College Physics (4th Edition)

The depth of the barge below the waterline is $1.5~m$
If the barge is floating in the water, then the buoyant force is equal in magnitude to the weight of the barge. According to Archimedes' principle, the buoyant force is equal to the weight of the water that is displaced. Let $V$ be the volume of water that is displaced by the barge. Let $m_b$ be the mass of the barge. Let $m_w$ be the mass of the water that is displaced by the barge. We can find the depth $h$ of the barge below the waterline: $m_w~g = m_b~g$ $m_w = m_b$ $\rho_w~V = m_b$ $\rho_w~(l\times w\times h) = m_b$ $h = \frac{m_b}{\rho_w~(l\times w)}$ $h = \frac{3.0\times 10^5~kg}{(10^3~kg/m^3)(20.0~m)(10.0~m)}$ $h = 1.5~m$ The depth of the barge below the waterline is $1.5~m$.