Archimedes Principle and hanging block

[pat666]

Homework Statement

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?

Homework Equations

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 ??

 

[Doc Al]

not sure on a) i found the up thrust to be 8.2KG?

Show how you determined the buoyant force.

 

[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

 

[Doc Al]

first i found the mass of just the block that was read on scale A which was (7.5-1-1.8)

That won't give you the mass of the block.

then i added the tension that was supported by the spring which was 3.5kg. that came out to 8.2Kg

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?

 

[pat666]

yeah i understand that it won't 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 ??

 

[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 that's 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 I am doing is right.. thanks

 

[Doc Al]

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 that's 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 I am doing is right.. thanks

Much better!

 

[pat666]

ok thank for your help
so the spring balance will read 8.2Kg and the scale will read 2.8Kg?

 

[Doc Al]

Right.

 

[pat666]

cool thanks again

 

[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

 

[Doc Al]

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

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.)

 

[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 I am missing something.

 

[Doc Al]

for this to be true the values used for this question must be wrong... quittteeee wrong...

Why is that?

for this to be true the volume of liquid in the beaker would be 1.45*10^-3, much smaller than the cube...

The volume of the cube is 4.7/1.8 = 2.6 times the volume of the liquid in the beaker. So?

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 I am missing something.

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.

 

[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 a lot wrong with the values. Perhaps u can explain to me how this scenario could ever exist in reality?

 

[Doc Al]

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 a lot wrong with the values. Perhaps u can explain to me how this scenario could ever exist in reality?

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 into fill the beaker.

 

[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.

 

[Doc Al]

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.

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.

 

[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?

 

[Doc Al]

yerrr but the waters not there so the buoyant force is not there so the reading on the scale never happened.

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"

 

[ombudsmansect]

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

 

[Doc Al]

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

That's Archimedes' principle. See: http://hyperphysics.phy-astr.gsu.edu/hbase/pbuoy.html#arch3"

 

[ombudsmansect]

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

 

[Doc Al]

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

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.

 

[ombudsmansect]

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

 

[Doc Al]

Much about fluid pressure is counterintuitive. You might like to take a look at this old thread: https://www.physicsforums.com/showthread.php?t=127942"