So we all know a 1000lb steel ball will sink in water, but if you were to leave the center of the ball hollow and fill with air, if the steel ball was big enough in diameter it should float correct? What size would the steel ball need to be to float?
convert lb to kg = 454 kg = 454 litres = 454,000 cm2 volume (454,000) of a sphere = 4/3 pi * r ^3 calculate r double it to get the diameter of the sphere in cm
sure, if the volume of air is about 7 times the volume of steel, the specific weight of the steel/air system would be below 1.0 and it will float. How do you think ships float?
specific gravity of a steel is approximately 7.82 as compared to water. so, the hollow sphere(filled with air) is having volume more by 7.82 as compared to solid sphere, then the 1000 lb hollow steel ball float in the water (Why Do Ships Float? The Greek Mathematician and inventor Archimedes lived during the 3rd century BC. According to history he was in the bath one day when he discovered the principle of buoyancy which is the reason why huge Greek ships weighing thousands of pounds could float on water. He noticed that as he lowered himself into the bath, the water displaced by his body overflowed the sides and he realised that there was a relationship between his weight and the volume of water displaced. It is said that he ran naked into the street yelling heurEka which is where we get our word eureka! (I found it), Greek heurEka I have found, from heuriskein to find. The Buoyancy Principle: Archimedes continued to do more experiments and came up with a buoyancy principle, that a ship will float when the weight of the water it displaces equals the weight of the ship and anything will float if it is shaped to displace its own weight of water before it reaches the point where it will submerge. This is kind of a technical way of looking at it. A ship that is launched sinks into the sea until the weight of the water it displaces is equal to its own weight. As the ship is loaded, it sinks deeper, displacing more water, and so the magnitude of the buoyant force continuously matches the weight of the ship and its cargo.)
Say The radius of steel ball is R. Its shell is t thick so the air filled cavity if of radius r where r = R - t Steel has a density p The mass of the ball m is given m = 4/3 pi (R^3 - r^3) p If the ball is just to float then the mass dived by the must equal the density fo water d d = m / 43/ pi R^3 Equating m in the above equations yields (r/R)^3 = (p - d)/p or r = c R where c =((p-d)/p))^1/3 Using above leads to m = 4/3 pi R^3 (1 - c^3) p and hence R can be solved for So taking p (density of steel) 490 lb/ft^3 and d (density of water) at 62 lb/ft^3 yields c= 0.9557 Applying the equation above gives R = 1.565 ft or a diameter of 37.55 in The shell thickness is R (1 - c) or 0.06938 ft or 0.8327 in Note I do all my work in metric units so I had to do some conversions on the run. I hope you will check things to make sure the maths is correct. Anyway good luck
