# Pressure & Force

1. Apr 28, 2008

### bcjochim07

1. The problem statement, all variables and given/known data
How much force does the fluid exert on the end of the cylinder at A?

2. Relevant 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?

2. Apr 28, 2008

### apelling

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.

3. Apr 28, 2008

### bcjochim07

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?

4. Apr 28, 2008

### alphysicist

Hi bcjochim07,

I think you need to account for the atmospheric pressure that acts above the piston.

5. Apr 28, 2008

### bcjochim07

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 ?

6. Apr 28, 2008

### bcjochim07

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?

7. Apr 28, 2008

### alphysicist

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.

8. Apr 28, 2008

### bcjochim07

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?

9. Apr 29, 2008

### bcjochim07

Oh ok.... scientific notation of course. Does my answer for part b seem logical?

10. Apr 29, 2008

### apelling

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.

11. Apr 29, 2008

### bcjochim07

Does my answer for part b seem ok?

12. Apr 29, 2008

### bcjochim07

Any thoughts anybody?

13. Apr 29, 2008

### alphysicist

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:

$$P_B = P_A +\rho g h$$

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:

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

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