petrophysics
Fb/m + (-Fg)/m = a Are you seriously suggesting that "a" is always equal to -9.8m/sec^2 ?
In this instance the acceleration by the force of gravity is matched by the force of buoyancy and the net acceleration reduces to zero.
The reason is simply that the buoyancy is measured in the same way as the gravity
Force of buoyancy is Fb=mg m equal to the mass of water displaced and g equal to gravitational acceleration while force of gravity is also Fg=mg m equal to mass of body and g equal to acceleration of gravity.
Thus,folowing your reasoning, we get Fb/m +{-}Fg/m = g. g = 9.8m/sec^2
so the answer to your question is Yes.
In accoradance with Newtons third law the force of buoyancy acts in a equal magnitude but opposite direction to the mass acceleration due to gravity. Since they are equal forces but applied in opposite directions the
net force becomes zero.
The acceleration due to gravity remains the same otherwise there would be an imbalanced force due to the opposing equal force of the water which would repel the mass and expel it from the water.The force of gravity remains constant thus it is always accelerating at 9.8m/sec^2 on earth.
This message has been edited by sidelined, Tue, 2006-04-18 09:36 AM