Give a symbolic expression for submerging a metal in water

1. Feb 18, 2012

Anna55

In the lab you place a beaker that is half full of water (density pw) on a scale. You now use a light string to suspend a piece of metal of volume V in the water. The metal is completely submerged, and none of the water spills out of the beaker.

Give a symbolic expression for the change in reading of the scale. Express your answer in terms of the variables pw, V, and appropriate constants.

relevant equations
Fb=pVg

solution
Ww+Wm-Ww=
Wm

The metal is submerged, thus the volume of the metal is the volume of the displaced water. The metal also experience buoyancy force.

2. Feb 18, 2012

I like Serena

Welcome to PF, Anna55!

In your solution you subtract Ww.
But shouldn't that be the buoyancy Fb?
And what happened to the tensional force in the string?

Can I assume that:
Ww is the weight of the water in the beaker,
Wm is the weight of the metal,
Fb is the buoyancy?

In that case I'm still missing the tensional force on the string.
Shall we call it Ft?

And to set up a system that is in equilibrium, we also need the upward force of the scale.
Shall we call that Fs?
This is the quantity that you would measure with a reading on your scale.

So before the submersion, the weight of the system was Ww.
For equilibrium that means that Fs=Ww.

Can you set up the equilibrium of the forces after the submersion of the metal?

3. Feb 19, 2012

Anna55

Yes, the symbols are correct.

Fs+Fb+Ft=Ww

Fs+Fb+Ft-Ww=0

Fs+Fb+Ft-Ww-(Fs-Ww)=
Fs+Fb+Ft-Ww-Fs+Ww=
Fb+Ft=
pwVdg+Ft=

Is this correct? How should I continue?

4. Feb 19, 2012

I like Serena

Hmm, this is not quite right yet, but I'm at a loss right now how to explain.

Let's try to look at this another way.

Suppose the metal had exactly the same density as the water.
Then the weight of the metal would equal the buoyancy force, and the tensional force in the string would be zero.
Furthermore, the level of the water in the beaker would rise, accounting for the amount of displaced water.

What would the reading of the scale indicate in this case?

5. Feb 19, 2012

Anna55

Fb=Wm
pw+Vd+g=mg
pm+Vd+g=mg

pm+Vd+g

6. Feb 19, 2012

LawrenceC

Draw a free body diagram of the beaker sitting on the scale with the metal suspended by the string fully submerged in the beaker.

Write an expression for the tension in the string in terms of densities and volumes.
Write another expression for the reading the scale would have based on your free body diagram of the situation. This equation should have 3 force terms that comprise the scale reading. One of the terms would be the beaker without the metal in it.

Rearrange the equation so that you have what you seek on one side of the equation.

7. Feb 19, 2012

I like Serena

So in this case (metal has equal density as the water) the added weight is $\rho_m V_d g = \rho_w V_d g$.

Now if the metal has a higher density, the buoyancy force remains the same, the weight of the metal increases, but the difference is compensated by the tensional force in the string.

8. Feb 19, 2012

Anna55

Before the metal is submerged in the water.
Ft=W

Submerged metal
Fb+Ft=W
Fb+Ft-W=
(They cancel out)
Fb=
pwVdg=
pwVmg

To summarize everything. Is this correct?

9. Feb 19, 2012

I like Serena

Assuming you meant Wm where you wrote W, this is true.

Again assuming you meant Wm where you wrote W, this is also true.
Note that Ft has a different value than before.

Assuming that with Vm you mean the volume of the metal, which is the same as Vd, the volume of the displaced water, this is also correct.

As yet, you have not summarized everything.
You did not specify the effect of the metal-on-a-string on the reading of the scale.

Last edited: Feb 19, 2012
10. Feb 19, 2012

Anna55

Sorry for not writing what the variables stand for in the beginning.
Yes, Wm is the weight of the metal.
Yes, Vm is the volume of the metal.

The change in reading of the scale will be pwVmg.

11. Feb 19, 2012

I like Serena

Okay! :)

12. Feb 19, 2012

Anna55

Thanks for all the help I have gotten! :) I really appreciate it. :)