Chapter 36 : Equilibrium of Floating Bodies 743
is less than the upthrust of the liquid, the body will float. This may be best understood by the
Archimedes’ principle as discussed below.
36.2.ARCHIMEDES’ PRINCIPLE
It states, “Whenever a body is immersed fully or partially in a fluid, it is buoyed up (i.e., lifted
up) by a force equal to the weight of fluid displaced by the body.” Or in other words, whenever a
body is immersed fully or partially in a fluid, the resultant force acting on it, is equal to the difference
between the upward pressure of the fluid on its bottom, and the downward force due to gravity.
36.3.BUOYANCY
The tendency of a fluid to uplift a submerged body, because of the upward thrust of the fluid,
is known as the force of buoyancy or simply buoyancy. It is always equal to the weight of the fluid
displaced by the body. It may be noted, that if the force of buoyancy is greater than the weight of the
body, it will be pushed up till the weight of the fluid displaced is equal to the weight of the body. Then
the body will float. But if the force of buoyancy is less than the weight of the body, it will sink down.
36.4.CENTRE OF BUOYANCY
The centre of buoyancy is the point, through which the force of buoyancy is supposed to act. It
is always the centre of gravity of the volume of the liquid displaced. In other words, the centre of
buoyancy is the centre of area of the immersed section.
Example 36.1. A wooden block 2 m × 1 m × 0·5 m and of specific gravity 0·76 is floating
in water. What load may be placed on the block, so that it may completely inmerse in water.
Solution. Given: Volume of block = 2 × 1 × 0·5 = 1 m^3 and specific gravity of wood = 0·76.
Let W= Weight placed on the block of wood.
We know that weight of the wooden block
= (9·8 × 0·76) 1 = 7·45 kN
∴ Total weight acting downwards
= 7·45 + W ...(i)and volume of water displaced when the block is completely immersed in it
=1 m^3
∴ Upward thrust when the block is completely immersed
= 9·8 × 1 = 9·8 kN ...(ii)
Now equating the total downward weight and upward thrust
7·45 + W= 9·8
W= 9·8 – 7·45 = 2·35 kN Ans.
Example 36.2. A block of wood 4 m long 2 m wide 1 m deep is floating horizontally in water.
If density of the wood is 7 kN/m^3 , find the volume of water displaced and the position of the centre of
buoyancy.
Solution. Given: Volume of block = 4 × 2 × 1 = 8 m^3 and density of wood = 7 kN/m^3.Volume of water displaced.
We know that weight of the block
= 8 × 7 = 56 kN