Question 4973Deck General70% to pass
A jack-up 210 feet in length is level during transit. The LCF is 140 feet aft of the bow. How much weight should be applied at the stern to re-level the jack-up if 75 kips is applied at the bow?
AI Explanation
The correct answer is D) 150 kips.
To re-level the jack-up, the weight applied at the stern needs to be equal to the moment created by the 75 kips applied at the bow. Since the LCF is 140 feet aft of the bow, the moment created by the 75 kips at the bow is 75 x 140 = 10,500 kip-feet. To counteract this moment, the weight applied at the stern needs to create an equal and opposite moment, which is 10,500 kip-feet. Dividing this moment by the 210-foot length of the jack-up gives the required weight at the stern of 10,500 / 210 = 150 kips.
The other options are incorrect because they do not provide the necessary moment to counteract the 10,500 kip-feet created by the 75 kips at the bow.
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