Question

An open cylindrical tank 350 mm in diameter and 1.8 m high is inserted vertically into a body of water with open end down and floats with a 1300 N block of concrete (SG = 2.4) suspended at its lower end. Neglecting the weight of the cylinder, determine the Buoyant force on the concrete, and the depth of water that in the cylinder? What are the pressures inside the cylinder before and after inserting in water?

Answer 1

First, let us calculate the buoyant force. We know that the force is equal to the weight of water in the volume of the body. The volume of block is

$V = \dfrac{P}{\rho g} = \dfrac{1300\,\mathrm{N}}{2400\,\mathrm{kg/m}^3\cdot 9.81\,\mathrm{N/kg}} = 0.055\,\mathrm{m}^3.$

The buoyant force will be $\rho_{\text{w}}gV = 541.7\,\mathrm{N}$ .

The system floats, so there should another force that is directed upwards and is equal to $1300-541.7 = 758.3\,\mathrm{N}.$ This force is equal to the buoyant force that acts on the volume of air under the water line. Therefore, this volume is

$V_1 = \dfrac{758.3\,\mathrm{N}}{1000\,\mathrm{kg/m}^3\cdot9.81\,\mathrm{N/kg}} = 0.077\,\mathrm{m}^3.$ The height of such a layer is $h_1 = \dfrac{V_1}{\pi D^2/4} = 0.8$ m.

The pressure of such a layer of water should be balanced by the pressure in air in the tank:

$p_1 = \rho_{\text{w}} g h_1 = 7882\,\mathrm{Pa}.$ We should take the atmospheric pressure $p_0$ into account to get the total pressure $10882$ Pa.

According to the ideal gas law, $p_0V_0 = p_1V_1, \;\; p_0\cdot\pi \dfrac{D^2}{4} h = p_1\cdot\pi \dfrac{D^2}{4} h_2$ , where $h_2$ is the height of air column. $h_2 = h \cdot{p_0}{p_1} = 1.67\,\mathrm{m}.$ Therefore, the depth of water will be only $1.8-1.67=0.13$ m