Question:
A ship of 3300 tonne displacement is undergoing an inclining experiment in seawater of density 1025 kg/m\(^3\). A mass of 6 tonne is displaced transversely by 12 m as shown in the figure. This results in a 0.12 m deflection of a 11 m long pendulum suspended from the centerline. The transverse metacenter of the ship is located at 7.25 m above the keel.
The distance of the center of gravity from the keel is ________ m (rounded off to two decimal places).
A ship of 3300 tonne displacement is undergoing an inclining experiment in seawater of density 1025 kg/m\(^3\). A mass of 6 tonne is displaced transversely by 12 m as shown in the figure. This results in a 0.12 m deflection of a 11 m long pendulum suspended from the centerline. The transverse metacenter of the ship is located at 7.25 m above the keel.
The distance of the center of gravity from the keel is ________ m (rounded off to two decimal places).
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In inclining experiments, the deflection of the pendulum helps in calculating the distance of the center of gravity by using the transverse metacentric height and the mass displaced.
Updated On: Apr 25, 2025
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Solution and Explanation
Step 1: Understanding the formula for the inclining experiment.
The formula that relates the deflection \( \delta \), the mass displaced \( m \), the length of the pendulum \( L \), the transverse metacentric height \( GM \), and the distance of the center of gravity \( KG \) is: \[ \delta = \frac{w \cdot GM}{m \cdot L} \times KG \] Where:
\( w = {weight of displaced mass} = m_{{displaced}} \times g \),
\( m = {total mass of the ship} = m_{{ship}} \),
\( g = 9.81 \, {m/s}^2 \) (gravitational acceleration).
Rearranging to solve for \( KG \): \[ KG = \frac{\delta \cdot m \cdot L}{w \cdot GM} \] Step 2: Substituting the known values.
Substitute all known values into the formula to calculate \( KG \): \[ KG = \frac{0.12 \cdot (3300 \times 10^3) \cdot 11}{(6 \times 10^3) \cdot 7.25} \] Step 3: Simplifying the expression. Simplify the expression to calculate: \[ KG = \frac{0.12 \cdot 3300 \cdot 11}{6 \cdot 7.25} = \frac{43560}{43.5} = 5.20 \, {m} \] Final Answer: The distance of the center of gravity from the keel is \( \boxed{5.20} \, {m} \).
The formula that relates the deflection \( \delta \), the mass displaced \( m \), the length of the pendulum \( L \), the transverse metacentric height \( GM \), and the distance of the center of gravity \( KG \) is: \[ \delta = \frac{w \cdot GM}{m \cdot L} \times KG \] Where:
\( w = {weight of displaced mass} = m_{{displaced}} \times g \),
\( m = {total mass of the ship} = m_{{ship}} \),
\( g = 9.81 \, {m/s}^2 \) (gravitational acceleration).
Rearranging to solve for \( KG \): \[ KG = \frac{\delta \cdot m \cdot L}{w \cdot GM} \] Step 2: Substituting the known values.
Substitute all known values into the formula to calculate \( KG \): \[ KG = \frac{0.12 \cdot (3300 \times 10^3) \cdot 11}{(6 \times 10^3) \cdot 7.25} \] Step 3: Simplifying the expression. Simplify the expression to calculate: \[ KG = \frac{0.12 \cdot 3300 \cdot 11}{6 \cdot 7.25} = \frac{43560}{43.5} = 5.20 \, {m} \] Final Answer: The distance of the center of gravity from the keel is \( \boxed{5.20} \, {m} \).
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