Buoyancy with the Cross-Sectional Area of a Rectangle

AI Thread Summary
The discussion revolves around calculating the buoyant force on a rectangular object submerged in water. The correct approach involves determining the volume of water displaced, which is found by multiplying the cross-sectional area by the depth of immersion. The formula for buoyant force is given as Fb = Vs * D * g, where Vs is the volume displaced, D is the water density, and g is the acceleration due to gravity. A participant initially miscalculated the volume but later corrected it, leading to the final buoyant force calculation of approximately 29,400 N. The conversation highlights the challenges of remote learning and the need for clearer instructional support from teachers.
Joe3502
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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
 
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Joe3502 said:
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?
 
Joe3502 said:
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.
 
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nrqed said:
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?
 
haruspex said:
Yes.
In that case, then 6m^2 * 0.5m = 3m^3 of water displaced.
 
Joe3502 said:
In that case, then 6m^2 * 0.5m = 3m^3 of water displaced.
Yes.
 
Joe3502 said:
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.
 
Joe3502 said:
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.
 
Adesh said:
This is the correct value for ##V##, that is the volume of water displaced.
How so?
 
  • #10
Joe3502 said:
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.
 
  • #11
haruspex said:
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?
 
  • #12
Joe3502 said:
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.
 
  • #13
Thank you guys so much for helping me with this problem! I really appreciate it!
 
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