# Buoyant force doubt

In summary: The pseudo force (mg) is not a real force. It is an artifact of your choice of frame of reference.In summary, the buoyant force acting on an object in a vessel with a liquid is equal to the product of the volume of the object, the effective gravity (which includes both actual gravity and any acceleration), and the density of the liquid. This is true even when considering the pseudo force in the accelerating frame of reference.

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 textbook: ##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

Here's the doubt. Why is F(buoyant force) = Vp(g+a) when I have already considered pseudo force?
Buoyant force is due to the pressure of the liquid acting on the object. When the vessel is accelerating, the fluid pressure actually increases.