Question

oil having a density of 926 kg/m3 floats on water. a rectangular block of wood 4.63 cm high and with a density of 972 kg/m3 floats partly in the oil and partly in the water. the oil completely covers the block. how far below the interface between the two liquids is the bottoms of the block?

Answer 1

buoyancy force should be equal to the weight of the object.
\[B_o+B_w=W\]
assume height in water is h_1; height in oil h_2; thus
\[h_1+h_2=4.63cm\]
weight of the object
\[W=mg=d_{wood}Vg=d_{wood}A(h_1+h_2)g\]
buoyancy in water
\[B_w=V_1d_{water}g=h_1Ad_{water}g\]
buoyancy in oil
\[B_o=h_2Ad_{oil}g\]
insert all 1st equation, cancel g's, A's
\[d_{wood}(h_1+h_2)=d_{water}h_1+d_{oil}h_2\]
insert densities in this equation you will get relation between h_1 and h_2, and using second equation above you can get h_1 and h_2 seperately