Pressure on a vertical rectangular plate submerged in water

[Kqwert]

The question goes as follows..:

A car driver has an accident and drives into a lake. The car sinks to the bottom. The distance from the car door to the water surface is 10m. The car door is 0.9m wide and 1.3m tall. Assume that the car door can be approximated as a vertical rectangular plate and that the car is filled with air.

A) Calculate the pressure force acting on the door
--------
The solution manual solves the problem by calculating the hydrostatic force halfway down the car door,

i.e. F = rho*g*(10+0.5*1.3) / 1.3*0.9

Isn't this wrong? My idea was to calculate the force by using the "prism" method, described here: https://en.wikipedia.org/wiki/Pressure_prism

 

[BvU]

Hello K,

Well, do you get a different answer ?

 

[Kqwert]

Yeah, I get a different answer. I get ~122kN while the above mentioned formula gives ~89kN

 

[BvU]

Could you show the steps ?

 

[Kqwert]

After looking at the solution manual I believe that the author has just mixed up the definition of Pressure. By multiplying and not dividing the area on the hydrostatic force halfway down the car door I get the same answer as mine. I get 122kN by the way. I assume that is the correct answer? :-)

 

[BvU]

I get 122 kN too. A dimension check would have shown the / 1.3 in the expression can't be right.

Is it clear to you why the prism expression gives the same result as taking the pressure halfway ?

 

[Kqwert]

Nope, could you explain it? :-)

 

[BvU]

$$F = \int p\,dA = \int \rho g h \;d(w\times h) = \rho g w \int h \,dh = {1\over 2} \rho g w \left . h^2 \displaystyle \right |_{bot}^{top} =\\ {1\over 2 } \rho g w (h_{top}^2 - h_{bot}^2) = \rho g w (h_{top} - h_{bot}) {1\over 2 } (h_{top} + h_{bot}) = \rho g A\, h_{\rm average} $$
with $A$ the area, $w$ the width.

 

[Kqwert]

Thank you. In these expressions, what is htop and hbot?

 

[BvU]

10 and 11.3 m, respectively

 

[Kqwert]

Thanks! I guess there generally will be no difference if you integrate the other way around - i.e. from top to bottom?

 

[BvU]

Correct. If you integrate from bottom to top $dA = w\,dh$, but if you integrate from top to bottom, then $dA = -w\,dh$

 

[Kqwert]

So you get a negative answer regardless?

 

[haruspex]

10 and 11.3 m, respectively

Should that be -10 and -11.3 ?

 

[BvU]

Should that be -10 and -11.3 ?

As always, Haru is right ! I used $dh>0$ for #8, so I chose up as positive. As a consequence, $g < 0$

So you get a negative answer regardless?

Good remark -- My sloppiness.. (although with $g<0$ the answer is positive)

Leaves us the question: what is the meaning of a plus or minus sign.

$p=\rho g h$ comes from $\Delta p=\rho g \Delta h$, so $p - p_{\rm surface} = \rho g (h - h_{\rm surface})$ where we take $h$ and $g$ in the same direction ( you take the depth and use $g > 0$ or you take the height and use $g < 0$.

As you know, $F$ is a vector $\vec F$ and $p$ is a scalar (a number without direction). Therefore $dF = p \;dA$ should in fact be written as $\vec F = p\;d\overrightarrow {A}$ and we need to establish coordinates. That means work and I try to avoid that ( -- except sometimes in the PF context --) . I know the force is inwards so I don't pay attention to the sign. I also know that the other door experiences the same force, but in the opposite direction. In this exercise, $\| \vec F\|$ is the answer.

In #8 I should have written $\vec F = \displaystyle \iint \ p \;d\overrightarrow {A}$ and the established coordinates would hopefully tell me the two force vectors for the two doors both point inwards.

I've forgotten the exact definition of $d \overrightarrow A$, though...

 

[Kqwert]

I guess dh becomes negative because you are integrating from a bigger number to a smaller number?