How do you calculate Apparent Weight and Bouyant Force?

Question

Given the substance, it's density, volume, mass, and weight, how do you calculate the apparent weight and Bouyant force, assuming the substance is totally submerged in water?

Example:
Substance - Steel
Density - 7.8 g/cm^3
Volume - 1 m^3
Mass - 7.8 kg
Weight - 76.44 N

Answer 1

Ok, let's remember Arquimides principle :

Every substance totally submerged in water experiments a force against its weight, this force is called the "push", push that the water exerts onto the substance :

E = push of the water, then :

Apparent weight = Weigh - push

R = W - E

W = 76.44 Newtons

E = (density of water)*gravity*(volume of the substance in water)

in this case the volume sumerged is the total volume, then :

E = 10^3*9.8*1 = 9800 Newtons

I think there is something wrong with your problem, because the push of the water, or in other terms, the bouyant force is higher than the weight of the block, so it cannot be sumerged in water.

Anyway, that's the procedure, hope that helps

Answer 2

Your weight is in error by a factor of 10^3:

W = ρVg = (1 m^3)(7.8 g/cm^3)(10^6 cm^3/m^3)(1 kg/10^3 g)(9.80662 m/s^2) = 76,491.636 N
BF = ρVg = (1 m^3)(1 g/cm^3)(10^6 cm^3/m^3)(1 kg/10^3 g)(9.80662 m/s^2) = 9,806.62 N
Apparent Weight = W - B = 66,685.016 N