How Does Buoyancy Affect Object Stability in Water?

AI Thread Summary
Buoyancy is central to understanding object stability in water, as demonstrated in the floating object problem where a 1 m³ object, with a third submerged, weighs approximately 326.6 grams based on buoyant force calculations. The second problem involves a hollow plastic box with negligible mass and a 3.5 kg copper piece; the box will sink since the total weight exceeds the buoyant force. In the third scenario, Bernoulli's principle explains that blowing air between two hanging bowling balls creates lower pressure, causing them to move closer together. The discussions highlight the importance of understanding buoyant force and fluid dynamics for solving related physics problems. Overall, grasping these concepts is crucial for accurately determining stability and movement in fluid environments.
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1.) 1 m3 object is floating in water. A third of the object is submerged. How much does the object weigh?


2.) Consider a floating hollow plastic box that has negligible mass. The volume of the box is 0.002 m3. If a 3.5 kg piece of copper is placed in the box, will the box sink or float? Justify your answer with calculations.

3.) Two bowling balls are hung from the ceiling such that both are the same distance from the floor. At their closest point the two balls are separated by only 2 cm. If you were to take a leaf blower operating at its maximum level and aim the nozzle of the blower between the two balls, which way would the balls move. Explain using the appropriate scientific concept.

4.) An inflated balloon occupies a volume of two (2) liters. The balloon is tied with a string and weighted down with a heavy stone. What is its volume when it reaches the bottom of a pond 5.2 meters deep? Assume the atmospheric pressure is equal to 1 atmosphere and that one atmosphere will support a column of water 10.4 meters high.






I'm doing a take home quiz and skipped these four problems. I can begin setting them up, but then I get stuck and can't follow through. The first two are hardest for me because I just can't grasp the concept of buoyancy. I understand the weight displaced = the force of buoyancy. But neither of the problems give me the weight of the water displaced. And the second two problems are the last two problems on the quiz, but we never covered in class. I need a lot of direction.
Helpp?
:eek:
 
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panther2041 said:
1.) 1 m3 object is floating in water. A third of the object is submerged. How much does the object weigh?

Is that supposed to be 1 m^{3} (1 meter cubed)?

I understand the weight displaced = the force of buoyancy. But neither of the problems give me the weight of the water displaced

Good, you understand the physical situation. Write out the equation for the bouyant force. That's a good place to start. This may help you see how you could calculate the weight of the water displaced.

Can you write out the set up for the remainder of the problems or do you have no idea? Do this for problems 2 through 3 if you can. Write out the relevant equations and what the knowns and unknowns are so we can create a strategy to solve the problems. And write out what is physically happening in each case, the physics behind the problem if you will.
 
CanIExplore said:
Is that supposed to be 1 m^{3} (1 meter cubed)?

Yes, it is. It's given me that the volume is 1 meter cubed. But I thought I had to know how much of the water that's displaced to find out the buoyant force. If the buoyant for is the amount of the object immersed, then it would be 1/3 meter cubed. But then all I know is that the weight is equal to the buoyant force. But weight can't be measured in meters cubed. So, then I just move around when circles.

2.) Consider a floating hollow plastic box that has negligible mass. The volume of the box is 0.002 m3. If a 3.5 kg piece of copper is placed in the box, will the box sink or float? Justify your answer with calculations.

Here I combined the box and the copper to create density of the total object. But I learned that the ratio of the submerged density to the total density is equal to the ratio of the volume of the total density to the submerged density. If they are equal, then the object is submerged but it does not sink. Again, I'm not given any information on the fluid, so I have no idea how to find the answer, when not given the fluids volume or density.

3.) Two bowling balls are hung from the ceiling such that both are the same distance from the floor. At their closest point the two balls are separated by only 2 cm. If you were to take a leaf blower operating at its maximum level and aim the nozzle of the blower between the two balls, which way would the balls move. Explain using the appropriate scientific concept.

Bernoulli's principle.. When the speed of a fluid increases, the pressure drops. If I have this one right, the speed of the air of both sides of the balls are moving at a slower speed compared to the air blown between the two balls. This reduces the force between the two balls and the stronger forces from the outside of the balls(slower air) push the balls together. (?)
 
panther2041 said:
Yes, it is. It's given me that the volume is 1 meter cubed. But I thought I had to know how much of the water that's displaced to find out the buoyant force. If the buoyant for is the amount of the object immersed, then it would be 1/3 meter cubed. But then all I know is that the weight is equal to the buoyant force. But weight can't be measured in meters cubed. So, then I just move around when circles.

Assume the density of water is 1g/ml=1g/cm3.

The equation for the bouyant force is FB=\rhoVa. Where \rho is the density of water, V is the volume of water displaced, and a is acceleration which in this case is just g. If you know a third of the object is submerged and you know the object's volume, what can you calculate? Using this can you now calculate the bouyant force?

You are exactly right about number 3. Good physical intuition.

Consider what I said about number 1 and see if it helps in understanding number 2.
 
1.)

So I would have to convert the meters to cm?

buoyant force= 1g/cm cubed x 100 cm cubed x 980 cm/s squared
buoyant force = 980 gcm/s squared

So, the weight equals 980 g?
 
Oh wait, a third is displaced.
So instead of 100 cm it's 33.3 repeated.

So it's weight should be roughly around 326.6 g?
 
panther2041 said:
1.)

So I would have to convert the meters to cm?

buoyant force= 1g/cm cubed x 100 cm cubed x 980 cm/s squared
buoyant force = 980 gcm/s squared

So, the weight equals 980 g?

To convert from cm3 to m3:
1cm3[\frac{1m}{100cm}]3. So you're dividing by 1003. 1m3 is not equal to 100cm3.

So first do the conversion and see what you get.
 
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