Recent content by mwalter

Thanks gneill!

The person is standing on top of the ice without getting their feet wet, so the volume of the ice should be equal to the volume of the water displaced? The way I sort of solved it was: 49.0kg * 9.8m/s^2 + 920kg/m^3*9.8m/s^2*V_ice = 1000 kg/m^3*9.8m/s^2*V_water 480.2 N + 9016kg/m2s2*V_ice =...

Hi gneill, Thinking more about this, and that the weight of the fluid displaced is equal to the buoyant force, I believe that: W_person + W_ice = W_water displaced, so m_person*g + p_ice*V_ice*g = p_water*V_water*g and that answers both of your questions. It probably answers mine as...

Homework Statement A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 49 kg person to be able to stand on it without getting his or her feet wet? (use 920kg/m^3 as density of ice and 1000 kg/m^3 for the density of freshwater) Homework Equations...