# Gauge pressure due to a floating body

## Homework Statement:

Both liquid columns stand to a height ##h## in the diagram below, but the second has a block of wood floating in it. How does the gauge pressure at the point marked ##\mathbf A## differ from the gauge pressure at ##\mathbf B##?

## Relevant Equations:

Gauge pressure at a depth ##h## inside a liquid is given by : ##P_G = \rho_L gh##, where ##\rho_L## is the density of the liquid.

My answer : Both pressures are equal, i.e. ##\boxed{P_A = P_B}##.

Reason : (1) The block of wood displaces an amount (mass) of liquid equal to its weight (archimedes' principle for floating bodies, or law of floatation). Hence we can imagine removing the block in the second case and filling it up by water equal to the block's weight. Water would stand to the same height in both the liquid columns and the pressures at A and B would be the same.

The same can be shown in a different way, keeping the block in place.

(2)
Let us divide the total height in the second case into two parts : ##h_1 \rightarrow## the height of the water column only and ##h_2 \rightarrow## the height of the block of wood. The pressure due to water column ##P_1 = \rho_W gh_1##. The pressure due to block ##P_2 = \rho_B g h_2##. [Pressure = (Force) / (area) = (m_B g) / (width_B) = (density_B volume_B g) / (width_B) = density_B h_2 g]. By archimedes' principle, ##P_2 = \rho_W g h'_2## where ##h'_2## is the height of water from the top to the base of the block (clearly ##h'_2 < h_2##). Hence the pressure ##P_B = \rho_W g h_1+ \rho_W g h'_2 = \rho g (h_1 + h'_2) = \rho g h = P_A##.

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