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# Fluid Mechanics #2

Posted on 2006-05-20
Medium Priority
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Block A in Figure (14.36) (see: http://www.carbosoft.com/1464.jpg) hangs by a cord from a spring balance D and is submerged in a liquid C contained in a beaker B. the mass of the beaker is 1.00kg; the mass of the liquid is 7.80kg. Balance D reads 3.50kg and balance E reads 13.5kg. Whe volume of the block A is 3.80 x 10^(-3) m^3.

a) what is the density of the liquid?
b) what will each balance read if block is pulled up out of the liquid?

I'm not sure how to do this problem, I would like to get the answer on my own, i'm not sure how to start it. I've created a free body diagram for the block a and found that T + B - mg = 0, T - tension, B - bouyancy. Not sure what to do here, also i know that B = (density of liquid)(volume of object)(gravity constant), How do I figure out the tension? I've assume its 9.8m/s/s * 3.5kg? Also I do not know the mass or density of the object A, I only know its volume...not really sure how to proceed.

For part (b) i'm assuming at this point that the balance E will decrease by 3.5kg? Also i'm going to predict that D will show the true mass of object A. Again both predictions not sure how to start.

-Brian
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Question by:BrianGEFF719
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Expert Comment

ID: 16726641
It may be easier to answer b) first, since E will be unaffected by A or D
and E + D together must account for all the mass
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Author Comment

ID: 16726645
>>and E + D together must account for all the mass

So what you're saying is that the reading on E while the object A is submerged plus the reading on D while the object A is submerged must be the total mass of all the objects?
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Author Comment

ID: 16726650
>>It may be easier to answer b) first, since E will be unaffected by A or D

I dont understand how this could be so: the object A is being pushed up by the liquid, so that means the object is pushing down the liquid, causing a greater force on object E (the bottom scale).

-Brian
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Expert Comment

ID: 16726655
> How do I figure out the tension?
tension is measured by balance D
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Expert Comment

ID: 16726662
b) what will each balance read if block is pulled up out of the liquid?

if the block is pulled up out of the liquid, what can affect the reading on E?
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Author Comment

ID: 16726669
>>if the block is pulled up out of the liquid, what can affect the reading on E?

In this case then E should read Mass of Beaker + Mass of Liquid.

>>b) what will each balance read if block is pulled up out of the liquid?

This should read the TRUE mass of object A, but how do I determine what this is?
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Expert Comment

ID: 16726683
if the block is pulled up out of the liquid, will (E+D) change?
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Author Comment

ID: 16726686
>>if the block is pulled up out of the liquid, will (E+D) change?

I dont see why the masses of those two objects would change.

Here is what i'm thinking though, since A has some weight and pushes downward on the liquid, this must increase the reading on object E. Is this incorrect?
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Author Comment

ID: 16726688
I'm sorry:

>>if the block is pulled up out of the liquid, will (E+D) change?

I'm not sure if the readings of the masses would change or not.

-Brian
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Expert Comment

ID: 16726699
The reading on E can change, and the reading on D can change, but what does
represent?

Or maybe it would be easier to write down all the equations you know, and see if there's any way to combine them to isolate the unknowns.
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Author Comment

ID: 16726705
>>represent?

This should represent the total mass of the system?
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Expert Comment

ID: 16726709
> the object A is being pushed up by the liquid, so that means the object is pushing down the liquid, causing a greater force on object E (the bottom scale).
When A is in the liquid, that is correct.
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Author Comment

ID: 16726719
>>> the object A is being pushed up by the liquid, so that means the object is pushing down the liquid, causing a >>greater force on object E (the bottom scale).
>>When A is in the liquid, that is correct.

So if I was to answer part (b), I need to find what E and D would read. I know what E reads when A is submerged, and E + D should not change when A is out, so I know the masses of the liquid and beaker so I can figure out what E should read, then I can figure out D ( which is the real mass of A ), Is this all correct?

-Brian
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Accepted Solution

ozo earned 2000 total points
ID: 16726736
Yes
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Author Comment

ID: 16726739
Alright thanks ozo, i'm done for the night, I'll come back to this in the morning and see if I can figure it out now that I know the mass of object A.

Thanks Again.
Brian
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Author Comment

ID: 16730247
Hey Ozo, after trying this again, i've found that the mass of object A is 8.2kg, and scales D and E read 8.2kg and 8.8kg.

Then my answer for part A was the density of the liquid was equal to:

density = (m_a * g - T)/(V_a * g)

m_a = mass of a
V_a = volume of a

I've found the density was 1240 kg/m^3, is this correct?

Brian
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Author Comment

ID: 16730687
I dont think that 1240kg/m^3 is correct.
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