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Pressure & Force

  • Thread starter bcjochim07
  • Start date
374
0
1. Homework Statement
How much force does the fluid exert on the end of the cylinder at A?

apressure.jpg


2. Homework Equations



3. The Attempt at a Solution

p= po + density*g*d
Can I say the pressure is (10kg)(9.80)/pi*(.02m)^2 + (900 kg/m^3)*(9.80)(.70m) and set that equal to F/pi*(.1m)^2 to solve for force? Does the tube enter the container at halfway up?
 

Answers and Replies

70
0
I cannot see anything wrong with your answer assuming as you have that the tube enters the container at halfway up. Without that assumption its not possible to give a final answer.
Normal atmospheric pressure can be ignored because its effect will be zero on the final force exerted.
 
374
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I did Pressure= (10kg)*(9.80)/pi*(.02m)^2 + (900kg/m^3)(9.80)(.70m) = 84159.92 N/m^2

84159.92 = F/pi * (.1m)^2 F= 2600 N rounded off with two significant figures because the question tells me to, but this is wrong. What's not quite right?
 
alphysicist
Homework Helper
2,238
1
Hi bcjochim07,

I think you need to account for the atmospheric pressure that acts above the piston.
 
374
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ok, I didn't understand why apelling told me I didn't

so Pressure = (10kg)(9.80)/pi*(.02m)^2 + 101300N/m^2 + (900kg/m^3)(9.80)(.70) = 185460

185460 = F/pi* (.1m)^2 F= 5800 N ?
 
374
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also the second part of the question asks for force at b, so I would do the same sort of thing only with a depth of 1.3m?
 
alphysicist
Homework Helper
2,238
1
As long as it's safe to assume that these pipes and things are circular in cross section (and that your original assumption about the halfway mark is correct), I think that's right.
 
374
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Ok so here's what I did for the force at b:

(10kg)(9.80)/pi*(.02m)^2 + 101300 + (1.3m)(9.80)(900 kg/m^3) = P

P= 190752 = F/ pi * (.1m)^2 F= 5992.65 N. This is mastering physics homework, and it says to round it to two significant figures, anybody know how I would do that on a computer?
 
374
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Oh ok.... scientific notation of course. Does my answer for part b seem logical?
 
70
0
The reason why I suggested ignoring atmospheric pressure is because I was thinking in terms of the resultant force on the cylinder end, whereas you required the force from the liquid only. Sorry to have mislead you.
 
374
0
Does my answer for part b seem ok?
 
374
0
Any thoughts anybody?
 
alphysicist
Homework Helper
2,238
1
If you're having doubts about it, you can use the pressures at A and B to check that they are consistent, using the same equation:

[tex]
P_B = P_A +\rho g h
[/tex]

where h is the depth of B relative to A.

Or you can compare the forces at A and B. The force at B is the force at A (that is the force the top of the container is pusing down on the fluid) plus the weight of the fluid above point B:

[tex]
\begin{align}
F_B &= F_A + m g \nonumber\\
&= F_A + \rho g (\pi r^2) h\nonumber
\end{align}
[/tex]

because [itex]\pi r^2 h[/itex] is the volume of the liquid above point B. (Be sure to use the full number for [itex]F_A[/itex], though, not the amount after you rounded off to two digits.)
 

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