# Buoyancy with the Cross-Sectional Area of a Rectangle

• Joe3502
In summary, the conversation is about a student struggling with a physics problem assigned by their teacher during the pandemic. They discuss the formula for buoyant force and how to calculate the volume of water displaced. Eventually, they arrive at the correct answer of 2.94 x 10^4 N for 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.
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

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.

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.

This is the correct value for ##V##, that is the volume of water displaced.
How so?

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.

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?

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.

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

## 1. How does the cross-sectional area of a rectangle affect buoyancy?

The cross-sectional area of a rectangle plays a crucial role in determining the buoyancy of an object. The larger the cross-sectional area, the more water the object will displace, resulting in greater buoyant force.

## 2. What is the formula for calculating buoyancy with the cross-sectional area of a rectangle?

The formula for calculating buoyancy with the cross-sectional area of a rectangle is Fb = ρVg, where Fb is the buoyant force, ρ is the density of the fluid, V is the volume of the displaced fluid, and g is the acceleration due to gravity.

## 3. How does the shape of a rectangle affect its buoyancy?

The shape of a rectangle does not directly affect its buoyancy. However, the cross-sectional area, which is determined by the length and width of the rectangle, does have an impact on buoyancy.

## 4. Can a rectangle with a larger cross-sectional area sink?

Yes, a rectangle with a larger cross-sectional area can still sink if its weight is greater than the buoyant force acting on it. This can happen if the rectangle is made of a denser material or if it is filled with air or other materials that increase its weight.

## 5. How does the density of the fluid affect buoyancy with the cross-sectional area of a rectangle?

The density of the fluid has a direct impact on buoyancy with the cross-sectional area of a rectangle. A denser fluid will provide more buoyant force, while a less dense fluid will provide less buoyant force. This is why objects float more easily in water than in air, as water is more dense than air.

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