Force to hold a 100kg man underwater and buoyant force on a helium balloon

[Sam]

I have two homework problems that I am unable to answer related to buoyancy/force. Please help! Thanks!

1. How much force (in Newtons) does it take to hold a 100kg man completely under water in the ocean? His density is 903 kg/m^3, and the specific gravity of sea water is 1.07.

They give the following hint: Buoyant force is greater than the weight: B.F. = Weight + Force Down

2. What is the buoyant force in N on a 6.6 m^3 helium balloon in air at standard conditions? The density of air is 1.3 kg/m^3

They give the following hint: According to Archimedes you only need to know the weight of the air displaced.

 

[arcnets]

Sam,
what have you got so far and where are you stuck? If you haven't started yet, then I think Archimedes' law is a good starting point.

 

[Sam]

Here's what I have so far...

I have the following, but I know my answer isn't correct because it's an online course and will let you know when you are correct:

Let:

Mm = mass of man in kg (given at 100 kg)
Vm = volume displaced by man in m^3 = Mm/Dm
Sg = specific gravity of seawater (given at 1.07)
Dw = Density of pure water 1 gm/cm^3 = 1000kg/m^3
Ds = Density of seawater
Fb = Buoyancy force
g = acceleration of gravity in m/s^2 = (9.8 m/s^2)

Here's how I tried to solve:

Fb = Vm*Dw*g
= (Mm/Dm)*Sg*Dw*g
= (100 kg/903kg/m^3)*1.07*(1000kg/m^3)*9.8m/s^2 N

Ok, so the buoyant force exceeds his weight. To keep him submerged you'd have to push him down with the difference force F=1161 - 980 or about 181 N.

What did I do wrong?

 

[arcnets]

Hmmm... looks OK to me. With the following changes, which don't change the result:

Fb = Vm*Sg*Dw*g
= (Mm/Dm)*Sg*Dw*g
= (100 kg/903kg/m^3)*1.07*(1000kg/m^3)*9.8m/s^2 no N

Since they say B.F. = Weight + Force Down, maybe they expect the answer to have a minus sign. Suggest you try that.