In Fig. , a rectangular block floats in a liquid . A block of mass m = 0.500 kg and density `rho = 700 kg//m^(3)` floats face down in a liquid of density `rho_(f) = 1200 kg//m^(3)` . What are the magnitude of the buoyant force on the block from the liquid , the weight of the liquid displaced by the block (liquid that would be in the space now occupied by the block ) , and the mass of that displaced liquid ? What fraction of the block's volume is submerged ? What fraction of its height is submerged ?

Explanation

Because the block is floating , the magnitude of the (upward) buoyant force on it must match the magnitude of the (downward) gravitational force on it :
`F_(b) = mg " " (14-22)`
`= (0.500 kg) (9.8 m//s^(2)) = 4.9 N`
Regardless of whether the block is floating , the magnitude of the buoyant force must equal the weight `m_(f) g` of the liquid displaced by the block . Thus , we have `m_(f) g = F_(b) " " (14-23)`
= 4.9 N
The mass of the displaced liquid is then
`m_(f) g = 0.500` kg the same as the block .
To find the fraction of the block.s volume that is submerged , let us combine Eqs. 14-22 and 14-23 to write `m_(f) g = mg " " (14-24)` and then relate the masses to the densities and volume using the general relation `rho = m//V` . Here , we have the submerged volume `V_(f)` and the full block volume V . We also have the liquid.s density `rho_(f)` and the block.s density `rho` . Equation 14-24 then becomes
`rho_(f) V_(f) g = rho Vg`.
or `(V_(f))/(V) = (rho)/(rho_(f))`
`= (700 kg//m^(3))/(1200 kg//m^(3)) = 0.58. " " (14-24 a)`
Thus , 58 % of the block.s volume is submerged .
To find the submerged height h , we note that the submerged volume `V_(f)` and the area of the bottom face are related by `V_(f) = Ah`.
Similarly , the full volume and the full height H are related by `V = AH`.
Substituting these expressions into Eq. 14-24 a leads to `(h)/(H) = (rho)/(rho_(f)) = 0.58`.
Thus , 58 % of the block.s height is submerged .