Question

A wooden block of volume$\mathbf{5}\mathbf{.24}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathbf{\text{\hspace{0.17em}m}}}^{\mathbf{\text{3}}}$floats in water, and a small steel object of mass$\mathit{m}$is placed on top of the block. When$\mathit{m}\mathbf{=}\mathbf{0}\mathbf{.31}\mathbf{\text{\hspace{0.17em}kg}}$, the system is in equilibrium and the top of the wooden block is at the level of the water.

(a) What is the density of the wood?

(b) What happens to the block when the steel object is replaced by an object whose mass is less than$\mathbf{0}\mathbf{.31}\mathbf{\text{\hspace{0.17em}kg}}$?

(c) What happens to the block when the steel object is replaced by an object whose mass is greater than$\mathbf{0}\mathbf{.31}\mathbf{\text{\hspace{0.17em}kg}}$?

Short answer

(a) The density of wood is $408.4\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$.

(b) If the steel object is replaced by an object whose mass is less than $0.31\text{\hspace{0.17em}kg}$the block will not be completely submerged and some portion of it will be above the water.

(c) If the steel object is replaced by an object whose mass is greater than $0.31\text{\hspace{0.17em}kg}$the block-steel system will sink.

Step-by-step solution

Identification of the given data:

The given data can be listed below as,

• Volume of the wooden block, $V=5.24\times {10}^{-4}{\text{\hspace{0.17em}m}}^{\text{3}}$
• Mass of the steel object, $m=0.31\text{\hspace{0.17em}kg}$

Buoyancy and gravitational force:

The buoyant force from a liquid of density $\mathit{d}$when volume of the liquid displaced$\mathit{v}$is,

$\mathit{B}\mathbf{=}\mathit{d}\mathit{g}\mathit{v}$ ..... (1)

Here, is the acceleration due to gravity having value

The gravitational force on a body of mass $\mathit{m}$is,

$\mathit{F}\mathbf{=}\mathit{m}\mathit{g}$ ..... (2)

(a): Determining the density of wood:

Since the wood is totally submerged, the volume of displaced water is the volume of the wood. The system is in equilibrium, that is the net force is zero. Downward force is due to the gravitational force on the steel and the wood. Upward force is the buoyancy. Let the density of the wood be $d_{wd}$.

Density of water, role="math" localid="1663753591824" $d_w={10}^3\raisebox{1ex}{${\displaystyle \text{kg}}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.$

Then, from equations (1) and (2),

$$\begin{gathered} d_{w}Vg=d_{wd}Vg+mg \\ {d}_{wd}=d_w-\frac{m}{V} \end{gathered}$$

Substitute all the value in the above equation,

role="math" localid="1663753611096" $\begin{array}{rcl}{d}_{wd}& =& {10}^3\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.-\frac{0.31\text{\hspace{0.17em}kg}}{5.24\times {10}^{-4}{\text{\hspace{0.17em}m}}^{\text{3}}}\\ & =& 408.4\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.\end{array}$

Thus, the required density is $408.4\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.$.

(b) Determining the state of the block if the steel has mass less than 0.31 kg:

If the mass of the steel is less than $0.31\text{\hspace{0.17em}kg}$, the buoyant force will be greater than the gravitational force. To achieve equilibrium the buoyant force has to be less. Thus the block of wood will not be completely submerged.

(c) Determining the state of the block if the steel has mass greater than 0.31 kg:

If the mass of the steel is greater than $0.31\text{\hspace{0.17em}kg}$, the buoyant force will be less than the gravitational force. To achieve equilibrium the buoyant force has to be increased. Thus the steel will also start sinking. If the mass of the steel is increased beyond a certain value, the wood-mass system will completely sink and move downward.