To check stability, a weight of 40 tons is...

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Question 1 of 2066636eec28f7522a1c51c0cc

To check stability, a weight of 40 tons is lifted with the jumbo boom, whose head is 40 feet from the ship's centerline. The clinometer shows a list of 6.5° with the weight suspended. Displacement including weight is 16,000 tons. What would be the GM while in this condition?

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Question 1 of 20

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🔍 Key Concepts

• Use the transverse metacentric height (GM) formula for a ship heeled by a known transverse moment: ( GM = \dfrac{w \times d}{\Delta \times \tan \phi} ) • Convert the heel angle in degrees to radians or use tan(°) correctly, and keep all units consistent (feet, tons) • Understand that displacement (Δ) must include the lifted weight, and that the horizontal distance is from the ship's centerline to the line of action of the lifted weight

💭 Think About

• What is the heeling moment created by the jumbo boom when it lifts a 40‑ton weight 40 feet off the centerline? How do you express that in ton‑feet? • How does the observed list angle (6.5°) relate to the ratio ( \dfrac{w \times d}{\Delta \times GM} )? Rearrange this relationship to solve for GM. • After you compute GM using your calculator, which option is closest to your numerical result (pay attention to rounding to two decimal places)?

✅ Before You Answer

• Be sure you are using tan(6.5°) (degrees mode), not sine or cosine, and not radians unless you convert properly. • Confirm that displacement Δ = 16,000 tons already includes the 40‑ton weight; do NOT add it again. • Double‑check the moment calculation: 40 tons × 40 feet = ? ton‑feet, and plug carefully into ( GM = \dfrac{w d}{\Delta \tan \phi} ).

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