Question

Three children, each of weight $\mathbf{356}\mathbf{}\mathbf{N}$, make a log raft by lashing together logs of diameter $0.30\mathrm{m}$and length $1.80\mathrm{m}$. How many logs will be needed to keep them afloat in fresh water? Take the density of the logs to be $800\raisebox{1ex}{$\mathrm{kg}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{m}}^3$}\right.$.

Short answer

The number of logs required to keep the children afloat in freshwaters $5$.

Step-by-step solution

The given data

i) Weight of each child,$\mathrm{W}=356\mathrm{N}$

ii) Diameter of log,$\mathrm{d}=0.30\mathrm{m}$

iii) Radius of log,$\mathrm{r}=0.15\mathrm{m}$

iv) Length of log,$\mathrm{l}=1.80\mathrm{m}$

v) Density of log,${\rho }_{\log }=800\mathrm{kg}/{\mathrm{m}}^3$

Understanding the concept of Archimedes Principle

Using Archimedes' principle, we can state that for the object to float, the buoyant force should be equal to the weight of the object. We can find the weight of the water, which should be displaced by the logs for children to float using the total weight of children. Using the weight and density of the logs, we can find the number of logs.

Formula:

The buoyant force exerted on a body by the fluid, ${\mathrm{F}}_{\mathrm{b}}={\mathrm{\rho }}_{\mathrm{W}}\times \mathrm{g}\times {\mathrm{V}}_{\text{submerged}}$(i)

The volume of log, $\mathrm{V}={\mathrm{\pi r}}^2\mathrm{h}$(ii)

Calculation of the number of logs

$\mathrm{N}$Let’s assume that there is N number of logs. So, the total required force of buoyancy for logs and three children to float is given by:

${\mathrm{F}}_{\mathrm{b}}=3\left(356\right)+\mathrm{N}\left({\mathrm{\rho }}_{\text{wood}}\right)\times \mathrm{g}\times \mathrm{V}\dots \dots \left(1\right)$

Using equation (ii), the volume of log is given as:

$\mathrm{V}=3.14(0.15{)}^2\times (1.80)$

$V=0.127{\mathrm{m}}^3$

Comparing equation (i) and equation (1), using the formula of total submerged volume of logs, ${\mathrm{V}}_{\text{submerge}}=\mathrm{N}\times \left(\text{volume of one single log}\right)$we get,

$3\left(356\right)+\mathrm{N}\left({\mathrm{\rho }}_{\text{wood}}\right)\mathrm{gV}={\mathrm{\rho }}_{\mathrm{w}}\mathrm{g}(\mathrm{N}\times \mathrm{V})$

$1068+\mathrm{N}\left(800\right)\times \mathrm{g}\times \mathrm{V}=1000\mathrm{g}(\mathrm{N}\times \mathrm{V})$

$\mathrm{N}=\frac{1068}{\mathrm{gV}(1000-800)}$

$=4.29$

It means $5$logs are needed.