Iceberg floating in the ocean

Let

$V$ be the total volume of the iceberg,

$V_1$ be the volume of the iceberg above water surface,

$V_D$ be the volume displaced,

$BF$ be the bouyant force

$W_{ice}$ be the weight of ice

$\gamma_{ice}$ be the unit weight of iceberg

$\gamma_{sw}$ be the unit weight of seawater

$s.g$ be the specific gravity

Solution:

$W_{ice}=\gamma_{ice}V=9.81(0.92)V=9.0252V$

$BF=\gamma_{sw}(V_D)=9.81(1.03)(V-1000)=10.1043(V-1000)$

From the free-body diagram,

$\sum F_v=0$

$W_{ice}=BF$

$9.0252V=10.1043(V-1000)$

$\dfrac{9.0252}{10.1043}V=V-1000$

$\color{#D61F06}\boxed{V=9364~m^3}$