Some fish have a density slightly less than that of water and must exert a force to stay submerged. An 85.0 kg grouper exerts such force to stay submerged in saltwater. Its body density is 1015 kg/m³? (Pseawater = 1025 kg/m³) a) Draw a free-body diagram of the fish when it is submerged in the sea water, (Enough to draw on the work.) b) What force must the fish exert to be submerged in the sea water?

 

 

Answers :

Part (a) Answer : See the attached file for the free-body diagram of the fish.

Part (b) Answer : Force exerted by the fish : F = 8.22 N

Step-by-step explanation

Here we have given :

Mass of the fish : m = 85 kg

Density of the body of the Fish : ρ = 1015 kg/m³

Since, Density : ρ = m / V

Thus, Volume displaced by the body of the fish will be : V = m / ρ = (85 kg) / (1015 kg/m³) = 0.0837 m³

Here, Density of the sea-water is : ρseawater = 1025 kg/m³

Free-body diagram :

Let the force exerted by the Fish be F.

Here, Upward Buoyant force acting on the fish will be given by : F_buoyancy = ρ_seawatwerg V

∴ F_buoyancy = (1025 kg/m³)(9.81 m/s²)(0.0837 m³) = 842.06 N

And, The weight of the Fish : W = m g = (85 kg)(9.81 m/s²) = 833.85 N

From the Free-body diagram : If the fish has to be submerged : Net force acting on the fish must be zero.

∴ F_buoyancy - F - W = 0

∴ F_buoyancy = F + W

∴ F = F_buoyancy - W

∴ F = (842.06 N) - (833.85 N)

∴ F = 8.22 N

Please see the attached file for the complete solution

Therefore, Force exerted by the fish : F = 8.22 N