Glass has a density of about 2500 Kg/m^3. fresh water has a density of 1000 kg/m^3. if a 5 kg block of glass is sitting on the bottom of a freshwater fish tank, what is the normal force on the floor of the tank?
30N is what my professor says is correct, but im confused how?
weight on bottom= mg-weightwaterdisp
= 5g-1g/cm^3*volume
but volume of the glass=5/2.5 =2cm^3
normal force=5*9.8-1g/cm^2 *2*9.8
normal forceabout 50-20=30N