Bouyant force

Relevant wiki: Fluid Mechanics

Consider the diagram. Because the system is in equilibrium, summation of forces acting vertical must be zero. We have

$W_C+W_{L}=BF_{C}+BF_{L}$

$3.825+W_{L}=\dfrac{\pi}{4}(1)^2(1.5)(9.81)+V_{L}(9.81)$

However, $\gamma_{L} =\dfrac{W_{L}}{V_{L}}$ or $V_{L}=\dfrac{W_{L}}{110}$

So we have

$3825+W_{L}=\dfrac{3.1416}{4}(1)^2(1.5)(9.81)+\dfrac{W_{L}}{100}(9.81)$

$W_{L}\approx \boxed{8.5~kN}$

Symbols:

$W_{L}$ = weight of lead

$W_{C}$ = weight of cylinder

$BF_{L}$ = bouyant force of lead

$BF_{C}$ = bouyant force of cylinder

$\gamma_{L}$ = unit weight of lead

$V_{L}$ = volume of lead

$W_{L}$ = weight of lead

Note:

Bouyant Force = weight of the fluid displaced = volume of object displaced x unit weight of fluid