A body of dimensions 1.5 m x 1.0 m x 2 m, weighs 1962 N in water. Find its weight in air. What will be its specific gravity?

Body Buoyancy Problem

Problem Statement

Given Data

Body Dimensions: 1.5 m x 1.0 m x 2.0 m
Weight in water, $ W_{water} = 1962 \, \text{N} $
Fluid is water, $ \rho_{water} = 1000 \, \text{kg/m}^3 $

Solution

1. Buoyant Force

First, calculate the volume of the body, which is equal to the volume of water it displaces when fully submerged.

$$ V_{body} = 1.5 \times 1.0 \times 2.0 $$ $$ V_{body} = 3.0 \, \text{m}^3 $$

Next, calculate the weight of the displaced water, which is the buoyant force acting on the body.

$$ \text{Buoyant Force } (F_B) = \rho_{water} \times g \times V_{body} $$ $$ F_B = 1000 \times 9.81 \times 3.0 $$ $$ F_B = 29430 \, \text{N} $$

(i) Weight in Air

The apparent weight of the body in water is its true weight in air minus the buoyant force.

$$ W_{water} = W_{air} - F_B $$ $$ 1962 = W_{air} - 29430 $$ $$ W_{air} = 1962 + 29430 $$ $$ W_{air} = 31392 \, \text{N} $$

(ii) Specific Gravity

First, find the density of the body. We can get its mass from its weight in air.

$$ M_{body} = \frac{W_{air}}{g} $$ $$ M_{body} = \frac{31392}{9.81} $$ $$ M_{body} = 3200 \, \text{kg} $$

$$ \rho_{body} = \frac{M_{body}}{V_{body}} $$ $$ \rho_{body} = \frac{3200}{3.0} $$ $$ \rho_{body} \approx 1066.67 \, \text{kg/m}^3 $$

Specific gravity is the ratio of the body's density to the density of water.

$$ S.G._{body} = \frac{\rho_{body}}{\rho_{water}} $$ $$ S.G._{body} = \frac{1066.67}{1000} $$ $$ S.G._{body} \approx 1.067 $$

Final Results:

Weight in air: $ 31392 \, \text{N} $ or $ \approx 31.39 \, \text{kN} $.

Specific gravity of the body: $ \approx 1.067 $.

Explanation of Concepts

Archimedes' Principle: This principle states that the buoyant force on a submerged object is equal to the weight of the fluid it displaces. This buoyant force is also the reason for the apparent loss of weight when the object is submerged.

Apparent Weight: An object's weight in a fluid ($W_{fluid}$) is its true weight in air ($W_{air}$) minus the upward buoyant force ($F_B$). By rearranging this relationship ($W_{air} = W_{fluid} + F_B$), we can find the object's true weight.

Specific Gravity: Specific gravity is a dimensionless quantity that compares the density of a substance to the density of a reference substance (usually water). A specific gravity of 1.067 means the body is slightly denser than water, which is consistent with the fact that it has a positive weight when submerged (i.e., it sinks).

A body of dimensions 1.5 m x 1.0 m x 2 m, weighs 1962 N in water. Find its weight in air. What will be its specific gravity?

Problem Statement

Given Data

Solution

1. Buoyant Force

(i) Weight in Air

(ii) Specific Gravity

Explanation of Concepts

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