Question

A rod of length $6m$ has specific gravity $\rho (=25/36)$. One end of the rod is tied to a $5m$ long rope, which in turn is tied to the floor of a pool $10m$ deep, as shown. Find the length (in $m$) of the part of rod which is out of water.

Answer 1

The forces on rod are:

$F_B=bouyantforce$ and $Mg$

Calculation of bouyant force on rod:

$F_B=weightofwaterdisplaced={\rho }_w{A}_c\frac{5}{sin\theta }gMg=\rho A_c\times 6g$

Taking moments about O.

Total clockwise moment = total anti-clockwise moment

Total clockwise moment=

$Mg\times 3cos\theta =\rho A_c\times 6g\times 3cos\theta ...\left(1\right)$

total anti-clockwise moment.

$={\rho }_w{A}_c\frac{5}{sin\theta }g\times OAcos\theta ={\rho }_w{A}_c\frac{5}{sin\theta }g\times \frac{5}{2sin\theta }cos\theta ...\left(2\right)$

equating (1) and (2)

${\rho }_w{A}_c\frac{5}{sin\theta }g\times \frac{5}{2sin\theta }cos\theta =\rho A_c\times 6g\times 3cos\theta$

${sin}^2\theta =\frac{25}{36}\times \frac{{\rho }_w}{\rho }=1$

$\theta ={90}^0$

Hence the length of the rod outside water=$1m$