Physics Problem on Tension in Water

[vdance]

Homework Statement

Out of interest, I encountered a physics problem and am unsure how to calculate it.
As shown in the figure, an outer cylinder and an inner cylinder are connected by a vertical sliding seal. When the top of the outer cylinder descends under gravity from 6 meters to 5.5 meters, how is the tension force Fcalculated? (Ignore friction).

Relevant Equations

G=mg=ρshg

F₁=(5000+200)g=5200*10=52000N --> F₂=(5000+100)g=5100*10=51000N ?

 

[haruspex]

Homework Statement: Out of interest, I encountered a physics problem and am unsure how to calculate it.
As shown in the figure, an outer cylinder and an inner cylinder are connected by a vertical sliding seal. When the top of the outer cylinder descends under gravity from 6 meters to 5.5 meters, how is the tension force Fcalculated? (Ignore friction).
Relevant Equations: G=mg=ρshg

F₁=(5000+200)g=5200*10=52000N --> F₂=(5000+100)g=5100*10=51000N ?

You might find it clearest to think in terms of energy. If the cylinder descends a distance $\Delta y$, what is the change in PE? Draw the region where the water goes from and where it goes to.

 

[vdance]

You might find it clearest to think in terms of energy. If the cylinder descends a distance $\Delta y$, what is the change in PE? Draw the region where the water goes from and where it goes to.

If calculated based on the gravitational force of water, the mass of water inside the tube decreased by 100 kg after descent. If calculated based on liquid pressure, should the force-bearing area be the top area (1 - 0.8 = 0.2 m²) or the bottom drainage area (1 m²)? I don't understand.

 

[haruspex]

the mass of water inside the tube decreased by 100 kg after descent

Yes, but sketch where that water has come from. And where it has gone to. What is the change in PE?

Btw, the force will not be constant.

 

[vdance]

Yes, but sketch where that water has come from. And where it has gone to. What is the change in PE?

100 kg of water descends from the top and flows out from the bottom of the outer cylinder into a pool. According to the formula F=ρghS, should the value of Sbe 0.2 m² or 1 m²? I am unclear about this. I believe the force-bearing area at the bottom drainage outlet is 1 m², so it is reasonable to use S=1m². I wonder if this is correct.

 

[haruspex]

According to the formula F=ρghS

Yes, but I am giving you another way to figure this out that's clearer.
I asked how much water disappears from the top if the cylinder only descends a small distance $\Delta y$, not the whole 0.5m. There is a reason for that, as you will see.
Looking at where extra water appears at the bottom, how far, in effect, has that water descended? How much PE has it lost?

 

[vdance]

Yes, but I am giving you another way to figure this out that's clearer.
I asked how much water disappears from the top if the cylinder only descends a small distance $\Delta y$, not the whole 0.5m. There is a reason for that, as you will see.
Looking at where extra water appears at the bottom, how far, in effect, has that water descended? How much PE has it lost?

I can't tell whether it goes from A->B+C->D or from E->F.

 

[haruspex]

View attachment 365044
I can't tell whether it goes from A->B+C->D or from E->F.

You just have to draw the before and after pictures. You can consider the water as having gone from the volume that is occupied in the before picture but not in the after picture to the volume occupied in the after picture but not in the before picture.

 

[jbriggs444]

can't tell whether it goes from A->B+C->D or from E->F.

The author has not given us the radius of the basin at the bottom. This is a hint that the radius is large enough to be irrelevant. From this we can conclude that the fluid level in the basin will rise slightly, but not significantly as a result of the process.

 

[kuruman]

The author has not given us the radius of the basin at the bottom. This is a hint that the radius is large enough to be irrelevant. From this we can conclude that the fluid level in the basin will rise slightly, but not significantly as a result of the process.

I interpret the two numbers 1 m² and 0.8 m² (left top) to be the areas of the cylinders. One may calculate radii from these. Units are useful after all.

One would think that a detailed figure like this would be drawn to scale. There appears to be no change in the fluid level in the basin (see figure below.) That could be confusing as one might legitimately ask, where did the water go?

 

[Lnewqban]

... how is the tension force F calculated?

F₁=(5000+200)g=5200*10=52000N --> F₂=(5000+100)g=5100*10=51000N ?

Isn't the force vector represented in the OP diagram pointing in the wrong direction?

The outer cylinder would naturally tend to fall induced by the weight of the water that it contains, unless a lifting force is keeping it at the original height.
In that case, the calculated value of that force should be negative, it seems.

Once the outer cylinder reaches its lowest position, resting on the inner cylinder and central support frame, no lifting external force would be necessary any longer.

 

[haruspex]

I interpret the two numbers 1 m² and 0.8 m² (left top) to be the areas of the cylinders. One may calculate radii from these. Units are useful after all.

Yes, I think we all read it that way. @jbriggs444's point is that we have no info re the surface area of the reservoir, other than that the level is not shown as having risen.

One would think that a detailed figure like this would be drawn to scale. There appears to be no change in the fluid level in the basin (see figure below.) That could be confusing as one might legitimately ask, where did the water go?

Maybe it's not circular - infinite into the page perhaps .

 

[haruspex]

Isn't the force vector represented in the OP diagram pointing in the wrong direction?

I agree that is confusing, but since the question asks for a tension there was, presumably, some mention of a rope in the original text. Thus, F might be illustrating the force on the rope, not on the cylinder.

 

[vdance]

The author has not given us the radius of the basin at the bottom. This is a hint that the radius is large enough to be irrelevant. From this we can conclude that the fluid level in the basin will rise slightly, but not significantly as a result of the process.

For the sake of analysis, the water level in the pool can be considered constant, much like the water level in a lake.

 

[vdance]

I agree that is confusing, but since the question asks for a tension there was, presumably, some mention of a rope in the original text. Thus, F might be illustrating the force on the rope, not on the cylinder.

I do want to know the tensile force exerted on the rope when the outer cylinder descends; F can be considered as the tensile force of the rope.

 

[haruspex]

I do want to know the tensile force exerted on the rope when the outer cylinder descends; F can be considered as the tensile force of the rope.

Ok, so can you do what I suggested in post #8? Do I need to explain more?

 

[vdance]

Ok, so can you do what I suggested in post #8? Do I need to explain more?

F=ρghS=1000*10*5.5*1=55000N ？
I believe the result you should calculate is F=G=ρghS=1000*10*5.5*0.2=11000N.​

 

[haruspex]

I believe the result you should calculate is F=G=ρghS=1000*10*5.5*0.2=11000N.

Yes, except that, as I wrote, the force is not constant. What you have calculated is the average. Is that what you are asked for?

 

[vdance]

Yes, except that, as I wrote, the force is not constant. What you have calculated is the average. Is that what you are asked for?

Yes, that's the answer I was looking for. You've taught me enough to understand it at this point. Thank you so much for helping me solve this problem.

 

[vdance]

Thank you all very much for your attention and assistance in answering my questions.