Question

A solid cylinder of height h and mass m is floating in a liquid of density $\rho$ as shown in the figure. Find the acceleration of the vessel (in $m/s^2)$ containing liquid for which the relative downward acceleration of the completely immersed cylinder to one-third of that of the vessel. (Take $g=10m/s^2)$

Answer 1

The weight of the cylinder must be balanced by buoyancy force in upward direction.
${\rho }^{\prime }Ahg=\rho A\frac{3}{4}hg\Rightarrow {\rho }^{\prime }=\frac{3}{4}\rho$
Now assume vessel is moving upward with an acceleration of $a_{1}m/s^2$.
Given the relative acceleration $a_{rel}$ of cylinder is 1/3rd of vessel in downward direction.
applying pseudo force $m{a}_1$ in downward direction
Writing equation of motion for cylinder $mg+m{a}_1-B_1=m{a}_{cylinder}$...(i)
where $B_1=(P_2-P_1)A$ and $a_{cylinder}=\frac{a_1}{3}$
$mg+m{a}_1-(P_2-P_1)A=m\frac{a_1}{3}$...(ii)
now taking an imaginary cylinder of water mass having same dimensions of cylinder
writing equation of motion for imaginary cylinder
$B_2-m_{liquid}g=m_{liquid}{a}_1$...(iii) where $B_2=(P_2-P_1)A$
$(P_2-P_1)A-m_{liquid}g=m_{liquid}{a}_1$
Substituting value of $(P_2-P_1)A$ in equation (ii)
$mg+m{a}_1-m_{liquid}(g+a_1)=m\frac{a_1}{3}$...(iv)
The ratio $\frac{m}{m_{liquid}}=\frac{{\rho }^{\prime }Ahg}{\rho Ahg}=\frac{3}{4}$ substituting this value in equation (iv)
$\frac{3}{4}g+\frac{3}{4}{a}_1-(g+a_1)=\frac{3}{4}\frac{a_1}{3}$...(v)
$\frac{a_1}{2}=-\frac{g}{4}=-5m/s^2$
i.e. the cylinder should move in downward direction with acceleration of $5m/s^2$