Find the weight of a partially submerged cylinder

In summary, the conversation discusses finding the weight of a cylinder using Archimedes' principle. The cylinder has a cross-sectional area of 10cm2 and a length of 50cm, with a portion of it submerged in water. The given values for acceleration due to gravity and water density are 10 m/s2 and 1000 kg/m3, respectively. Using the formula for upthrust and the known values, the volume of the submerged part of the cylinder is calculated to be 500cm3. However, the mass and volume of the entire cylinder remain unknown. To determine these values, the upthrust is set equal to the weight of the cylinder for equilibrium. The conversation also discusses the forces acting on the cylinder
  • #1
Richie Smash
293
15

Homework Statement


Hello, there is a cylinder with cross sectional area 10cm2.
The cylinder has a length of 50cm partially submerged in water and floats upright.
Taking acceleration due to gravity as 10 m s-2 and the density of water to be 1000 kg m-3
Find the weight of the cylinder using archimedes principle.

Homework Equations


Upthrust = volume *density *gravity
Upthrust = mass of water displaced *gravity

The Attempt at a Solution


So, I found the volume of the part of the cylinder that is submerged by multiplying the cross sectional area by 50cm and got 500cm3

From here I converted that that to metres and then used the formula for upthrust involving density and gravitational acceleration and I got 0.05 N

Now I use the knowledge that the upthrust = mass of water displaced * gravity

so 0.05N= Mw*10
Mw= 0.05/10
Mw = 0.005Kg

So now know the mass for that part of the cylinder, but the problem is they didnt inform me of what fraction was submerged, so I don't know how to find the mass for the whole thing.

If I did I would use F=ma and solve.
 
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  • #2
I think this is a statics problem. You need to find the point at which the buoyancy is equal to weight of the cylinder. There has to be a sweet spot where the cylinder is not going up or down.
 
  • #3
Richie Smash said:
The cylinder has a length of 50cm partially submerged
I gather you mean that it has unknown length, but 500cm of it is submerged.
Richie Smash said:
I got 0.05 N
Doesn't seem enough. Check that.
Richie Smash said:
Now I use the knowledge that the upthrust = mass of water displaced * gravity
You already used that. What else must the upthrust equal to achieve steady state?
 
  • #4
The question says that yes the height is unknown but 50 cm is submerged, not 500.

Well to figure out that value I multipled the height that do know, 50cm, by the cross sectional area, giving me 500cm cubed.

I converted that to metres then used the upthrust formula, so did get that tiny value, but its only for piece of the cylinder
 
  • #5
Richie Smash said:
The question says that yes the height is unknown but 50 cm is submerged, not 500.

Well to figure out that value I multipled the height that do know, 50cm, by the cross sectional area, giving me 500cm cubed.

I converted that to metres then used the upthrust formula, so did get that tiny value, but its only for piece of the cylinder
I stand by saying that whatever the buoyant force is, the entire weight of the cylinder. Set a variable and solve for the entire weight, then you wil know all the details.

Richie Smash said:
I got 0.05 N
And be careful when converting units when the units are cubed.
 
  • #6
Richie Smash said:
50 cm is submerged, not 500.
Sorry, typo.
Richie Smash said:
I converted that to metres then used the upthrust formula, so did get that tiny value, but its only for piece of the cylinder
Yes, but I am querying the accuracy of that calculation. Please show the details.

You did not answer this question:
haruspex said:
What else must the upthrust equal to achieve steady state?
 
  • #7
The upthrust must also equal to the volume of water displaced, my calucation was this, 10cm2 is 0.001m2
Multiply that by 50 cm or 0.05 m and you will get 0.0005 so multiply that by 1000 you get 0.5 multiply by 10 you get 5

SO the upthrust is actually 5 Newtons and I'm guessing the volume is 5mcubed for that part of the clinder
 
  • #8
Richie Smash said:
50 cm or 0.05 m
50 cm is 0.5 m
 
  • #9
yes sorry typo
 
  • #10
Richie Smash said:
I'm guessing the volume is 5mcubed for that part of the clinder
You already calculated the volume to find the weight of water displaced, and hence the upthrust.
What else must the upthrust equal for the system to be in equilibrium?
 
  • #11
I'm afraid I'm unsure, perhaps the mass?
 
  • #12
Richie Smash said:
yes sorry typo
First of all, not a typo. You made an error in the entire calculation because of that.
Richie Smash said:
I'm afraid I'm unsure, perhaps the mass?
Second, I have literally told you and @haruspex has repeatedly hinted at it, but you yet keep asking the same question.
 
  • #13
Richie Smash said:
I'm afraid I'm unsure, perhaps the mass?
What are the forces on the cylinder?
 
  • #14
The forces on the cylinder are the force of grvity acting upon it, and the upthrust.

Also I am sure I fixed this calculation now, it should be 50N is the upthrust
 
Last edited:
  • #15
Richie Smash said:
The forces on the cylinder are the force of grvity acting upon it, and the upthrust.
So what equation can you write?
 
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  • #16
I included a diagram here.

The equation I can write is

Mass x gravity = volume * density*gravity

However as you can see in the diagram, this would only be for the section of the cylinder underwater, my problem all along is how do I know figure out the weight, or mass or volume whichever one of the rest on top, because according to my calculation the upthrust is only equal to that section of the cylinder, and my thinking is, I need the height of the entire thing because I only calculated the volume of one section of the cylinder, but they don't give any fractions or hints in this question.
 

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  • #17
Richie Smash said:
The equation I can write is

Mass x gravity = volume * density*gravity
No, just an equation connecting the forces on the cylinder.
 
  • #18
Ok it will be,

Fb=mg
 
  • #19
Richie Smash said:
Ok it will be,

Fb=mg
It is however the force of buoyancy (which is only on part of the cylinder) compared to the ##mg## which is of the whole cylinder.
 
  • #20
So put it all together:
Richie Smash said:
Find the weight of the cylinder
Richie Smash said:
50N is the upthrust
Richie Smash said:
Fb=mg
 
  • #21
Richie Smash said:

Homework Statement


Hello, there is a cylinder with cross sectional area 10cm2.
The cylinder has a length of 50cm partially submerged in water and floats upright.
Taking acceleration due to gravity as 10 m s-2 and the density of water to be 1000 kg m-3
Find the weight of the cylinder using Archimedes Principle.
2. Homework Equations

...
Please state Archimedes Principle .
 
  • #22
Archimedes principle states that the upthrust on a partially or fully submerged object in a fluid is equal to the weight of the fluid that is displaced.
Well I can figure out that using
Fb=mg

The mass of the cylinder I have found would be 5kg. I know I'm missing something vital here... but it just is not adding up to me.

As Lekh stated in this equation, the ''Fb'' term would be the force only on part of the cylinder, so using the formula, would the 5kg mass I have computed = to the entire mass or just part of the mass, if it the entire mass then problem solved but if not...
 
  • #23
Richie Smash said:
Well I can figure out that using
Fb=mg
That is ambiguous because there are two masses.

To be clear, you successfully used Archimedes' principle (some numerical errors aside) to arrive at
Fb = mwater displaced g

The equation it took so long to add was Newton's ΣF=ma:

Fb - mfloating object g = 0

Richie Smash said:
The mass of the cylinder I have found would be 5kg
Yes, but you are asked for the weight, not the mass.
 
  • #24
SO the weight is 50N?
 
  • #25
Richie Smash said:
SO the weight is 50N?
Assuming the mass is ##5 kg## and the acceleration due to gravity is ##10m/s^2##
 
Last edited:
  • #26
SammyS said:
Units?
Sorry, I fixed it.
 

Related to Find the weight of a partially submerged cylinder

1. What is the equation for finding the weight of a partially submerged cylinder?

The equation for finding the weight of a partially submerged cylinder is W = ρVg, where W is the weight, ρ is the density of the liquid, V is the volume of the cylinder, and g is the acceleration due to gravity.

2. How can I determine the volume of a partially submerged cylinder?

You can determine the volume of a partially submerged cylinder by multiplying the cross-sectional area of the cylinder by its height above the liquid level. This will give you the volume of the portion of the cylinder that is submerged in the liquid.

3. Is the weight of a partially submerged cylinder affected by the density of the liquid?

Yes, the weight of a partially submerged cylinder is affected by the density of the liquid. The weight is directly proportional to the density, so a higher density liquid will result in a greater weight for the cylinder.

4. Can I use the same equation to find the weight of a fully submerged cylinder?

No, the equation for finding the weight of a partially submerged cylinder cannot be used for a fully submerged cylinder. This is because a fully submerged cylinder will displace its entire volume of liquid, whereas a partially submerged cylinder only displaces a portion of its volume.

5. How does the shape of the cylinder affect its weight when partially submerged?

The shape of the cylinder does not affect its weight when partially submerged. As long as the cross-sectional area and height of the cylinder are the same, the weight will remain constant regardless of the shape of the cylinder.

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