Iron is submerged in water question

Question

A scale reads 234 N when a piece of iron is hanging from it. What does it read (in N) when it is lowered so that the iron is submerged in water?

Answer 1

When a piece of iron is in the water, it has two forces acting on it that determine its apparent weight: gravity from the earth (acting down) and buoyancy from water (acting up).

The gravitational force is equal to the mass of the iron multiplied by g=9.81m/s², or more specifically

Weight_iron = (m)g = (ρ_ironV)g = 234Newtons (Equ.1)

where we plugged in ρ_iron aka the density of iron (which you can look up) and V aka the volume of our iron piece. We can determine this volume explicitly by re-arranging equation 1

V = 234N/(ρ_irong) (Equ.2)

Buoyancy is determined by how much water is displaced by the object: the volume of water displaced multiplied by its density.

F_Buoyancy = ρ_H20V*g = 234N × (ρ_H20 /ρ_iron) (Equ.3)

Where we plugged equation 2 into V and ρ_H20 is the density of water.

If we want to find the apparent weight, we just need to subtract these two quantities (gravity pointing down, so it's negative)

Weight_apparent = F_Buoyancy - Weight_iron = 234N [(ρ_H20 /ρ_iron) - 1] (Final Eqn)

All that needs to be done now is plug in the values of ρ_H20/iron and solve. Be sure to use the same units for each density, meaning if you use ρ_H2O = 1g/cm³ , make sure the units of ρ_iron are also in g/cm³