Buoyancy with the Cross-Sectional Area of a Rectangle

[Joe3502]

Homework Statement

Find the buoyant force acting on a rectangular object with a cross-sectional area of 6m^2 when 0.5m of the object is below the surface of the water.

Relevant Equations

Fnet = Fb-Fg

Hi all,

My teacher assigned us a problem to do a few days ago and have attempted it many times, often leaving and coming back to see if I could figure it out. I imagine that you would take the cross-sectional area and multiply it by how far under the surface of the water the rectangular object is to get how much water it displaced. I am not sure if that is the right way to go about it, though.

Obviously, this is even harder because of the stay-at-home thing, but what's even worse is that my teacher hardly answers his email since the coming of this pandemic nor has he send any example problems or instructional videos. He does, however, post daily work but isn't mindful of those that need help.

Thank you very much for your time and stay safe,
Joe

 

[nrqed]

Homework Statement:: Find the buoyant force acting on a rectangular object with a cross-sectional area of 6m^2 when 0.5m of the object is below the surface of the water.
Relevant Equations:: Fnet = Fb-Fg

Hi all,

My teacher assigned us a problem to do a few days ago and have attempted it many times, often leaving and coming back to see if I could figure it out. I imagine that you would take the cross-sectional area and multiply it by how far under the surface of the water the rectangular object is to get how much water it displaced. I am not sure if that is the right way to go about it, though.

Obviously, this is even harder because of the stay-at-home thing, but what's even worse is that my teacher hardly answers his email since the coming of this pandemic nor has he send any example problems or instructional videos. He does, however, post daily work but isn't mindful of those that need help.

Thank you very much for your time and stay safe,
Joe

What is the formula for the buoyant force?

 

[haruspex]

I imagine that you would take the cross-sectional area and multiply it by how far under the surface of the water the rectangular object is to get how much water it displaced.

Yes.

 

[Joe3502]

What is the formula for the buoyant force?

Fb = Vs*D*g

So if that is the case, then:
Fb = 0.5m * 1000kg/m^3 * 9.81m/s^2
Fb = 4905?

 

[Joe3502]

Yes.

In that case, then 6m^2 * 0.5m = 3m^3 of water displaced.

 

[haruspex]

In that case, then 6m^2 * 0.5m = 3m^3 of water displaced.

Yes.

 

[Adesh]

Fb = Vs*D*g

So if that is the case, then:
Fb = 0.5m * 1000kg/m^3 * 9.81m/s^2
Fb = 4905?

Your value for $V$ is wrong.

 

[Adesh]

In that case, then 6m^2 * 0.5m = 3m^3 of water displaced.

This is the correct value for $V$, that is the volume of water displaced.

 

[Joe3502]

This is the correct value for $V$, that is the volume of water displaced.

How so?

 

[haruspex]

How so?

In the formula Fb = Vs*D*g, Vs is the volume displaced. You calculated this in post #5. But in post #4 you plugged in the depth of immersion, 0.5m, instead.

 

[Joe3502]

In the formula Fb = Vs*D*g, Vs is the volume displaced. You calculated this in post #5. But in post #4 you plugged in the depth of immersion, 0.5m, instead.

Ahhhhhhh thank you for catching my mistake.

Ok, so with that in mind:
Fb = 3m^3 * 1000kg/m^3 * 9.81m/s^2
Fb = 2.94 x 10^4 N?

 

[haruspex]

Ahhhhhhh thank you for catching my mistake.

Ok, so with that in mind:
Fb = 3m^3 * 1000kg/m^3 * 9.81m/s^2
Fb = 2.94 x 10^4 N?

Yes.

 

[Joe3502]

Thank you guys so much for helping me with this problem! I really appreciate it!