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.

Cylinder Reaction Force Problem

Problem Statement

Given Data

Diameter of Cylinder, $ D = 3 \, \text{m} $ (Radius, $ R = 1.5 \, \text{m} $)
Length of Cylinder, $ L = 4 \, \text{m} $
Weight of Cylinder, $ W = 196.2 \, \text{kN} = 196200 \, \text{N} $
Density of Water, $ \rho = 1000 \, \text{kg/m}^3 $
Acceleration due to Gravity, $ g = 9.81 \, \text{m/s}^2 $

Diagram of Supported Cylinder

Diagram of the supported cylinder retaining water

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.

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

Final Result:

The horizontal reaction at A is $ R_A = 176580 \, \text{N} $ or $ 176.58 \, \text{kN} $.

The vertical reaction at B is $ R_B = 57516 \, \text{N} $ or $ 57.52 \, \text{kN} $.

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, $R_B$).

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.

Problem Statement

Given Data

Diagram of Supported Cylinder

Solution

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

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

3. Reaction Forces (Equilibrium)

Explanation of Concepts

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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.

Problem Statement

Given Data

Diagram of Supported Cylinder

Solution

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

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

3. Reaction Forces (Equilibrium)

Explanation of Concepts

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