Archimedes' principle allows us to solve this problem, find:
a) the submerged part h = 2.04 m
b) the downward force W = 3.9 10⁹ N
c) the upward force B = 4 10⁸ N
d) The free-body diagram attach
The parameters given
* The tanker mass with oil M = 400000 Tonnes (10³ kg/1 tonn) = 4 10⁸ kg
* The length and width of the tanker l = 500m and w= 40m
To find
* the submerged part of the tanker
* The force in the tanker
* free body diagram
Fluid mechanics exercises, the Archimedes' principle says that the hydrostatic thrust of a fluidIn on a body is equal to the weight of the desalted liquid.
B = [tex]\rho \ g \ V_{body}[/tex]
Where ρ is the density of the fluid in this case water (ρ = 1000 kg /m³), g is the acceleration of gravity (g = 9.80 m / s²) and V_{body} is the volume of the tanker.
a) In the attached diagram we can see the forces involved, we apply the equilibrium equation
B - W = 0
The volume of the submerged tanker is
V_{body} = l w h
Where l and w are the length and width of the tanker and h is the submerged quantity
We substitute
ρ g (l w h) = W
[tex]h = \frac{W}{\rho \ g\ l \ w}[/tex]
We calculate
h = 4 10⁸ / (10³ 9.8 500 40)
h = 2.04 m
The submerged quantity of tanker using Archimedes' principle is
h = 2.04m
b) The downward force on the tanker's weight
W = m g
W = 4 10⁸ 9.80
W = 3.9 10⁹ N
c) the upward force is the thrust
B = ρ g
B = 1000 9.8 (500 40 2.04)
B = 4 10⁸ N
d) in the attachment we can see a free body diagram with the two forces on the tanker: the weight down and the fluid push up
Archimedes' principle allows finding the submerged part of the tanker
h = 2.04 m
Learn more about Archimedes' principle here:
brainly.com/question/13106989
