# Homework Help: Archimedes Principle and hanging block

1. May 28, 2010

### pat666

1. The problem statement, all variables and given/known data
9. Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.00 kg; the mass of the liquid is 1.80 kg. Balance D reads 3.50 kg, and balance E reads 7.50 kg. The volume of block A is 3.80  10-3 m3.
(a) What is the density of the liquid?
(b) What will each balance read if block A is pulled up out of the liquid?
2. Relevant equations

3. The attempt at a solution
not sure on a) i found the up thrust to be 8.2KG?????? then that is equal to the mass t of liquid displaced. rho=m/v i found the density to be 2158Kg/m^3
Part b was easier- still need someone to check
m_D=7.5-1-1.8
m_D=4.7Kg

m_A=7.5-4.7
m_A=2.8Kg ?????????????????????????????

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2. May 29, 2010

### Staff: Mentor

Show how you determined the buoyant force.

3. May 29, 2010

### pat666

first i found the mass of just the block that was read on scale A which was (7.5-1-1.8) then i added the tension that was supported by the spring which was 3.5kg. that came out to 8.2Kg

4. May 29, 2010

### Staff: Mentor

That won't give you the mass of the block.
Hint: The amount of the block's weight that the scale reads equals its mass minus the amount supported by the spring. What does that represent?

5. May 29, 2010

### pat666

yeah i understand that it wont give the actual weight of the block (thats why my part b) has a different answer) I thought that what i did would give the "apparent weight" but now i think that that is wrong... should it be
7.5-2.8-3.5=1.2Kg ???????

6. May 30, 2010

### pat666

the actual weight of the block is D + E = Mbeaker + M liquid + Mblock
Mblock = 11 - 2.8 = 8.2kg
that should give the actual weight of the block??
so then the upthrust is 8.2kg-3.5kg=4.7Kg
then thats equal to the weight of the water displaced
form there rho = m/v
rho=1236.84kg/m^3?????????????????? please reply when you have a chance i need to know if what im doing is right.. thanks

7. May 30, 2010

### Staff: Mentor

Much better!

8. May 30, 2010

### pat666

so the spring balance will read 8.2Kg and the scale will read 2.8Kg?

9. May 30, 2010

### Staff: Mentor

Right.

10. May 30, 2010

### pat666

cool thanks again

11. May 31, 2010

### ombudsmansect

Hey guys,

can someone explain to me how the weight or liquid displaced is 4.7kg when the total weight of the liquid is 1.8kg?

cheers

12. May 31, 2010

### Staff: Mentor

The volume of liquid displaced equals the volume of the object. What's the weight of that volume of displaced liquid? (The actual amount of liquid in the beaker is not relevant. The diagram may not be to scale.)

13. May 31, 2010

### ombudsmansect

for this to be true the values used for this question must be wrong... quittteeee wrong...for this to be true the volume of liquid in the beaker would be 1.45*10^-3, much smaller than the cube...i know this solution must be correct unless anything pops out at u...I am just ata loss as to how such a huge conceptual error could b made...makes me think im missing something.

14. May 31, 2010

### Staff: Mentor

Why is that?
The volume of the cube is 4.7/1.8 = 2.6 times the volume of the liquid in the beaker. So?
Had I written the problem, I would not have chosen those values. Nonetheless, there's nothing wrong with them. You are thinking that to 'displace' a given volume of liquid means you must have at least that much liquid to start with. No, that's not what is meant by 'displace'.

Imagine the beaker was empty. Dangle the object into the empty beaker, then pour the liquid around it.

15. May 31, 2010

### ombudsmansect

I think for it to be a valid set question the answer should at least be possible. I guess the real answer to this question is 'This is not possible'. How can the cube displace a volume of liquid that is equal to its own volume when there is not that much liquid to begin with? The question specifically states it does this and specifically states all other values that make it impossible. The set parameters do not exist nor can ever exist given the values and cirumstances quoted. Having said this there is alot wrong with the values. Perhaps u can explain to me how this scenario could ever exist in reality?

16. May 31, 2010

### Staff: Mentor

There's nothing 'not possible' about this scenario. You just need to understand what "amount of fluid displaced" means when discussing Archimedes' principle and buoyancy.

Example. Imagine the beaker was a cube with sides of length 10 cm. I lower a solid cube with sides of length 9.5 cm into the beaker, dangling it from a string. Now I pour liquid into the beaker, filling it to the top. The amount of water "displaced" by the solid will equal the volume of the solid. That's clearly much greater than the volume of liquid that you had to pour in to fill the beaker.

17. May 31, 2010

### ombudsmansect

then how can that amount of water be displaced since it does not physically exist in the given scenario. i understand that it does 'displace' the liquid by a certain volume. but it does not displace a certain volume of liquid tht is equal to itself since it is not there. so i think that by displace they mean it in the latter sense unless u can tell me otherwise? please do if u knw it is true but i do not think so so far.

18. May 31, 2010

### Staff: Mentor

Think of "displace" as follows. When the object is submerged, the liquid level is at a certain height. If you now remove the object, how much liquid do you have to add to bring the level back to where it was when the object was submerged? That's the amount of liquid displaced.

This is the meaning of "displaced liquid" in Archimedes' principle.

19. May 31, 2010

### ombudsmansect

yerrr but the waters not there so the buoyant force is not there so the reading on the scale never happened. if your sure about that meaning then ok. but i reckon the molecules actually have to be there to create the pressure and that those molecules are the ones physically present. i see what u mean however and perhaps u can post a link to a source that verifies this?

20. May 31, 2010

### Staff: Mentor

Buoyant force is due to fluid pressure. Fluid pressure depends on the depth of the fluid, not on how much fluid there is.

See: http://hyperphysics.phy-astr.gsu.edu/hbase/pflu.html#fp"

Last edited by a moderator: Apr 25, 2017
21. May 31, 2010

### ombudsmansect

then why is the buoyant force also equal to density*gravity*volume?

22. May 31, 2010

### Staff: Mentor

Last edited by a moderator: Apr 25, 2017
23. May 31, 2010

### ombudsmansect

being the buoyant force how can these two different methods give the same units. one is by height, the other volume.

24. May 31, 2010

### Staff: Mentor

Pressure at some depth below the surface of a fluid is given by ρgh. You can use the pressure surrounding the object to calculate the buoyant force. Archimedes' principle tells us that the answer will equal ρgV.

25. May 31, 2010

### ombudsmansect

interesting. thanks for taking the time to explain. it is just hard to picture a skin tight beaker around the cube to accomodate the tiny amount of water present. it annoys me when unrealistic diagrams accompany a question.