Question

(b) A piece of wood of volume 0.6 m² floats in water. Find the volume

exposed. What force is required to immerse it completely under water?
(Density of wood = 600 kg/m3, water = 1000 kg/m3)
[8]​

Answer 1

Step-by-step explanation:

Let the volume below water be v . Then

buoyant force = v d g where d is density of water , g is acceleration due to gravity

= v x 1000 x g

weight of wood piece = volume x density of wood x g

= .6 x 600 x g

for equilibrium while floating

buoyant force = weight

= v x 1000 x g = .6 x 600 x g

v = .36 m²

volume above water or volume exposed = .6 - .36

= .24 m²

When immersed completely ,

buoyant force = .6 x 1000 x 9.8

= 5880 N

weight of wood

= .6 x 600 x g

= 3528 N

buoyant force is more than the weight . In order to equalise them for floating with full volume in water

weight required = 5880 - 3528

= 2352 N.