Question:
A ship of length 180 m has a displacement of 14400 tonnes and is floating on an even keel in sea water of density 1025 kg/m\(^3\). The trim changes by 0.18 m when a weight of 120 tonnes that is already onboard, is shifted 24 m forward. The longitudinal metacentric height is .................... m.
A ship of length 180 m has a displacement of 14400 tonnes and is floating on an even keel in sea water of density 1025 kg/m\(^3\). The trim changes by 0.18 m when a weight of 120 tonnes that is already onboard, is shifted 24 m forward. The longitudinal metacentric height is .................... m.
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The formula \( GM_L = \frac{w \times d \times L}{\Delta \times \delta t} \) is a direct and powerful tool for solving weight-shifting and trim problems. Ensure all units are consistent (e.g., weights in tonnes, lengths in meters) before calculating.
Updated On: Aug 30, 2025
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Solution and Explanation
Step 1: Understanding the Concept:
This problem involves calculating the longitudinal metacentric height (\(GM_L\)) of a ship. \(GM_L\) is a key measure of a ship's longitudinal stability, which governs its response to changes in trim. The problem provides data from a weight shifting experiment, which allows for the calculation of \(GM_L\).
Step 2: Key Formula or Approach:
1. Trimming Moment: When a weight \(w\) is shifted a longitudinal distance \(d\), it creates a trimming moment \( M_t = w \times d \).
2. Change in Trim (\(\delta t\)): This trimming moment causes the ship to trim (change its drafts forward and aft). The change in trim is related to the trimming moment and the "Moment to Change Trim by one Centimetre" (MCTC). However, a more direct formula relates the trimming moment to \(GM_L\):
\[ M_t = \Delta \times GM_L \times \tan(\theta) \] where \(\Delta\) is the displacement and \(\theta\) is the angle of trim.
3. Small Angle Approximation: For small angles of trim, \( \tan(\theta) \approx \theta \approx \frac{\delta t}{L} \), where \(\delta t\) is the total change in trim and \(L\) is the length of the ship (Length Between Perpendiculars, LBP, is typically used).
Substituting this into the moment equation gives: \[ w \times d = \Delta \times GM_L \times \frac{\delta t}{L} \] 4. We can rearrange this formula to solve for \(GM_L\).
Step 3: Detailed Calculation:
Given values:
- Length, \(L = 180\) m.
- Displacement, \(\Delta = 14400\) tonnes.
- Shifted weight, \(w = 120\) tonnes.
- Longitudinal shift distance, \(d = 24\) m.
- Change in trim, \(\delta t = 0.18\) m.
Rearrange the formula to solve for \(GM_L\):
\[ GM_L = \frac{w \times d \times L}{\Delta \times \delta t} \] Substitute the given values into the formula: \[ GM_L = \frac{120 \text{ tonnes} \times 24 \text{ m} \times 180 \text{ m}}{14400 \text{ tonnes} \times 0.18 \text{ m}} \] \[ GM_L = \frac{120 \times 24 \times 180}{14400 \times 0.18} \] \[ GM_L = \frac{518400}{2592} \] \[ GM_L = 200 \text{ m} \] Step 4: Final Answer:
The longitudinal metacentric height is 200 m.
This problem involves calculating the longitudinal metacentric height (\(GM_L\)) of a ship. \(GM_L\) is a key measure of a ship's longitudinal stability, which governs its response to changes in trim. The problem provides data from a weight shifting experiment, which allows for the calculation of \(GM_L\).
Step 2: Key Formula or Approach:
1. Trimming Moment: When a weight \(w\) is shifted a longitudinal distance \(d\), it creates a trimming moment \( M_t = w \times d \).
2. Change in Trim (\(\delta t\)): This trimming moment causes the ship to trim (change its drafts forward and aft). The change in trim is related to the trimming moment and the "Moment to Change Trim by one Centimetre" (MCTC). However, a more direct formula relates the trimming moment to \(GM_L\):
\[ M_t = \Delta \times GM_L \times \tan(\theta) \] where \(\Delta\) is the displacement and \(\theta\) is the angle of trim.
3. Small Angle Approximation: For small angles of trim, \( \tan(\theta) \approx \theta \approx \frac{\delta t}{L} \), where \(\delta t\) is the total change in trim and \(L\) is the length of the ship (Length Between Perpendiculars, LBP, is typically used).
Substituting this into the moment equation gives: \[ w \times d = \Delta \times GM_L \times \frac{\delta t}{L} \] 4. We can rearrange this formula to solve for \(GM_L\).
Step 3: Detailed Calculation:
Given values:
- Length, \(L = 180\) m.
- Displacement, \(\Delta = 14400\) tonnes.
- Shifted weight, \(w = 120\) tonnes.
- Longitudinal shift distance, \(d = 24\) m.
- Change in trim, \(\delta t = 0.18\) m.
Rearrange the formula to solve for \(GM_L\):
\[ GM_L = \frac{w \times d \times L}{\Delta \times \delta t} \] Substitute the given values into the formula: \[ GM_L = \frac{120 \text{ tonnes} \times 24 \text{ m} \times 180 \text{ m}}{14400 \text{ tonnes} \times 0.18 \text{ m}} \] \[ GM_L = \frac{120 \times 24 \times 180}{14400 \times 0.18} \] \[ GM_L = \frac{518400}{2592} \] \[ GM_L = 200 \text{ m} \] Step 4: Final Answer:
The longitudinal metacentric height is 200 m.
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