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What % of an object’s mass is above the water line if the object’s density is 0.82g/ml?

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To develop this problem we will apply the Archimedes model. As well as the definitions of Weight based on mass and acceleration. The first in turn will be considered under the relationship of Density and Volume. From the values given we have to:


\rho_w =1 g/mL \rightarrow \text{Water Density}


\rho_o = 0.82g/mL \rightarrow \text{Object density}

Since it is in equilibrium, the weight of the object will have a reaction from the water, which will cause the sum of forces between the two objects to be zero, therefore


\sum F= 0


F_w-F_o = 0


F_w = F_o


m_w g = m_og


\rho_w V_w g = \rho_o V_o g

The value of gravity is canceled because it is a constant


(V_w)/(V_o) = (\rho_o)/(\rho_w)


(V_w)/(V_o) = (0.82)/(1)


(V_w)/(V_o) = 0.82

The portion of the object that is submerged corresponds to 82%, while the portion that is visible, above the water level will be 18%

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