Question

A ship with a displacement of 10,000 tonnes has the center of gravity at 4 m above the keel and 1.5 m forward of midship. If 2,000 tonnes of cargo is placed at 10 m above the keel and 1.5 m aft of midship, then the new position of the center of gravity is:

• A. \( 5 \, \text{m above the keel and 1 m aft of midship} \)
• B. \( 6 \, \text{m above the keel and 1 m forward of midship} \)
• C. \( 6 \, \text{m above the keel and 1 m aft of midship} \)
• D. \( 5 \, \text{m above the keel and 1 m forward of midship} \)

Hint:

To calculate the new center of gravity after adding or shifting weights, use the weighted average formula for both vertical and longitudinal positions. Ensure the directions (forward, aft, above) are accounted for correctly.

Answer 1

The Correct Option is D

Step 1: Determine the vertical position of the new center of gravity (\( KG_{\text{new}} \)).
The new vertical center of gravity is calculated using the formula: \[ KG_{\text{new}} = \frac{(W_{\text{ship}} \cdot KG_{\text{ship}}) + (W_{\text{cargo}} \cdot KG_{\text{cargo}})}{W_{\text{total}}}, \] where: - \( W_{\text{ship}} = 10000 \, \text{tonnes} \), - \( KG_{\text{ship}} = 4 \, \text{m} \), - \( W_{\text{cargo}} = 2000 \, \text{tonnes} \), - \( KG_{\text{cargo}} = 10 \, \text{m} \), - \( W_{\text{total}} = W_{\text{ship}} + W_{\text{cargo}} = 12000 \, \text{tonnes}. \) Substitute the values: \[ KG_{\text{new}} = \frac{(10000 \cdot 4) + (2000 \cdot 10)}{12000}. \] Simplify: \[ KG_{\text{new}} = \frac{40000 + 20000}{12000} = \frac{60000}{12000} = 5 \, \text{m}. \] Step 2: Determine the longitudinal position of the new center of gravity (\( LCG_{\text{new}} \)).
The new longitudinal center of gravity is calculated using the formula: \[ LCG_{\text{new}} = \frac{(W_{\text{ship}} \cdot LCG_{\text{ship}}) + (W_{\text{cargo}} \cdot LCG_{\text{cargo}})}{W_{\text{total}}}, \] where: - \( LCG_{\text{ship}} = 1.5 \, \text{m forward of midship} \), - \( LCG_{\text{cargo}} = 1.5 \, \text{m aft of midship} \) (negative direction). Substitute the values: \[ LCG_{\text{new}} = \frac{(10000 \cdot 1.5) + (2000 \cdot -1.5)}{12000}. \] Simplify: \[ LCG_{\text{new}} = \frac{15000 - 3000}{12000} = \frac{12000}{12000} = 1 \, \text{m forward of midship}. \] Conclusion: The new position of the center of gravity is \( 5 \, \text{m above the keel and 1 m forward of midship} \).