# Buoyant force doubt

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1. Dec 9, 2014

### AdityaDev

If I have an object of mass m tied to the lower surface of a vessel having a liquid and the vessel accelerates upwards...
From FBD of object, Buoyant force acts upwards, mg down, pseudo force downwards (frame of reference is vessel) T down.
Here's the doubt. Why is F(buoyant force) = Vp(g+a) when I have already considered pseudo force?
Given in text book: $Vp(g+a)-mg-T=ma$
But when I take the vessel as the frame of reference,
Shouldn't this be the equation: $Vpg-mg-T-ma=0$
They say $F=Vpg1$
g1=effective gravity

2. Dec 9, 2014

### Staff: Mentor

Buoyant force is due to the pressure of the liquid acting on the object. When the vessel is accelerating, the fluid pressure actually increases.

3. Dec 9, 2014

### Staff: Mentor

To see this more clearly, consider just the vessel and liquid alone. (Leave out the object for now.) With no acceleration, how does the pressure vary with depth? Now give the vessel an upward acceleration and figure out the new relationship between pressure and depth.

4. Dec 9, 2014

Il try

5. Dec 9, 2014

### jbriggs444

To add to Doc Al's point, buoyancy is a real force. It is the name we use for the net force arising from the higher fluid pressure on the bottom surfaces of an object and the relatively lower fluid pressure on its top surfaces. Since it is a real force, it exists regardless of whether you adopt an inertial or an accelerating frame of reference.

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