# Buoyant force in an accelerating elevator

I can't understand why buoyant force exerted by the liquid on a ball attached to a rope in a box having liquid of density $$p$$ moving in an elevator moving with acceleration $$a$$ has buoyant force as,

$$B= Vp(g+a)$$ (V is the volume immersed ,i.e. the volume of ball itself)

Why its not $$B=Vpg$$?

Please try explaining mathematically/by using equations please.

• If the elevator accelerates upward, the formula should have g+a in it. The force would be greater than from gravity alone. g-a is for a downward acceleration Dec 26, 2015 at 15:18
• Related demonstration: Buoyancy Paradox - University of Maryland May 19, 2020 at 9:54

This becomes clearer if you think about the liquid, not the ball.

The buoyant force is caused by the pressure of the liquid. The pressure is caused by the weight of the liquid. The weight is cause by gravity.

When an elevator accelerates downward, it is just as if gravity had become weaker. The weight of the fluid decreases. The pressure decreases. The buoyant force decreases.

This question may also help. How does buoyancy work?

Ask yourself why it was $$pVg$$ in case of non accelerated (inertial) frame. Because the pressure gradient is $$pg$$ or $$p$$ times the acceleration due to the force that is pulling it downwards.

Now in an accelerated frame, the pressure gradient is not really $$pg$$ but $$p(g+a)$$.

If you had to write pressure at a depth $$d$$, would you write $$pVg$$ or $$pV (g+a)$$?

If the elevator goes up the water might concave slightly at the surface but remain in constant state because of the box it is contained by.If the elevator goes down the water can not move due to the suction pressure to the bottom of the box,and the pressure remains the same and the state is unchanged.So the medium of water did not deiferated enough to show drastic movement of the ball.

Since the water is accelerating, the pressure distribution within the water must change to accommodate Newton's 2nd law locally. The net result of this is that the vertical pressure gradient becomes $\rho (g+a)$ rather than $\rho g$. It is as if gravity had increased.