W = 0 $$ $$ R_B = W - F_y $$ $$ R_B = 196200 - 138684 = 57516 \, \text{N} $$

Explanation of Concepts

Equilibrium of Bodies: For an object to be stationary (in equilibrium), the sum of all forces acting on it in any direction must be zero (\(\sum F = 0\)). In this problem, we apply this principle separately to the horizontal and vertical directions.

• Horizontal Equilibrium: The push from the water (\(F_x\)) to the right must be balanced by the push from the support at A (\(R_A\)) to the left.
• Vertical Equilibrium: The downward force (Weight of the cylinder, \(W\)) must be balanced by the sum of all upward forces (the buoyant force from the water, \(F_y\), and the support reaction at B, \(R_B\)).