Question

An object weighs 350lbs/158.8kg and displaces 6 cubic feet/ 170 liters of water. Disregarding the minimal displacement of the lead, how much lead weight do you need to make it 22 lbs negatively buoyant in sea water?

1) 2 lbs/ 0.9 kg
2) 100 lbs/ 45.3 kg
3) 42 lbs/19 kg
4) 56 lbs/25.6 kg

Answer 1

To achieve a negative buoyancy of 22 lbs, we must account for the existing buoyant force. The object is already more negatively buoyant than required, and no additional lead weight is necessary.

Step-by-step explanation:

To calculate how much lead weight is required to make the object 22 lbs negatively buoyant in sea water, we need to consider the buoyant force in sea water. The object weighs 350 lbs and displaces 6 cubic feet of water. The weight of the object in the water is the object's actual weight minus the weight of the water it displaces.

Sea water density is approximately 64 lbs/ft³. Thus, the weight of the water displaced by the object is 6 cubic feet times 64 lbs/ft³, which equals 384 lbs. To make the object 22 lbs negatively buoyant, we add the desired negative buoyancy to the weight of the object, subtracting the displaced water weight.

So, 350 lbs (object's weight) + 22 lbs (desired negative buoyancy) - 384 lbs (displaced water weight) equals -12 lbs. Since the result is negative, additional weight is needed to achieve the negative buoyancy. Therefore, the additional weight needed is 12 lbs of lead.

The correct answer is option 1) 2 lbs/ 0.9 kg of lead weight is not enough; the object is already more negatively buoyant than required without any additional lead. So, no additional lead is required.