Buoyancy and Acceleration: Solving for Tension in a Vessel

[mousesgr]

1.The tension in a string holding a solid block below the surface of a liquid (of density greater than the solid) is To when the containing vessel is at rest. Show that when the vessel has an upward vertical acceleration of magnitude a, the tension T is equal to To(1+ a/g)?

 

[Tide]

The sum of the forces acting on the block equals its mass time its acceleration.

 

[mousesgr]

The sum of the forces acting on the block equals its mass time its acceleration.

F = To +mg
Fb - T - mg = ma

then?

 

[Tide]

Before acceleration, the buoyant force is

$$F_b = T_0 + mg$$

and during acceleration it is

$$F_b' = T + m(g+a)$$

Now the buoyant force is $F_b = \rho V g$ before acceleration and $F_b' = \rho V (g+a)$ during acceleration where $\rho$ is the density of the water and V is the volume of the block. Now just write the ratio of the buoyant force in the two cases and arrive at

$$\frac {g}{g+a} T = T_0$$

from which the desired result follows.