Archimedes Principle and steel bar question

[Ben_Walker1978]

Homework Statement

Can someone tell me if i have done this correct please?

A steel bar of length 0.6m and diameter 70mm is suspended from a chain and lowered into a tank of liquid whose relative density is 0.9. Determine the tension in the chain when the bar is fully immersed. The density of the water is 1000kg/m³ and the density of steel is 7870kg/m³.

Homework Equations

Fb = ΡF Vd g

T = mg - Fb

The Attempt at a Solution

 

[BvU]

Hi Ben,

Can someone tell me if i have done this correct please?

That's not really what PF is about. Your teacher is the one who grades the homework (and he/she gets paid a little bit for doing that).
What we can do is help you build sufficient self-confidence, so you can convince yourself that's your work is correct.

My question in return, therefore, is: do you see any weaknesses in your working ?

But PF IS the place for comments, like the following:
You have been given values in one , two or three digits accuracy. In such a case it does not make sense to answer in 10 digits accuracy.

 

[Ben_Walker1978]

Hello,

Sorry i thought it was for things like this.

I thought it was correct.

I was just unsure with the relative density of 0.9. Was not to sure what to do with the relative density.

I thought the relative density could affect the final Tension.

Thank you for the replys.

 

[BvU]

Excellent !
$\rho_F$ is the density of the fluid.
What is the density of the fluid that you used in the calculation ?

 

[Ben_Walker1978]

1000Kg/m^3 which is water.

 

[stockzahn]

1000Kg/m^3 which is water.

The relative density gives a relation between the density of the used liquid $\rho_f$ and a reference liquid (which probably is water, which is why the density of water is stated in the question). The relationship reads

$$rel\;\rho =\frac{\rho_f}{\rho_{ref}}$$

Therefore I probably misread the question the first time, my bad, I'm sorry. Your method to calculate is basically correct, but the density of the liquid in which the bar is immersed is different to the one of water. You probably have to calculate it with the relation given above.

 

[Ben_Walker1978]

Hello,

Thats ok.

So instead of using 100Kg/m^3 in, Fb = ΡF Vd g.

Use 1111.11 ? I got this by 1000/0.9.

Thanks.

 

[stockzahn]

So instead of using 100Kg/m^3 in, Fb = ΡF Vd g.
Use 1111.11 ? I got this by 1000/0.9.

I don't think so. Water should be the reference liquid with the reference density $\rho_{ref}$.

 

[Ben_Walker1978]

Ok. I don't really understand what you mean.

Thanks.

 

[BvU]

If the relative density is 0.9, that means the density is 0.9 times that of water, so 900 kg/m³

 

[Ben_Walker1978]

If the relative density is 0.9, that means the density is 0.9 times that of water, so 900 kg/m³

So times not divide like i did?

So does that mean i put 900Kg/m^3 instead of 1000Kg/m^3 ?

I have watched a few YouTube videos, but none show how to use relative density, like in my question.

This is why i got stuck.

Thank you.

 

[BvU]

So times not divide like i did? Yes

So does that mean i put 900Kg/m^3 instead of 1000Kg/m^3 ? Yes

 

[Ben_Walker1978]

Thank you. Really appreciate your help.

I had a feeling i had got that part wrong. But could not find anywhere that shown me how to apply relative density.

Thanks.