Fluid mechanics

Question

The highest natural concentration of salts in water is found in the evaporating remnants of old oceans, such as the Dead Sea in Israel. Suppose a swimmer with a volume of 0.75m³ is able to float just beneath the surface of water with a density of 1.02 x 10³ kg/m³. How much extra mass can the swimmer carry and be able to float just beneath the surface of the Dead Sea, which has a density of 1.22 x 10³ kg/m³?

Answer 1

Hi John!

 

This problem uses Archimedes' principle, that an object placed in a fluid will be supported by a buoyant force equal to the weight of fluid displaced.  One consequence of this is that a floating object must displace a weight of fluid equal to its own weight (since it is supported by the same buoyant force that used to support the fluid) and thus must displace a mass of fluid equal to its own mass.

 

So, if the person floats in the "regular" water of density ρ_w = 1020 kg/m³ with his or her entire volume submerged (which is what I take to mean that he or she floats "just beneath the surface), the person displaces 0.75 m³ of water.  Therefore, since the person must displace a mass of water equal to his or her own mass to float, the person must have a mass equal to 0.75 m³ of "regular" water.

 

Let's then calculate the person's mass, by calculating the mass of 0.75 m³ of "regular" water.  If density, ρ, is known, then mass is just m = ρV, where V is the volume of fluid.  So the mass of 0.75 m³ of "regular" water is given by, and thus the mass of the floating person, is given by:

 

m_p = m_w = ρ_wV = (1020 kg/m³)(0.75 m³) = 765 kg

 

[NOTE:  I suspect the volume number is too large, because this swimmer would weigh nearly 1700 lbs.  I think the volume number is meant to be 0.075 m³, which is closer to what I can find for the average human body.  I will continue with the given number, but the procedure can be repeated for the adjusted number]

 

Now, if the person were to "just" float in the Dead Sea, the person would also have to remove 0.75 m³ of Dead Sea water, at ρ_DS=1220 kg/m³.  What mass would this amount of Dead Sea water have?

 

m_DS = ρ_DSV = (1220 kg/m³)(0.75 m³) = 915 kg.

 

So the swimmer would have to have 915 kg of mass to remove 0.75 m³ of Dead Sea water, and thus "just" float in the Dead Sea.  But the swimmer him or herself only has 765 kg, as we calculated above.  Can you see how to get to the amount of mass the swimmer could add to him or herself and "just" float?

 

Let me know if you want to go over any of the details, or check an answer.  I hope this gets you on your way!