Buoyant Force!!!?

Question

What buoyant force is exerted on the bottom of a flat 5 m long, 2 m wide, and .2 m under the surface of pure water? the desnsity of water is 1000 kg/m^3

Answer 1

Archimedes says:
The force is simply the weight of the displaced water.

Calculate the volume, multiply by density to get mass, and multiply by g to get the weight.

Easy as that.

Answer 2

Buoyancy = p * V * g

p = density of the Liquid
V = Volume of the submerged object
g = Gravitational force

p = 1000 kg/m^3
V = (5*2*0.2) = 2 M^3
g = 9.81 m/s^2

Thus Buoyant force = 1000 Kg/m^3 * 2m^3 * 9.81 m/s^2 = 19620 kgm/s^2

1 Newton = 1 kgm/s^2

thus the Buoyant Force is 19620 Newtons.

Answer 3

P=phg,where p=density
h=height
g=acceleration of free fall
P=pressure
P=1000x2x10
=20000Pa
P=F/A ,where A=area
F=PA
=20000x5x2
=200000N

Answer 4

First calculate the volume of displaced water
5m*2m*.2m=2m^3
Now calculate the mass of the water displaced using
density = mass/volume
1000kg/m^3*2m^3=2000kg
By assuming the gravitational acceleration constant to be 9.81m/s^2 you can write:
F=ma
F=2000kg*9.81m/s^2
F=19620N