Freeboard of a Soup Can in a Pool

  • Thread starter waterwalker10
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In summary, Willie takes a soup can and adds rusty ball bearings to it before putting it into the family pool. Dad is not pleased and warns Willie that he'll have to retrieve it if it doesn't float. The can, with the added weight, weighs 0.5 pounds and has a volume of 282.74 cm3. Using Archimedes Principle and converting to the MKS system, it is determined that the can will float because its density is less than that of water. The specific gravity of the can is calculated to be 0.805 cm3. To find the volume of water displaced, the formula diss = m_c/p_w = v_w is used. The result is then compared to the volume
  • #1
waterwalker10
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Willie took a soup can from the kitchen trash, added a handful of rusty ball bearings from the bottom of his Dad's junk drawer, and is now about to put it into the deep end of the family pool that Dad has just finished cleaning. Dad is not amused. "Don't worry," Willie says, "It'll float." "It had better," his Dad says, "or you'll be going in after it."

The can, plus bearings, weights exactly 0.5 pounds (force). Measurements are shown on the right. Will the can float? If so, how much freeboard will it have? (Freeboard, a nautical term, is distance between the waterline of a ship, and the ship's deck. In this case, it's the distance from the water level to the top of the can.) Show your calculations. (Hint: Convert pounds (force) to Newton’s and work in the MKS system.)



Homework Equations



Archimedes Principle?

lbs to Newtons

Vol=Pi*h*r^2

Freeboard = distance from the water level to the top of the can

The Attempt at a Solution



Ok so here is where I get confused. I believe the can will float because the can is less than 1 kg/l which is the density of water. So far here is what I have...any suggestions on where to start?

Weight = .5
Converted to Newtons
.5lbs*4.4482216 = 2.224 N

Calculated Volume of Soup Can

3.14159*10cm*32
V = 282.74
 
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  • #2
Here is the image. BTW this is an old homework assignment that I never could figure out...any help is much appreciated. Thanks.
 

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  • #3
OP did you ever figure this out... I have this exact same problem and can't figure it out!

As a matter of fact I got the exact same answers as you did above with the add info below:

density of can (pc)=0.805cm3

after that I get just as stuck you were.
 
Last edited:
  • #4
It's an old post; the OP may no longer be around or be interested in this particular question. Why don't you show your own work on it?

Hint: How about if you were to consider how much of the can, with the given base area, would have to be submerged in order to displace the can's weight in water?
 
  • #5
ok, well after looking at this for a VERY long time and using a friends help this is what I have come up with...

Vc=282.74cm3
Mc=0.227kg
Finding the specific gravity of the can (pc):
pc=m/V
pc=0.227/0.2827
pc=0.805m3

Taking this and the mass of the can I *think* that I can find the volume of displaced water

Displacement(diss)=mass/pc
(diss)=227/0.805
(diss)=281.99

I then took the volume of the can (282.74cm3) and subtracted the displacement (281.99cm3) which game me 0.75cm3

Knowing this I reversed the area equation used to find the volume of the can Vc=[itex]\pi[/itex]r2h [itex]\rightarrow[/itex] h = [itex]\pi[/itex]r2/Vc to find the height of the can out of water
which I found to be .0283cm [itex]\rightarrow[/itex] 28.3[itex]\mu[/itex]m

If this is wrong SOMEONE please point this out to me! I have no clue if I did this right or not.
 
  • #6
Count_Chockul said:
ok, well after looking at this for a VERY long time and using a friends help this is what I have come up with...

Vc=282.74cm3
Mc=0.227kg
Finding the specific gravity of the can (pc):
pc=m/V
pc=0.227/0.2827
pc=0.805m3
You've lost some accuracy here, possibly due to rounding and/or using a not too accurate value for g. The number should be closer to .80213 g/cm3 (or perhaps kg/L, since you seem to have converted the volume units to liters for your calculation).

Taking this and the mass of the can I *think* that I can find the volume of displaced water

Displacement(diss)=mass/pc
(diss)=227/0.805
(diss)=281.99
That would be the volume of displaced water if the WHOLE CAN was submerged. Effectively you've done the calculation:
[tex] p_c = m_c/v_c [/tex]
[tex] diss = m_c/p_c = v_c [/tex]
Which doesn't produce any new information...

The difference in value between this result and the volume that you calculated previously for the volume of the can is due to the rounding error mentioned above.
I then took the volume of the can (282.74cm3) and subtracted the displacement (281.99cm3) which game me 0.75cm3

Knowing this I reversed the area equation used to find the volume of the can Vc=[itex]\pi[/itex]r2h [itex]\rightarrow[/itex] h = [itex]\pi[/itex]r2/Vc to find the height of the can out of water
which I found to be .0283cm [itex]\rightarrow[/itex] 28.3[itex]\mu[/itex]m

If this is wrong SOMEONE please point this out to me! I have no clue if I did this right or not.

I'm not sure why you would want to calculate the specific gravity (actually the density) of the can, except as a check to see if it is less than the density of water and thus can float. Once that is ascertained you don't really need the can density.

Instead, use the density of the water (the thing that's being displaced) to determine what volume of water needs to be displaced in order to equal the weight of the can. Then look at how much of the can needs to be submerged in order to displace that volume of water.
 
  • #7
[tex] diss = m_c/p_c = v_c [/tex]

Instead, use the density of the water (the thing that's being displaced) to determine what volume of water needs to be displaced in order to equal the weight of the can. Then look at how much of the can needs to be submerged in order to displace that volume of water.
I see where I might have gone wrong, but...

so should I use this formula to find the water displaced?
[tex] diss = m_c/p_w = v_w [/tex]

If so that doesn't seem right to me... would I take this new displacement and try to convert it back to Newtons so that I can find the weight difference? After that... what?

and how would I find how much needs to be submerged? The previous attempt was the best that I had...I'm so lost...
 
  • #8
Count_Chockul said:
I see where I might have gone wrong, but...

so should I use this formula to find the water displaced?
[tex] diss = m_c/p_w = v_w [/tex]
Yes. This is the amount of water that the can must displace if it is to float; the can floats when it displaces a volume of water whose weight (or mass) is equal to the weight (or mass) of the can.
If so that doesn't seem right to me... would I take this new displacement and try to convert it back to Newtons so that I can find the weight difference? After that... what?
This "new displacement", as you put it, is the amount of the can that needs to be submerged; the portion of the can that is underwater and displacing water.
and how would I find how much needs to be submerged? The previous attempt was the best that I had...
You now have the volume of the portion of the can that needs to be submerged. You should be able to work out the length of the can cylinder that corresponds to that volume. What's the formula for the volume of a cylinder?
 
  • #9
OK So I took your equation and I think that I did the math correct, but I think that some of my conversions are incorrect in places.

Anyway: here is what I got

[tex] diss = m_c/p_w = v_w [/tex]
[tex] diss = 227/1 = 227cm^3 [/tex]

Taking that and subtracting the original volume of the can:
[tex]New Vol = 282.74-227 [/tex]
[tex]New Vol = 55.74 cm^3 [/tex]

Now taking this and plugging it back into the volume eqation:
[tex] V = \pi r^2h[/tex]
[tex] H = \pi r^2/V[/tex]
[tex] H = \pi 3^2/ 55.74[/tex]
[tex] H = 1.97cm[/tex]

So I *think* that my answer should be 1.97cm of freeboard?

Like I said I know that somewhere in that my conversions went a little wacky but I think that my answer is right this time.
 
  • #10
Looks good to me! :smile:
 
  • #11
Thank you so much for all your help!
 

FAQ: Freeboard of a Soup Can in a Pool

What is the purpose of measuring the freeboard of a soup can in a pool?

The freeboard of a soup can in a pool is measured to determine the buoyancy of the can, which is a measure of its ability to float on the surface of the water. This measurement is important for understanding the stability and safety of objects that are designed to float in water, like boats and other watercraft.

How is the freeboard of a soup can in a pool measured?

The freeboard of a soup can in a pool is typically measured by placing the can in the water and measuring the distance between the waterline and the top of the can. This can be done with a ruler or measuring tape.

What factors can affect the freeboard of a soup can in a pool?

The freeboard of a soup can in a pool can be affected by several factors, including the weight and shape of the can, the density of the liquid in the can, and the temperature and salinity of the water in the pool.

Why is it important to consider the freeboard of a soup can in a pool when designing boats or other watercraft?

The freeboard of a soup can in a pool is an important factor to consider when designing boats and other watercraft because it affects the stability and safety of the vessel. A low freeboard can make a boat more prone to capsizing, while a high freeboard can make it more difficult to board or maneuver.

Is there a standard or recommended freeboard for boats?

There is no standard or recommended freeboard for boats, as it can vary depending on the type and size of the boat, as well as the conditions in which it will be used. However, most boats are designed with a freeboard that provides a good balance of stability and maneuverability.

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