A scale reads 380 N when a piece of nickel is hanging from it. What does it read (in N) when it is lowered so that the nickel is submerged in water?

Question

A scale reads 380 N when a piece of nickel is hanging from it. What does it read (in N) when it is lowered so that the nickel is submerged in water?

Answer 1

Please note: I use the rather unconventional "B" to denote the buoyant force in place of F_B

Let's start with Newton's second law:

∑F = T + B - F_g = ma = 0

T + B - F_g = 0 => T = F_g - B

Now, let's define buoyancy

=> B ≡ ρ_fluidV_fluidg

Because the nickel is completely submerged into the liquid, we can say that the volume of the fluid displaced is equal to the volume of the nickel object

=> V_fluid = V_nickel

Now, what we want to do is define the volume in terms we know: the density of the nickel and its mass

=> ρ_nickel = mass_nickel*V_nickel^-1 => V_nickel = mass_nickel*ρ_nickel^-1

Let's go ahead and substitute these into our original buoyancy definition

=> B = ρ_fluidV_fluidg = ρ_fluid*V_nickelg = ρ_fluid(mass_nickel*ρ_nickel^-1)g

Now that we this all into terms we know, we can substitute this into our Newton's second law expression

T = F_g - B = m_nickelg - ρ_fluid(mass_nickel*ρ_nickel^-1)g = F_{g nickel}(1 - ρ_fluid*ρ_nickel^-1)

=> ρ_fluid = 1000kg*m^-3

=> ρ_nickel = 8902kg*m^-3

<=> 380N[1 - (1000kg*m^-3/8902kg*m^-3)]

B = 337.3N