Imagine we shape 0.5x103 kg of solid steel into a small boat model which can displace 0.8 m3 of water at its maximum capacity. What is the additional buoyant force we get from this ship compared to the buoyant force we would get by submerging this chunk of solid steel directly into water, in unit of N? ρ(steel) = 7.8x103kg/m3. ρ(water) = 1.0x103kg/m3. Use g = 9.8 m/s2.
If it can displace 0.8 m3 of water by archimedes principle buoyancy = rho_water*V*g = 1000*0.8*9.8 = 78400 N
in only steel was submerged
volume of steel = mass/density = 500/7800 = 0.0641
buoyant force = rho_water*V*g = 1000*0.0641*9.8 = 628 N
Thus additional buoyancy = 78400 - 628 = 77772 N
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Imagine we shape 0.5x103 kg of solid steel into a small boat model which can displace...
Imagine we shape 0.1x103 kg of solid steel into a small boat model which can displace 0.1 m3 of water at its maximum capacity. What is the additional buoyant force we get from this ship compared to the buoyant force we would get by submerging this chunk of solid steel directly into water, in unit of N? ρ(steel) = 7.8x103kg/m3. ρ(water) = 1.0x103kg/m3. Use g = 9.8 m/s2