A bodybuilder is holding a 27 kg steel barbell above her head?

Question

How much force would she have to exert if the barbell were lifted underwater?

Normal water. If no information is given about salinity, one must assume we are talking about just plain ol' H20

Answer 1

You need to know the density of steel and the density of water in the MKS system. density of steel = 7850 kg/m3 ; density of water at 20 degrees C = 998.2 kg/m^3 (you could round this to 1000 kg/m^3

When a rigid object is submerged in a fluid (completely or partially), there exits an upward force on the object that is equal to the weight of the fluid that is displaced by the object.

You can find the weight of the water displaced by the steel barbell because you know its mass and density.

V=M / D = 27 kg / 7850 kg/m^3 = 0.0034 m^3
Buoyant F = (density)(V) (g) = (1000 kg/m^3)( 0.0034 m^3)(9.8m/s^2)= 33.32 Newtons

Since the weight on land is 27 kg x 9.8 N/kg =264.6
we subtract the buoyant force from this.

264.6 N - 33.3 Newtons = 231.3 Newtons is the force she needs to hold it up underwater.

Answer 2

The force she's have to exert equals the force due to gravity (F_g), minus the upward force due to buoyancy (F_b).

F_exerted = F_g − F_b

F_g is just the barbell's weight (in air); this is (m)(g).

F_b is the weight of the water that is displaced by the barbell; that is: the weight of a barbell-shaped mass of water. To get that, you need to know the volume (V) of the barbell, and the density (ρ_w) of water. Then the weight of the displaced water is:

(weight of water) = F_b = V(ρ_w)(g)

The problem is, they don't tell you V, the volume of the barbell. However, you can figure it out if you know the density (ρ_s) of steel. In that case:

V = m/ρ_s
(where "m" is the barbell's mass, 27 kg)

Substituting into the previous equation:

F_b = (m/ρ_s)(ρ_w)(g)

So then the exerted force is:

F_exerted = F_g − F_b
= mg − (m/ρ_s)(ρ_w)(g)
= mg(1 − ρ_w/ρ_s)

Answer 3

It depends on the salinity content of the water. If it's in the Dead Sea, then not very much.