Give a symbolic expression for submerging a metal in water

[Anna55]

In the lab you place a beaker that is half full of water (density p_w) 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
F_b=pVg

solution
W_w+W_m-W_w=
W_m

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

 

[I like Serena]

In the lab you place a beaker that is half full of water (density p_w) 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
F_b=pVg

solution
W_w+W_m-W_w=
W_m

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

Welcome to PF, Anna55! In your solution you subtract W_w.
But shouldn't that be the buoyancy F_b?
And what happened to the tensional force in the string?Let's start with the definitions of your symbols.
Can I assume that:
W_w is the weight of the water in the beaker,
W_m is the weight of the metal,
F_b is the buoyancy?

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

And to set up a system that is in equilibrium, we also need the upward force of the scale.
Shall we call that F_s?
This is the quantity that you would measure with a reading on your scale.So before the submersion, the weight of the system was W_w.
For equilibrium that means that F_s=W_w.
Can you set up the equilibrium of the forces after the submersion of the metal?

 

[Anna55]

Yes, the symbols are correct.

F_s+F_b+F_t=W_w

F_s+F_b+F_t-W_w=0

F_s+F_b+F_t-W_w-(F_s-W_w)=
F_s+F_b+F_t-W_w-F_s+W_w=
F_b+F_t=
p_wV_dg+F_t=

Is this correct? How should I continue?

 

[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?

 

[Anna55]

F_b=W_m
p_w+V_d+g=mg
p_m+V_d+g=mg

p_m+V_d+g

 

[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.

 

[I like Serena]

Yes, that looks about right, although you should multiply instead of add.
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.

 

[Anna55]

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

Submerged metal
F_b+F_t=W
F_b+F_t-W=
(They cancel out)
F_b=
p_wV_dg=
p_wV_mg

To summarize everything. Is this correct?

 

[I like Serena]

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

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

Submerged metal
F_b+F_t=W
F_b+F_t-W=
(They cancel out)

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

F_b=
p_wV_dg=
p_wV_mg

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

To summarize everything. Is this 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.

 

[Anna55]

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

The change in reading of the scale will be p_wV_mg.

 

[I like Serena]

Okay! :)

 

[Anna55]

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