Question

The center of buoyancy of a floating body is at x = 0, z = 0 and its center of gravity is at x = 0, z = 0.9 where (x, z) are the horizontal and vertical coordinates. The body is tilted through an angle θ = 5◦ so that the center of buoyancy shifts to x = 0.1, z = 0. Will the floating body (a) Return to its undisturbed configuration (b) Move further away from its undisturbed configuration ?

 

Answer 1

The question does not mention the axis about which the body is tilted. Therefore, we will consider two general cases.
In the above figure, the initial and final positions of the body is represented.
To find out the new coordinates of the CG, we will use:
Tan 5^o = x/0.9
⇒ x = 0.0787
This means that the center of gravity of the body shifts 0.0787 units towards right IF the body is tilted about the origin.
If the body, however is tilted about the CG, which is more general condition, then there will be no change in the position of the CG.
In both the cases, the Center of buoyancy shifts 0.1 units towards right.
Now, since the body is floating, Force of buoyancy = weight of the body.
Weight acts at the CG in the downward direction and buoyant force acts at CB in the upward direction.
Here, since the CB is towards the right of CG in both the cases, the buoyant force will produce a counter-clockwise moment which will try to turn back the body to it's original position. The body, therefore, is in stable equilibrium.
Therefore, the body will return to its undisturbed configuration.

(b) The body floats fully submerged. This means that the Buoyant force acting on the body uses up the entire volume of the body and is equal to the weight of the body. In simple terms, the average density of the body is equal to the average density of the fluid.
Let us assume a cubical object floating fully submerged in the fluid as shown below:

Consider a infinitesimally small cubical element of dimensions dx, dy and dz inside the object. The density of the body at this location is given by it's density function ρ(x,y,z). Therefore, volume of this small element = dxdydz
And mass = volume x density = ρ(x,y,z) dxdydz
Now, to get the mass of the entire object, we can integrate this small mass as
M = ∫∫∫ρ(x,y,z) dxdydz
Now, the total weight of the body = Mg, where g = acceleration due to gravity.
Weight = ∫∫∫ρ(x,y,z) dxdydz x g .............(1)
Now this will be equal to the total buoyant force on the body
Buoyant force = Weight of the volume of fluid displaced by the object.
Here, volume displaced by the object = Volume of the object (Since it is fully submerged)
Volume of the object = ∫∫∫dxdydz = V_b
Therefore, buoyant force = V_b x ρ_f x g............ (2)
Equating equations (1) and (2),
∫∫∫ρ(x,y,z) dxdydz x g = V_b x ρ_f x g
Cancelling out the acceleration due to gravity,

∫∫∫ρ(x,y,z) dxdydz = ρ_f V_b