# Mass submerged in oil and scales problem

1. Nov 9, 2012

### bdh2991

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

Fbuoyant = ρ*g*Vdisplaced

Fg = mg

T = tension

3. The attempt at a solution

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

ƩF = T + Fbuoyant - mblock*g = 0

substituting buoyant force I get: T + ρoil*g*Vdisplaced - 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 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.

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

ƩF = Fbuoyant + mbeaker*g + moil*g = bottom scale reading.

solving for that i got 48.804 N, but that is also assuming Fbuoyant = mblock*g

Any help?

2. Nov 9, 2012

### Simon Bridge

Nicely analysed.

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