What Is the Net Inward Force on a Sinking Barrel?

AI Thread Summary
To determine the net inward force on a sinking barrel at a depth of 1.61 km, the total pressure exerted by seawater must be calculated using the formula P_tot = P_surface + ρ_water x g x h. The calculated total pressure is 116 atm, but it is advised to use MKS units for accuracy. The net force on the barrel's surface is zero, as the pressure inside the barrel balances the external pressure, making the net inward force effectively meaningless. The discussion suggests that this problem may be a trick question, with the expected answer being zero. Understanding the vector nature of force is crucial in this context.
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Homework Statement


Suppose a cylindrical barrel falls off a ship and sinks to the bottom of the sea at a depth of 1.61 km. Assume that the seawater is incompressible, so that its density at the bottom of the sea is the same at the surface :1020 kg/m^3

If the pressure inside the barrel is 1.00 atm (it was sealed at sea level) and the total surface area of the barrel is 3.50 m^2 , find the NET INWARD FORCE acting on the surface of the barrel when it reaches the bottom of the ocean.

Homework Equations



P tot = Psurface + water x g x h
P= F/A

The Attempt at a Solution



For my attempt, I used the formula Ptot = P surface + ρ density of water x g x h.
The total pressure I got was 116 atm, then I plugged this pressure into P = F/A and got a force of 4. 67 x 10^6 N...

I am not sure what I am supposed to be doing first actually, any guidance would be helpful!
 
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Can you show explicitly how you came up with the 116 atm?
 
Sea surface pressure will balance the pressure inside the barrel, so you can ignore this for the 'net force'.
That in turn means there's no point in converting to atms. Do it all in MKS units. (I think you'll find you've made an arithmetic error.)

That said, I really don't like this question. Force is a vector; a net force involves performing a vector sum; the net force acting on the outside of the barrel is zero. Yes, you can take the pressure and multiply by the magnitude of the surface area, but the number that results has no physical meaning. Again, area is technically a vector here, so that multiplication should be done as a vector integral ∫P.dA = P∫dA = 0.
It is just possible that this is a trick question and the required answer is zero.
 
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