Mass submerged in oil and scales problem

[bdh2991]

Homework Statement

A 1.02 kg beaker containing 1.98 kg of oil (density = 916.0 kg/m3) rests on a scale. A 1.98 kg block of iron is suspended from a spring scale and completely submerged in the oil. Determine the equilibrium readings of both scales.

What does the top scale read?

What does the bottom scale read?

Homework Equations

F_buoyant = ρ*g*V_displaced

F_g = mg

T = tension

The Attempt at a Solution

For the top scale I did the sum of the forces, therefore i get:

ƩF = T + F_buoyant - m_block*g = 0

substituting buoyant force I get: T + ρ_oil*g*V_displaced - mg = 0

At first I thought I couldn't get the volume but then i figured since the density*volume = mass, that it would just be equal to m_block*g, however that would mean the T = 0 and the top scale reads zero. I do not believe this is the right answer however.

For the second part I also did sum of the forces for the beaker:

ƩF = F_buoyant + m_beaker*g + m_oil*g = bottom scale reading.

solving for that i got 48.804 N, but that is also assuming F_buoyant = m_block*g

Any help?

 

[Simon Bridge]

At first I thought I couldn't get the volume but then i figured since the density*volume = mass, that it would just be equal to mblock*g, however that would mean the T = 0 and the top scale reads zero. I do not believe this is the right answer however.

Nicely analysed.

However - the iron (Fe) is not floating.
$F_{bouy}=\rho_{oil}V_{Fe}g$ and $\rho_{oil}V_{Fe}\neq m_{Fe}$