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Strategy for (a)
To find the buoyant force, we must find the weight of water displaced. We can do this by using the densities of water and steel given in Table [insert table #] We note that, since the steel is completely submerged, its volume and the water’s volume are the same. Once we know the volume of water, we can find its mass and weight
First, we use the definition of density to find the steel’s volume, and then we substitute values for mass and density. This gives :
Because the steel is completely submerged, this is also the volume of water displaced, Vw. We can now find the mass of water displaced from the relationship between its volume and density, both of which are known. This gives:
By Archimedes’ principle, the weight of water displaced is m w g , so the buoyant force is:
The steel’s weight is 9.80×10 7 N , which is much greater than the buoyant force, so the steel will remain submerged.
Strategy for (b)
Here we are given the maximum volume of water the steel boat can displace. The buoyant force is the weight of this volume of water.
The mass of water displaced is found from its relationship to density and volume, both of which are known. That is:
The maximum buoyant force is the weight of this much water, or
Discussion
The maximum buoyant force is ten times the weight of the steel, meaning the ship can carry a load nine times its own weight without sinking.
Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/
For the root mean square value of the ripple voltage, the calculation is more involved as the shape of the ripple waveform has a bearing on the result. ... more
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(a) Calculate the buoyant force on 10,000 metric tons (1.00×10 7 kg) of solid steel completely submerged in water, and compare this with the steel’s weight.
(b) What is the maximum buoyant force that water could exert on this same steel if it were shaped into a boat that could displace 1.00×10 5 m 3 of water?