A large rectangular raft has a density of 725.5 kg/m3) and floating on a lake. The surface area of the top of the raft is 39.6 m 2 and its volume is 5.2 m3. Assume acceleration of gravity is 9.8 m/s2 and the density of the water is 1000 kg/m3. How many meters "h" is the raft above the water?

h=0.036 m

Explanation:

First we calculate the volume of the raft:

Volume of the raft = raft surface x A

where A is the dimension of the raft that we don't know but we know that A = h+d where d is the submerged edge from the raft and h the distance asked

Volume of the raft = raft surface x A

5.2 m3=39.6 m2 x A

A = (13/99) m

Now let's calculate de weight of the raft:

ρ=m/v

ρ (raft) = m (raft) / v (raft)

725.5 kg/m3 = m (raft) / 5.2 m3

m (raft) = 3772.6 kg

W=m x g

W (raft) = m (raft) x g

W = 3772.6 kg x 9.8 m/s2

W = 36971.48 N

The raft is floating so W (raft) = water thrust

Water thrust = water volume x water density x gravity

36971.48 N = (39.6 m2 x d) x (1000 kg/m3) x 9.8 m/s2

d = 0.0953 m

Finally A=h+d

h = A - d

h = (13/99) m - 0.0953 m

h = 0.036 m