Help with physics?

Question

1) A 1.2 kg beaker containing 1.89 kg of oil of density 906kg/m^3 rests on a scale. A 1.66kg block of iron is suspended from a spring scale and is completely submerged in the oil. The acceleration of gravity is 9.8m/s^2. Find the equilibrium reading of teh spring scale. The density of teh iron is 7860 kg/m^3. Answer in units of N.

thanks for any help!

Answer 1

rho (oil) = 906 kg/m3
m (iron) = 1.66 kg
rho (iron) = 7860 kg/m3

volume (iron) = volume (displaced oil) = m (iron)/rho(iron) = 1.66/7860 = 2.111E-4 m3

volume (displaced oil) x rho (oil) = d.m (oil) = 2.111E-4 x 906 = 0.191 kg

Buoyancy = d.m (oil) x gravity = 0.191 x 9.8 = 1.8718N

Reading on spring = Weight (iron) - buoyancy = 1.66 x 9.8 - 1.8718 = 14.39N

Answer 2

In a vacuum, the scale would read the total weight of the iron. SUbmerged, it weighs less. The VOLUME it displaces is the key.
So, convert from weight of iron to its volume. That volume of oil is what is displaced and so the weight of that volume of oil is what you would subtract from the weight of the iron. Converting from Kg to N is just by F=ma. In this problem you may use either F or m. I'd use Kg and convert afterwards to N. Any object which floats is by definition displacing its weight. (That is, its effective weight is zero when it is floating , when it is displacing its weight. Things which are denser than the fluid they are in can not float (unless you somehow make them displace more like turning a block into a bowl))