Tension in the string of a submerged object

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Homework Help Overview

The discussion revolves around calculating the tension in a string holding an aluminum object submerged in ethyl alcohol. The object has a given density and volume, and participants are exploring the implications of buoyancy and weight in this context.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of the object's weight and the buoyant force acting on it. Questions arise regarding the relationship between the weight of the displaced liquid and the apparent weight of the object. Some participants attempt to clarify the application of Archimedes' principle and the necessary calculations for tension.

Discussion Status

There is an ongoing exploration of the concepts involved, with some participants providing insights into the calculations and principles at play. While attempts have been made to derive the tension, there is no explicit consensus on the final approach or answer yet.

Contextual Notes

Participants note the importance of understanding the buoyant force in relation to the weight of the object and the displaced fluid. There is mention of specific values for densities and volumes, as well as references to educational materials that may provide additional context.

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..
 

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