Question

A container partially filled with water is moved horizontally with acceleration $a=\frac{g}{3}$ . A small wooden ball of mass m is tied to the bottom of the container using a string. The ball remains inside the water with the string inclined at an angle $\theta$ to the horizontal. Assuming that the density of wood is half the density of water, and if the tension in the string is $xmg$, find $x$.
(Answer upto two digit after the decimal point)

Answer 1

As the container is accelerating the liquid surface will be inclined such that

$tan\alpha =\frac{a}{g}=\frac{1}{3}$

Hence $sin\alpha =\frac{1}{\sqrt{10}}$ and $cos\alpha =\frac{3}{\sqrt{10}}$

In this case the buoyant force on the ball will be normal to water surface. The magnitude of buoyant force

$B=V_{dis}{\rho }_w{g}_{eff}$
$B=\frac{m}{\rho }{\rho }_w\sqrt{g^2+a^2}=2m\sqrt{g^2+\frac{g^2}{9}}$

which gives $B=\frac{2\sqrt{10}}{3}mg$
The force acting on the ball are shown in FBD (w.r.t container), then
$B=T+mgcos\alpha +masin\alpha$
$\frac{2\sqrt{10}}{3}mg=T+mg.\frac{3}{\sqrt{10}}+m\frac{g}{3}.\frac{1}{\sqrt{10}}$

$T=\frac{\sqrt{10}}{3}mg$