Find the volume of the water displaced and position of centre of buoyancy for a wooden block of width 2.5 m and of depth 1.5 m, when it floats horizontally in water. The density of wooden block is 650 kg/m³ and its length is 6.0 m.

Floating Wooden Block Buoyancy Problem

Problem Statement

Given Data

Block Width = 2.5 m
Block Depth (Total Height) = 1.5 m
Block Length = 6.0 m
Density of wood, $ \rho_{wood} = 650 \, \text{kg/m}^3 $
Density of water, $ \rho_{water} = 1000 \, \text{kg/m}^3 $

Diagram of the Floating Block

Diagram of a wooden block floating in water

Solution

According to the principle of flotation, the weight of the floating body is equal to the weight of the fluid it displaces.

1. Weight of the Wooden Block

First, calculate the total volume of the block.

$$ V_{block} = \text{Width} \times \text{Depth} \times \text{Length} $$ $$ V_{block} = 2.5 \times 1.5 \times 6.0 $$ $$ V_{block} = 22.5 \, \text{m}^3 $$

Next, calculate the weight of the block.

$$ W_{block} = \rho_{wood} \times g \times V_{block} $$ $$ W_{block} = 650 \times 9.81 \times 22.5 $$ $$ W_{block} \approx 143471 \, \text{N} $$

(i) Volume of Water Displaced

For equilibrium, the weight of water displaced is equal to the weight of the block.

$$ W_{displaced} = W_{block} \approx 143471 \, \text{N} $$

The volume of water displaced is this weight divided by the weight density of water.

$$ V_{displaced} = \frac{W_{displaced}}{\rho_{water} \times g} $$ $$ V_{displaced} = \frac{143471}{1000 \times 9.81} $$ $$ V_{displaced} = 14.625 \, \text{m}^3 $$

(ii) Position of Centre of Buoyancy

The centre of buoyancy is the centroid of the displaced volume. To find it, we first need to find the depth of submergence ($h$).

$$ V_{displaced} = \text{Width} \times h \times \text{Length} $$ $$ 14.625 = 2.5 \times h \times 6.0 $$ $$ 14.625 = 15 \times h $$ $$ h = \frac{14.625}{15} $$ $$ h = 0.975 \, \text{m} $$

The centre of buoyancy (CB) is at half of this submerged depth, measured from the base of the block.

$$ \text{Position of CB} = \frac{h}{2} $$ $$ \text{Position of CB} = \frac{0.975}{2} $$ $$ \text{Position of CB} = 0.4875 \, \text{m} $$

Final Results:

(i) Volume of water displaced: $ 14.625 \, \text{m}^3 $.

(ii) Position of Centre of Buoyancy: $ 0.4875 \, \text{m} $ from the base.

Explanation of Concepts

Archimedes' Principle of Flotation: This principle states that a floating object displaces a weight of fluid equal to its own weight. This is the key to solving the first part of the problem. We calculate the block's total weight, and this immediately tells us the weight of the water it must displace to float.

Centre of Buoyancy (CB): The buoyant force acts upwards through a single point called the centre of buoyancy. This point is the geometric center (centroid) of the *displaced volume of fluid*. For a simple rectangular shape like the submerged portion of this block, the centroid is located at half its height, half its width, and half its length. The vertical position is therefore at half the depth of submergence ($h/2$).

Find the volume of the water displaced and position of centre of buoyancy for a wooden block of width 2.5 m and of depth 1.5 m, when it floats horizontally in water. The density of wooden block is 650 kg/m³ and its length is 6.0 m.

Problem Statement

Given Data

Diagram of the Floating Block

Solution

1. Weight of the Wooden Block

(i) Volume of Water Displaced

(ii) Position of Centre of Buoyancy

Explanation of Concepts

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