The pole will come to a static position where
the sum of the moments about the point O is equal to zero.
ΣM_{o} = 0
Therefore, the buoyancy force multiplied by its lever arm must be equal
and opposite the gravitational force times its lever arm.
ΣM_{o} =
0 = (W_{pole}L_{pole}  F_{B}L_{B})cosθ
where
L_{pole} = L/2 (gravity acts uniformly along the entire
length), and
(buoyancy
force acts at the centroid of the submerged part of the pole). Note, θ is not needed since it factors out of the equation. Summing moments at O gives,
W_{pole}L_{pole} = F_{B}L_{B}
where
W_{pole} = ρ_{pole} V g = ρ_{pole} (π r^{2} L) g
F_{B} = ρ_{water} V g = ρ_{water} (π r^{2} L/3) g
thus,
ρ_{pole} (π r^{2} L) g (L/2) = ρ_{water} (π r^{2} L/3) g (5L/6)
Cancelling terms,
ρ_{pole} (1/2) =(1.94 slug/ft^{3}) (5/18)
The density of the pole is
ρ_{pole} = 1.078 slug/ft^{3}
