Tension in the string of a submerged object

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The discussion centers on calculating the tension in a string holding an aluminum object submerged in ethyl alcohol. Participants clarify that the tension in the string is determined by the difference between the gravitational force on the object and the buoyant force exerted by the displaced ethyl alcohol. The correct approach involves using Archimedes' principle, which states that the buoyant force equals the weight of the displaced fluid, not the weight of the object itself. One user successfully calculated the tension by applying the formula that incorporates the densities and volumes of both the object and the fluid. The conversation emphasizes the importance of understanding buoyancy in relation to submerged objects.
Liketothink
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Homework Statement


An aluminum object suspended from a string with density 2700 kg/m^3 is submerged into ethyl alcohol.

what is the tension in the string? V=142 cm^3

Homework Equations


vsub/vobj=(ro)object/(ro)ethyl alcohol
p=po+rogh

The Attempt at a Solution


I tried to multiply the volume 142 cm^3 (.000142 m^3) by the density 2700 kg/m^3 to get the mass. Then I multiplied the mass by gravity since Archimedes' principle tells us that Ethyl Alcohol pushes with a force equal to the weight. But that is wrong for some reason. Please help me. Thank you.
 
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What is the weight of the displaced ethyl alcohol?

Wouldn't you subtract it from the free weight of the object?
 
That is correct. But why would you subtract the weight of the liquid from the weight of the object? Is it because the weight I calculated was just the apparent weight?
 
Liketothink said:
I tried to multiply the volume 142 cm^3 (.000142 m^3) by the density 2700 kg/m^3 to get the mass. Then I multiplied the mass by gravity since Archimedes' principle tells us that Ethyl Alcohol pushes with a force equal to the weight.
You calculated the mass of the aluminum object, which you'll need but is not the final answer. Archimedes' principle says that the liquid pushes up with a buoyant force equal to the weight of the displaced liquid, not the weight of the object.
 
I have just tried the quesiton..just going to list what i did..
Fnet = Ftension+ Fbuoyancy-Fgravity=0
Ftension = Fgravity-Fbuoyancy
=> m(object)g - P(fluid)*V(fluid)*g
=>P(object)*V(object)*g - P(fluid)*v(fluid)*g
In my case the aluminum has same density,2700kg/m^3..but i has the volum of 100cm^3
and for completely submerged object V(fluid)= V(object)...which is 1 m^3 in my case
The average value of density for ethyl alcohol i found online is 790 kg/m^2
After plug in everythign that is given and know..i got an answer of 1.87 (round of to 1.9 since 2 sig fig)
Hope that help..
P.S this question is from my mastering physics as well..and i am in first year taking introductory physics...i am using physics for scientists and engineers 2nd edition..u can find some helpful informations from chapter 15(pg456) if you didn't get what i wrote..
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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