Question

A raft is constructed of wood having a density of 452.0 kg/m3. The surface area of the bottom of the raft is 5.4 m2, and the volume of the raft is 0.51 m3. When the raft is placed in fresh water having density 1.0 103 kg/m3, how deep is the bottom of the raft below water level?

Answer 1

A raft is constructed of wood having a density of $452.0\frac{kg}{m^3}$ . The surface area of the bottom of the raft is $5.4m^2$ , and the volume of the raft is $0.51m^3$ . When the raft is placed in fresh water having density $1.0 \cdot 10^3\frac{kg}{m^3}$ , how deep is the bottom of the raft below water level?

Solution.

Newton's second law in vector form:

$$m \vec {a} = \vec {B} + m \vec {g}.$$

A raft is at rest, then:

$$\vec {a} = 0.$$ $$0 = \vec {B} + m \vec {g}.$$

Projection on Y:

$$0 = B - m g;$$ $$B = m g.$$

$B$ - a buoyancy force.

$m$ - a mass of a raft.

$$m = \rho_ {r} V _ {r};$$

$\rho_{r}$ – the density of a raft.

$V_{r}$ – a volume of a raft.

$B=\rho_{w}Vg;$

$\rho_{w}$ – the density of a water;

$V$ – a part of volume of a raft below water level.

$V=Sh;$

$h$ - a high of the bottom of the raft below water level.

$\rho_{w}Vg=mg;$

$\rho_{w}V=m;$

$\rho_{w}Sh=\rho_{r}V_{r};$

$h=\frac{\rho_{r}V_{r}}{\rho_{w}S}.$

$h=\frac{452.0\cdot 0.51}{1.0\cdot 10^{3}\cdot 5.4}=0.043(m).$

Answer: $h=0.043m.$