Question

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

Answer 1

Forces acting on rod are force of buoyancy, weight of rod and tension due to string.
Let Rod makes an angle $\theta$ with horizontal


Balancing torque about O,
$F_b{r}_{\perp }=W{r}_{\perp }$
${\rho }_w(6-x)Ag\frac{(6-x)}{2}\cos \theta ={\rho }_b\left(6A\right)\left(3\right)\cos \theta$

$(6-x{)}^2=36\rho$
$6-x=5$
$x=1$