Archimedes Principle -- A crane lowers a vertical cylindrical pillar into a reservoir

  • #1

Homework Statement:

Tension in air, water and submerged

Relevant Equations:

Archimedes Principle
Hello,

Looking for help on the following:

A crane is used to lower a vertical cylindrical pillar into a reservoir. The pillar has a mass of 4 tonnes and is submerged to a depth of 2.5m. It has a diameter of 1.1m.

The specific gravity of fresh water is 1125kg/m^3, and 9.81m/s^2.

Use a free body diagram to show the forces acting on the pillar when submerged and calculate the upthrust, the tension in the rope in air and the tension in the cable when submerged.

I have this for Upthrust:

Fbuoy = Pbottom x A - Ptop x A
= Delta x P x A
= Pf x g x h x A
= 1.125 x 9.81 x 2.375 = 26.2N

Upthrust = 26.2N

Is this correct?

For the tension on the rope i am really stuck so could do with some help please.

I have that Tension = m x g

4 tonns = 4000kg

4000 x 9.81 = 39,240N

T = 39,240N

I know T = w - Fb

But Fb needs a volume. Which i can't work out because i don't have the height.

But now everything i look to learn so i can complete is says i need the height of the pillar. Can anyone help me please?

Thanks.
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
2019 Award
13,544
3,265
Hi,

What about
submerged to a depth of 2.5m
?

And: what units do you use ? Do you use them consistently ?
 
  • #3
@BvU is my upthrust correct? I have seen that volume = mass/density. So mass of the pillar/ density of the water. Is this correct?
So 4 x 1.1= 4.5

1.125 x 3.5 x 9.81 = 49.6N

T = w - Fb = 4 - 49.6 = 45.6N

Is this correct? If not where have i gone wrong?
 
  • #4
I have now realized 2.5 is my height.

This is now my answer.

Upthrust:

P = Pa + h x p x g

P1 = Pa + h1 x p x g
P2 = Pa + h2 x p x g

DeltaP = P2 - P1 = P x g (h2 - h1) = P x g Delta h

P = f/a
F = p x a

DeltaF = DeltaP x a = p x g delta h x a

p x g Delta h x a = Delta h x a = Volume

= mw x g = Upthrust

(mw) Mass of water = Volume of object x Density of water

Volume of object = pir^2h

3.14 x .55^2 x 2.5 = 25.95m^3

25.95 x 1.125 = 29.193N

Density of water = 1125kg/m^3

Mass of water = 25.95 x 1.125 = 29.193N

mw x g = 29.193 x 9.81 = 286.39N

Upthrust = 286.39N

Is this correct for upthrust??
 
  • #5
BvU
Science Advisor
Homework Helper
2019 Award
13,544
3,265
what units do you use ? Do you use them consistently ?
You work with numbers without mentioning the units. Bad idea. Very bad. Free advice: Instead, work with symbols and do not substitute numerical values and units until the very, very last.
At that moment:
  • check that the units indeed come out correctly
  • cancel things that can be cancelled
  • estimate order of magnitude of numerical result
  • roughly estimate value
  • do the calculation (probably on a calculator)
Great advantages!
  • much less chance of errors
  • more credit if calculator errors
More free advice: Do not use commas in numbers.

Relevant Equations:: Archimedes Principle
Write all equations needed, and a list of all variables with units, not the name of a principle.

A crane is used to lower a vertical cylindrical pillar into a reservoir. The pillar has a mass of 4 tonnes and is submerged to a depth of 2.5m. It has a diameter of 1.1m.

The specific gravity of fresh water is 1125kg/m^3, and 9.81m/s^2.
Does fresh water have two densities ? Or did you forget ##\bf g = ## :wink: ?
And the length of the pillar isn't mentioned. Nor the density. Nor the material. Did you forget something ? Otherwise the exercise is impossible to do ! Makes a difference if you tru to push under a 4000 kg balsawood pillar of 26 m long, or an Iron pillar of 0.6 m !

In contrast with my earlier post: the 2.5 m does not give you any relevant information if it is really stated exactly like this ! See below.


Fbuoy = Pbottom x A - Ptop x A
= Delta x P x A
= Pf x g x h x A
= 1.125 x 9.81 x 2.375 = 26.2N
Three dimensionless numbers. Two come out of the blue. The given was ##\rho = 1125 ## kg/m3. What is 1.125 ? What dimension is it ? And what is 2.375 ? What dimension is it ?


I have now realized 2.5 is my height.
Not if the problem statement is as you rendered it. But let's assume the density is indeed 1683 kg/m3 so that the height is $$ {m\over \rho\, A } = {4000\over 1683 \ 0.95 } \ {\sf {kg\over kg/m^3 \ m^2}} = 2.5 \ {\sf m}$$in that case
 
  • Like
Likes Ben_Walker1978
  • #6
Thank You @BvU . I am going to attempt this correctly and post here. Very much appreciate your help
 
  • Like
Likes BvU
  • #7
BvU
Science Advisor
Homework Helper
2019 Award
13,544
3,265
(My post became unmanageably long, so here is the continuation. I assure you I am not nitpicking or such; just trying to help)

I have now realized 2.5 is my height.

This is now my answer.
Let's assume that.

Upthrust:

P = Pa + h x p x g
That's not upthrust but pressure. Using both Pand p is asking for trouble. There is a list of symbols under the ##\sqrt x## button:

1588769694647.png
But using ##\LaTeX## is infinitely better.

P1 = Pa + h1 x p x g
P2 = Pa + h2 x p x g

DeltaP = P2 - P1 = P x g (h2 - h1) = P x g Delta h
There! an uppercase P - not to be confused with P o0)

P = f/a
F = p x a
You get my point by now, I hope...
DeltaF = DeltaP x a = p x g delta h x a

p x g Delta h x a = Delta h x a = Volume

= mw x g = Upthrust

(mw) Mass of water = Volume of object x Density of water

Volume of object = pir^2h

3.14 x .55^2 x 2.5 = 25.95m^3
No! 3 x 2.5 x something smaller than 1 can never exceed 10 ! And 3.14 x .55^2 x 2.5 = 2.376 so it's more than just a typo

25.95 x 1.125 = 29.193N
No ! 25.95 x 1.125 = 29.193 or 25.95 m3 x 1.125 kg/m3 = 29.193 kg but we already know the 1.125 is NOT in kg/m3
And don't forget the space between value and unit.

Density of water = 1125kg/m^3
Comical note: Problem statement said "fresh water" -- well, with such a density that water is anything but fresh !
 
  • #8
BvU
Science Advisor
Homework Helper
2019 Award
13,544
3,265
More comments: The exercise wants a free body diagram. Important tool/instrument. Did you make one and can you post it ?

For the record: to me "sumerged to a depth of 2.5 m" means that the top surface is 2.5 m below the water level.

Let's assume the density of the pillar is indeed 1683 kg/m3 and go through the motions.
 
  • #9
Hello @BvU

Thank you for all the effort you have put in to help me. Much appreciated.

I have read your comments, learn more and have now got it all correct (I think)

Upthrust:

Fbuoy = Pf x Vd x g

pi r^2 h

pi r^2 h = 3.14 x .55^2 x 2.5 = 2.375m^3

Pf x Vd x g = 1125 x 2.375 x 9.81 = 26,211.09N

Upthrust = 26,211.09N

Tension in air:


Tair = w x g

w x g = 4000 x 9.81 = 39,240N

Tension in air = 39,240N

Tension when submerged:


V = pi x (D^2/2) x h

pi x (D^2/2) x h = 3.14 x 0.3025 x 2.5 = 2.375m^3

Fb = Pf x Vd x g

Pf x Vd x g = 1125 x 2.375 x 9.81 = 26,211.09N

Tension when submerged = w x g - Fb

w x g - Fb = 4000 x 9.81 - 26,211.09 = 13,028N

Tension when submerged = 13,028N
 
  • #10
BvU
Science Advisor
Homework Helper
2019 Award
13,544
3,265
Still some comments -- if you want

Upthrust = 26,211.09N
  • Build a single expression with symbols, fill in numbers and dimensions only at the very last: $${\sf Upthrust}= \rho\, g\, V =\rho\, g \, \pi\, (D/2)^2\, h =\\ 1125 \times 9.81 \times \pi \times 0.55^2 \times 2.5 \ {\sf {kg/m}^3 \ m/s^2 \ m^2 \ {m} } = 26200 \ {\sf {kg m/s^2}} = 26200 \ \sf N $$
  • Round off once only, at the very last ( I get 26220.25)
  • Since we have the 1.1 and the 2.5 in two digits (but g and ##\rho## and ##\pi## in more), quoting more than two, three digits is suggesting non-existent precision
  • Don't use grouping commas in numbers ( that's for bean counters :wink: -- we use SI prefixes if necessary)
  • Don't forget the dimensions: both sides of the ##\ = \ ## should have the same dimension
  • Don't forget the space between number and dimension
The rest is just nitpicking:
  • symbol ##m## for mass is more customary. Then ##mg## is weight in N
  • Once you have weight (in N, 3 digits) and upthrust, you can simply subtract them to get tension when submerged
Welll done !

FBD 🤔 ?
 

Related Threads on Archimedes Principle -- A crane lowers a vertical cylindrical pillar into a reservoir

Replies
0
Views
991
Replies
7
Views
3K
Replies
12
Views
1K
Replies
7
Views
9K
Replies
2
Views
500
  • Last Post
Replies
16
Views
6K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
11
Views
4K
Top