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Zorluk: OrtaBuoyancy and Archimedes' Principle

A salvage crane is lifting a solid brass propeller from the bottom of a freshwater lake. The propeller has a true weight of 1,200 N1,200 \text{ N} in the air. When it is completely submerged in the water, it displaces 35 liters35 \text{ liters} of water.

If 1 liter1 \text{ liter} of water weighs exactly 10 N10 \text{ N}, what is the apparent weight of the propeller, in Newtons, while it is still completely underwater?

Cevap: 850 N

Cevap

850
The propeller experiences an upward buoyant force equal to the weight of the water it displaces. Since it displaces 35 liters35 \text{ liters} and each liter weighs 10 N10 \text{ N}, the buoyant force is 350 N350 \text{ N}. The apparent weight is the true weight (1,200 N1,200 \text{ N}) minus this buoyant force (350 N350 \text{ N}), resulting in 850 N850 \text{ N}.

Adım Adım Çözüm

1
Calculate the weight of the displaced water to find the buoyant force.
Buoyant force = 350 N350 \text{ N}
According to Archimedes' Principle, the buoyant force on an object is equal to the weight of the fluid it displaces. Here, 35 liters×10 N/liter=350 N35 \text{ liters} \times 10 \text{ N/liter} = 350 \text{ N}.
2
Subtract the buoyant force from the true weight to find the apparent weight.
Apparent weight = 850 N850 \text{ N}
The apparent weight of a submerged object is its actual weight in air minus the upward buoyant force exerted by the fluid (1,200 N−350 N=850 N1,200 \text{ N} - 350 \text{ N} = 850 \text{ N}).

Anahtar Kavram

Archimedes' principle states that the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object. The apparent weight of a submerged object is its true weight minus the buoyant force.
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Buoyancy and Archimedes' Principle

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