Calculating the force required to push a steel rod up 50 cm into a water filled steel tank

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Hello ,all
I'm calculating the force that required to push steel rod up 50 cm into water filled steel tank . pls see picture . If i exclude negligible friction force (which is friction between steel rod and water sealing gaskets )
Total Force = Force to lift 40 kg rod + Fluid Force exerted on horizontal surface of rod
so, 1. Force to lift 40 kg rod = 40 x 9.81 = 392.4 N
2 . Fluid Force exerted on horizontal surface of rod - F= PxA
- hydrostatic pressure of 15 m high water P = 147000 N/m2
- surface area of 10 cm diameter rod A = 0.00785 m2
F= 147000 x 0.00785 = 1154 N
Total Force = 392.4 +1154 = 1546.4 N
pls help me , is this correct ? any other forces missed into calculation ?

[Moderator's note: moved from a technical forum.]
 

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  • #2
haruspex
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Hello ,all
I'm calculating the force that required to push steel rod up 50 cm into water filled steel tank . pls see picture . If i exclude negligible friction force (which is friction between steel rod and water sealing gaskets )
Total Force = Force to lift 40 kg rod + Fluid Force exerted on horizontal surface of rod
so, 1. Force to lift 40 kg rod = 40 x 9.81 = 392.4 N
2 . Fluid Force exerted on horizontal surface of rod - F= PxA
- hydrostatic pressure of 15 m high water P = 147000 N/m2
- surface area of 10 cm diameter rod A = 0.00785 m2
F= 147000 x 0.00785 = 1154 N
Total Force = 392.4 +1154 = 1546.4 N
pls tell me , is this correct ? any other forces missed into calculation ?

[Moderator's note: moved from a technical forum.]
Looks right. Of course, the hydrostatic pressure on the top of the rod will reduce as the rod rises, but maybe it is short compared to the water depth, or maybe you only care about the max force.
 
  • #3
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Looks right. Of course, the hydrostatic pressure on the top of the rod will reduce as the rod rises, but maybe it is short compared to the water depth, or maybe you only care about the max force.
Thank you Haruspex, i still wonder about water displacement by inserted rod volume. almost 4 liter water displacement and water level rise up by 3 mm in 40 cm diameter water tube tank . and i confusing on , may be more force need to lift such 4 liter water up 15 meter high ?
 
  • #4
Lnewqban
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Thank you Haruspex, i still wonder about water displacement by inserted rod volume. almost 4 liter water displacement and water level rise up by 3 mm in 40 cm diameter water tube tank . and i confusing on , may be more force need to lift such 4 liter water up 15 meter high ?
The work done on the water (just like a piston pump would do) is dedicated to push the mass of water in 3.93 liters of water up only 31.25 mm rather than 15 m.
The rest of the mass of water remains where it was before inserting the rod; therefore, no work is done on it.

Welcome! :)

(Edit: Correcting values).
 
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  • #5
Steve4Physics
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Hi @ucody0911. Can I add to what @haruspex and @Lnewqban have said.

The wording in the question is unclear. The force needed changes as the rod is moved upwards. Maybe you are being asked what force is needed to keep the rod in its final equlilibrium position with 50cm of the rod in the water.

If so, that changes the pressure you need to use in your calculation (how deep is the rod’s top face?).

@Lnewqban, with 50cm inserted, I get the water level rising by about 3cm, not 3mm. If an approximate answer is acceptable, this rise in water level is still small enough to be ignored as it will only change the pressure slightly. (If an accurate answer is needed, the rise in water level can be used when calculating the pressure.)

@ucody0911, does it make sense to give the final answer to 5 significant figures? And a minor point: you don’t mean ‘exclude negligible friction’. That’s a sort of double negative!
 
  • #6
Lnewqban
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@Lnewqban, with 50cm inserted, I get the water level rising by about 3cm, not 3mm. If an approximate answer is acceptable, this rise in water level is still small enough to be ignored as it will only change the pressure slightly. (If an accurate answer is needed, the rise in water level can be used when calculating the pressure.)
Thank you, Steve.
Yes, the level in the tank should go up 31.25 mm when the rod is inserted those 500 mm.

To the OP: The amount of energy or work into the fluid only should be the weight of the displaced volume of water (3.93 dm^3) times that rising height (0.31 dm).
 
  • #7
Steve4Physics
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To the OP: The amount of energy or work into the fluid only should be the weight of the displaced volume of water (3.93 dm^3) times that rising height (0.31 dm).
Note that the problem can be solved wihtout using work/energy.

I assume the question wants the upwards force exerted on the rod to maintain equilibrium when 50cm of the rod is in the water.

Once we have the distance (d) between the surface of the water and the top face of the rod, the required force is simply the rod’s weight plus the hydrostatic force on the rod’s top face (which is at depth d).
 
  • #8
Lnewqban
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Note that the problem can be solved wihtout using work/energy.
Please, see quotation in post #4.
This seems to be a practical problem that the OP is trying to figure out rather than a homework.
 
  • #9
Steve4Physics
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The amount of energy or work into the fluid only should be the weight of the displaced volume of water (3.93 dm^3) times that rising height (0.31 dm).
It's worth noting that the displaced water is effectively shifted from near the bottom, to the surface (adding 31.25mm to the surface height). The height-rise of the displaced water's centre of gravity is therefore much more than 31.25mm.
 
  • #10
haruspex
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The work done on the water (just like a piston pump would do) is dedicated to push the mass of water in 3.93 liters of water up only 31.25 mm rather than 15 m.
The rest of the mass of water remains where it was before inserting the rod; therefore, no work is done on it.

Welcome! :)

(Edit: Correcting values).
I wrote that the hydrostatic pressure at the top of the rod reduces. Suppose the rod were inserted 16m. That would mean that it would just about reach the surface when fully inserted, so the gauge pressure reduces linearly to zero over that distance. Hence, inserting it 50cm should reduce it by about 3%.

You should get the same result using work done. The diameter ratio is 4:1, so the X-section ratio is 16:1. The rising rod reduces the water's available area by 1/16. Hence the water surface will rise up to 1m. And as @Steve4Physics notes, that 1m has been lifted by a lot more than 1m.
 
  • #11
jbriggs444
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To the OP: The amount of energy or work into the fluid only should be the weight of the displaced volume of water (3.93 dm^3) times that rising height (0.31 dm).
Wait, that cannot be right. We've raised that displaced water from its original position (centroid of the displaced cylindrical hole) to its new position (centroid of new location at 0.155 dm above prior surface).

Edit: Scooped by @Steve4Physics as noted by @haruspex
 
  • #12
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thank you all , question is not for force to keep the rod in its final equlilibrium position with 50cm of the rod in the water , it is for force required to push rod up ( 0-50 cm distance ) . so
1. Max force = 1546.4 N when start 0 cm distance ( as calculated above)
Min force = 1507.4 N when final 50 cm distance ,
is it right ?
2. 1546.4 N is for closed top end of water tank , if open end top, atmospheric pressure (is equal to 101325 pascal ) added to hydrostatic pressure and total pressure is 101325 + 147000 = 248325 pa , so
will be Max force = 2341.7 N ?

3. indicating as W1 - The work done to push the mass of water , 3.93 liters of water up 31.25 mm
indicating as W2 - The work done to push the rod up 50 cm
so total work = W1+W2 or W2-W1 ? or total work is just W2 ?
 
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  • #13
haruspex
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Min force = 1507.4 N when final 50 cm distance ,
My analysis in post #10 says it should reduce by the fraction 1/32. That would make it a bit under 1500. Please post the details of your calculation.
1546.4 N is for closed top end of water tank , if open end top, atmospheric pressure (is equal to 101325 pascal ) added
I wouid think the rod is also subject to atmospheric pressure below, so you only need to consider "gauge" pressure.
 
  • #14
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The work done on the water (just like a piston pump would do) is dedicated to push the mass of water in 3.93 liters of water up only 31.25 mm rather than 15 m.
The rest of the mass of water remains where it was before inserting the rod; therefore, no work is done on it.

Welcome! :)

(Edit: Correcting values).
if this force is dedicated to push the mass of water , 1546.4 N is too big for 3.93 liters of water up only 31.25 mm . i think , this force is to lift up 15 meter of 3.93 l water OR lift up 31.25 mm of total mass of water in tank . because :
accurately, water displacement is 3.92699 liter , total mass in water tank is 1884.96 kg
as W=fxd , work1 = 3.92699 x 9.8 x 15 m = 577.26 J
work2 = 1884.96 x 9.8 x 31.25 mm = 577.26 J
same amount of work .
but , by this formula , work done on pushed rod is : 1546.4 N x 0.5 m = 773 J , not equal to 577.26 J
again still confused :(
 
  • #15
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My analysis in post #10 says it should reduce by the fraction 1/32. That would make it a bit under 1500. Please post the details of your calculation.

I wouid think the rod is also subject to atmospheric pressure below, so you only need to consider "gauge" pressure.
calculated hydrostatic pressure of 14.5 meter , ( final position of 50 cm inserted rod ) as P = 142100 N/m2
- surface area of 10 cm diameter rod A = 0.00785 m2
Min Force = 142100 x 0.00785 = 1115.48 N
Total Min Force = 392.4 +1115.48 = 1507.88 N
i didnt understand force reduction by fraction 1/32 , pls explain and also calculate Min force
 
  • #16
haruspex
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i didnt understand force reduction by fraction 1/32
I was forgetting that you are adding the weight of the rod. It's only the hydrostatic pressure that should reduce by one thirtysecond.
When the rod is pushed up 50cm it displaces some water. The rod's cross section is one sixteenth of the tank's, so the water level will rise in the tank by 50/16cm, and the min hydrostatic force is more like 1118N.
 
  • #17
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I was forgetting that you are adding the weight of the rod. It's only the hydrostatic pressure that should reduce by one thirtysecond.
When the rod is pushed up 50cm it displaces some water. The rod's cross section is one sixteenth of the tank's, so the water level will rise in the tank by 50/16cm, and the min hydrostatic force is more like 1118N.
thank you . did you read my # 14 post ? pls share your opinion on work, work done by pushing rod and work amount of water level raised didnt match
 
  • #18
haruspex
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thank you . did you read my # 14 post ? pls share your opinion on work, work done by pushing rod and work amount of water level raised didnt match
Work done on the water is 1154N x 0.5m= 577J.
The rest of the work was in raising the rod.
 
  • #19
Lnewqban
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if this force is dedicated to push the mass of water , 1546.4 N is too big for 3.93 liters of water up only 31.25 mm . i think , this force is to lift up 15 meter of 3.93 l water OR lift up 31.25 mm of total mass of water in tank . because :
accurately, water displacement is 3.92699 liter , total mass in water tank is 1884.96 kg
as W=fxd , work1 = 3.92699 x 9.8 x 15 m = 577.26 J
work2 = 1884.96 x 9.8 x 31.25 mm = 577.26 J
same amount of work .
but , by this formula , work done on pushed rod is : 1546.4 N x 0.5 m = 773 J , not equal to 577.26 J
again still confused :(
The work of pumping the water is not for lifting a column of 15 meters all the way up.
Has it reaches the surface, the height of that idealized column of water moved up by that force (distance between top of piston and surface) gets gradually reduced.
In my opinion, the input energy is only used in increasing the potential energy of the piston (respect to its initial position) and in increasing the level of water in the tank.
 
  • #20
Steve4Physics
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@ucody0911 has already shown a lot of working. So I think it's OK to chip-in to show some details of the calculations.

1. When the rod is at it low position, its top face is at a depth of h =15m. The hydrostatic force on the rod’s top face is:

F = (ρgh)(πr²) =1000*9.81*15*π*0.05² = 1155.7N

@ucody0911, this differs slightly from your value of 1154N because you have introduced a small rounding-error by using the rod’s area to only 3 significant figures (0.00785m²) at intermediate steps.
____________

2. When the rod is at its highest position, we have established that the water level is raised by 0.03125m. The depth of the rod’s top face is now h =15+0.03125-0.5 = 14.53125m

You can work out the hydrostatic force on the rod's top face at this depth for yourself.
____________

3. The work done raising the 40kg rod (ignoring the water) a distance (d=0.5m) is
W = mgd = 40*9.81*0.5 = 196.2J.
That corresponds to the rod’s increase in potential energy.

We can deal with the energy required to raise the water-level in several ways, but I prefer the following:

We are displacing a 50cm long, [edi 5cm]10cm diameter cylinder of water, with its centre of gravity 0.25m above the container’s base. The mass of this water 3.927kg.

It is correct to think of this water as simply being relocated to the surface where its new shape is a disc of thickness 3.125cm and diameter [Edit 20cm] 40cm added to the top of the original surface. This disc’s centre of gravity is 3.125cm/2 = 1.5625cm above the old water level.

We have raised 3.927kg of water a distance of d = (15+0.015625) - 0.25 = 14.77m

The work done raising the water is therefore
W = mgd = 3.927*9.81*14.77 = 569.0J
That corresponds to the water’s increase in potential energy.
____________

4. Total work done on rod and water (= total increase in potential energy) = 196.2+569.0 = 765.2J
____________

5. For final answers, I would say values should be rounded to no more than 3 significant figures, since the values in the original question are given to only 1 or 2 significant figures.
 
  • #21
haruspex
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not for lifting a column of 15 meters all the way up.
No, but it is lifted a fraction over 14.5m.
and in increasing the level of water in the tank.
It is not clear to me how much work you believe that is. Are you sticking with your analysis in post #4, or do you now accept that was wrong?
 
  • #23
Tom.G
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Wouldn't a simpler approach be:
The Center-of-Gravity of the water was raised by ???.
The Center-of-Gravity of the rod was raised by ???

Cheers,
Tom
 
  • #24
haruspex
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Wouldn't a simpler approach be:
The Center-of-Gravity of the water was raised by ???.
The Center-of-Gravity of the rod was raised by ???

Cheers,
Tom
Not if the aim is to find the maximum force, or how the force varies in the process.
But even if it's only the work done that's needed, there's no need to deal with the COG of the whole body of water... just what volume of water is displaced by the rod and how far has that risen. That approach was discussed in posts #9 and #11.
 
  • #25
Steve4Physics
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Wouldn't a simpler approach be:
The Center-of-Gravity of the water was raised by ???.
The Center-of-Gravity of the rod was raised by ???
The shape of the water changes. It starts as a simple cylinder. It finishes as a cylinder with a short cylindrical hole at one end (where the rod has displaced some water).

The shape-change complicates the centre-of-gravity-of-the-whole-of-the-water method.

The hole's volume is relatively small so, to a decent approximation, you could ignore the hole. But for an accurate result it shouldn't be ignored.
 

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