Question

A uniform rod AB, $4m$ long and weighing $12xkg$, is supported at end A, with a $6xkg$ lead weight at B, The rod floats as shown in figure with one-half of its length submerged, The buoyant force on the lead mass is negligible as its volume is also negligible. Find the tension in the cord and the total volume of the rod.

Answer 1

$T+U=60+W$ (i)
$\sum$(Moments)about O $=U\left(\frac{1}{4}\right)+T\left(l\right)=W\left(\frac{1}{2}\right)$
or $U+4T=2W$ (ii)
$W=120N$ (iii)
Solving these three equations we get $T=20N$ and $U=160N$
Now,
$160\left(\frac{V}{2}\right){\rho }_{w}g=\left(\frac{V}{2}\right)\left(1000\right)\left(10\right)$
$\therefore$ $V=32\times {10}^{-3}$ m${}^3$