Question

A block of wood having specific gravity 0.85 floats on water. Some oil of specific gravity 0.82 is poured on the surface of water until the the wooden block is just immersed. calculate the fraction of block lying below the surface of water in the second case.

Answer 1

A block of wood having specific gravity 0.85 floats on water. Some oil of specific gravity 0.82 is poured on the surface of water until the wooden block is just immersed. Calculate the fraction of block lying below the surface of water in the second case.

Solution.

$$\frac {\rho_ {1}}{\rho_ {w}} = 0. 8 5, \frac {\rho_ {2}}{\rho_ {w}} = 0. 8 2;$$ $$\frac {V _ {1}}{V} -?$$

Newton's second law in vector form:

$$m \vec {a} = \overrightarrow {B _ {1}} + \overrightarrow {B _ {2}} + m \vec {g}.$$

A block is at rest, then:

$$\vec {a} = 0.$$ $$0 = \overrightarrow {B _ {1}} + \overrightarrow {B _ {2}} + m \vec {g}.$$

Projection on Y:

$$0 = B _ {1} + B _ {2} - m g;$$ $$B _ {1} + B _ {2} = m g.$$

$B_{1}$ - the buoyancy force in the water.

$B_{2}$ - the buoyancy force in the oil.

$m$ - the mass of a block.

$$m = \rho_ {1} V;$$

$\rho_{1}$ – the density of the wood.

$V$ – the volume of the block.

$$B _ {1} = \rho_ {w} V _ {1} g;$$ $$B _ {2} = \rho_ {2} V _ {2} g.$$

$V_{2}$ – the part of volume of the block above water level.

$$V _ {2} = V - V _ {1};$$ $$B _ {2} = \rho_ {2} (V - V _ {1}) g;$$

$\rho_w$ – the density of the water;

$\rho_{2}$ – the density of the oil;

$V_{1}$ – the part of volume of the block below water level.

$$\rho_ {w} V _ {1} g + \rho_ {2} (V - V _ {1}) g = \rho_ {1} V g.$$

Divide by $g$:

$$\rho_ {w} V _ {1} + \rho_ {2} (V - V _ {1}) = \rho_ {1} V.$$

Divide by $\rho_w$:

$$V _ {1} + \frac {\rho_ {2}}{\rho_ {w}} (V - V _ {1}) = \frac {\rho_ {1}}{\rho_ {w}} V;$$ $$V _ {1} + \frac {\rho_ {2}}{\rho_ {w}} V - \frac {\rho_ {2}}{\rho_ {w}} V _ {1} = \frac {\rho_ {1}}{\rho_ {w}} V.$$

Divide by $V$:

$$\frac {V _ {1}}{V} + \frac {\rho_ {2}}{\rho_ {w}} - \frac {\rho_ {2}}{\rho_ {w}} \frac {V _ {1}}{V} = \frac {\rho_ {1}}{\rho_ {w}};$$ $$\frac {V _ {1}}{V} \left(1 - \frac {\rho_ {2}}{\rho_ {w}}\right) = \frac {\rho_ {1}}{\rho_ {w}} - \frac {\rho_ {2}}{\rho_ {w}};$$ $$\frac {V _ {1}}{V} = \frac {\frac {\rho_ {1}}{\rho_ {w}} - \frac {\rho_ {2}}{\rho_ {w}}{\left(1 - \frac {\rho_ {2}}{\rho_ {w}}\right)}},$$ $$\frac {V _ {1}}{V} = \frac {0.85 - 0.82}{(1 - 0.82)} = 0.17.$$

Answer: $\frac{V_1}{V} = 0.17$.