4
$\begingroup$

enter image description here

$\rho, p_o, \alpha, F_g, h, b$ are given. $F_T$ is wanted. There is a plate that can rotate around $A$. The plate measures b perpendicularly to the screen surface.

Pressure gradient

$p(z) = \rho g (2h - z)$

I neglect $p_0$ since it it's the same pressure on both sides of the plate.

Pressure Force

$$F_p = \int_{l_0}^{l_1} \int_0^b \! p \, \mathrm{d}y\mathrm{d}l$$

With

$sin(\alpha)=\dfrac{dz}{dl}$

$dl=\dfrac{dz}{sin(\alpha)}$

$$F_p = \int_{l_0}^{l_1} \int_0^b \! p \, \mathrm{d}y\mathrm{d}z \frac{1}{sin(\alpha)}$$

Now I can integrate from $0$ to $h$ since I substituted $l$ with $z$ and I get

$$F_p = b \rho g \dfrac{3h^2}{2sin(\alpha)}$$

This is the pressure force that works on the plate. To calculate its torque it has to be multiplied with its lever $l_a$.

In order to find $l_a$ I have to regard the actual torque which is

$$F_{p} \cdot l_a = \int l \cdot p \, \mathrm{d}A$$

$$F_p \cdot l_a = \int_{l_0}^{l_1} \int_0^b \! l \cdot p \, \mathrm{d}y\mathrm{d}l$$

Again substitute $dl$ with $dz$ yields

$$F_p \cdot l_a = \int_{l_0}^{l_1} \int_0^b \! l \cdot p \, \mathrm{d}y\mathrm{d}z \frac{1}{sin(\alpha)}$$

I assume I can now insert insert $0$ for $l_0$ and $h$ for $l_1$ again. And at this point I'm unsure but I think since I integrate over $dz$ again I can insert $\frac{z}{sin(\alpha)}$ for $l$.

$$F_p \cdot l_a = \int_{0}^{h} \int_0^b \! z \cdot p \, \mathrm{d}y\mathrm{d}z \frac{1}{sin(\alpha)}$$

Inserting and integrating

$$b \rho g \dfrac{3h^2}{2sin(\alpha)} \cdot l_a = \int_{0}^{h} \int_0^b \! \frac{z}{sin(\alpha)} \cdot \rho g (2h-z) \, \mathrm{d}y\mathrm{d}z \frac{1}{sin(\alpha)}$$

$$b \rho g \dfrac{3h^2}{2sin(\alpha)} \cdot l_a = b \rho g \frac{1}{sin^2(\alpha)} \left[h^3-\frac{1}{3}h^3 \right]$$

$$\dfrac{3h^2}{2} \cdot l_a = \frac{1}{sin(\alpha)} \left[h^3-\frac{1}{3}h^3 \right]$$

$$l_a = \frac{1}{sin(\alpha)} \left[\frac{2}{3}h-\frac{2}{9}h \right]$$

$$l_a = \frac{4h}{9sin(\alpha)}$$

I'm very unsure with how I substituted and inserted. I would appreciate someone to review if I did it correctly and maybe elaborate on what I got wrong if possible.

$\endgroup$
  • 1
    $\begingroup$ @Air $\alpha$ is in right corner on the bottom, maybe green wasn't the best color to mark it. It's the angle between $\over{AB}$ and the x-axies $\endgroup$ – idkfa Jul 16 '15 at 18:35
  • $\begingroup$ Oh, I see it now. The compression of the thumbnail makes it almost invisible, but it's okay in the full resolution image. $\endgroup$ – Air Jul 16 '15 at 18:37
  • $\begingroup$ Updated once again, I think I figured it out myself. Since I got the question posted along with my calculations anyway, someone might would like to review it. $\endgroup$ – idkfa Jul 17 '15 at 8:55
  • $\begingroup$ Does this need a "homework-exercises" tag? $\endgroup$ – Dan Jul 18 '15 at 17:32
  • 1
    $\begingroup$ Cool, thanks. Some of the other SE sites have that as a pretty strict about that. Some folks get annoyed if they don't see it when they feel that they should. $\endgroup$ – Dan Jul 18 '15 at 17:41
2
$\begingroup$

It looks like, from your math, you attempt to find the total normal force due to the pressure on the incline $F_p$, and then try to find a distance $l_a$ from point A such that if $F_p$ was applied there the torque would be equivalent to the torque caused by the pressure.

I'm not sure what you would use $l_a$ for, but if you're trying to equlibrate torques, it's easier to just calculate the torque due to the pressure (which you do in order to find $l_a$ anyway)

So I'm going to start from your calculations and note the errors in bold.

Throughout your calculation you've used the center dot to indicate scalar multiplication, while also using implied multiplication. Generally if the center dot is not everywhere, then it implies the dot product of vectors, so I've removed all instances.

The actual torque is:

$$\tau = \int \vec{l} \times \vec{n} \, p \, \mathrm{d}A$$

Where $\vec{n}$ is the vector normal to the surface, and $\vec{l}$ is the vector to the integrating point from axis A. As these vectors will always be perpendicular we can simplify using $l$ as the distance to point A.

$$\tau = \int l \, p \, \mathrm{d}A$$

$$\tau = \int_{l_0}^{l_1} \int_0^b \! l \, p \, \mathrm{d}y\mathrm{d}l$$

At this point I'd integrate over y to get:

$$\tau = b\int_{l_0}^{l_1} \! l \, p \, \mathrm{d}l$$

But we can delay that as you did.

Now we'll change our integration variable from $l$ to $z$ This step requires substituting the differential and changing the bounds

$$z=l\,\sin(\alpha)$$ $$\mathrm{d}z=\mathrm{d}l\,\sin(\alpha)$$ $$\mathrm{d}z \frac1{\sin(\alpha)}=\mathrm{d}l$$

$$\tau = \int_{z_0}^{z_1} \int_0^b \! l \, p \, \frac{1}{\sin(\alpha)} \mathrm{d}y\mathrm{d}z $$

Now inserting $0$ for $z_0$ and $h$ for $z_1$ makes sense but we can't substitute $z$ for $l$.

$$\tau = \int_{0}^{h} \int_0^b \! l \, p \, \frac{1}{\sin(\alpha)} \mathrm{d}y\mathrm{d}z $$

Now we need to substitute $l$ for $z\frac1{\sin(\alpha)}$

$$\tau = \int_{0}^{h} \int_0^b \! \frac{z}{\sin(\alpha)} \, \rho g (2h-z) \frac{1}{\sin(\alpha)}\, \mathrm{d}y\mathrm{d}z $$

$$\tau = b \rho g \frac{1}{\sin^2(\alpha)} \left[h^3-\frac{1}{3}h^3 \right]$$

So while your substitutions weren't valid in some intermediate states the final torque was correct.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.