here is a copy of the problem statement. Problem Statement

Free Body Diagrams

I have attempted to do the problem, but some concepts I really dont understand. My professor showed us the answer and his free body diagram which I drew at the top and is in the middle. Here are the issues/concepts I don't understand.

Here is my work:

Force Balance $$R_1+R_2+F_p=W$$


$$R_1+R_1+F_p=W \rightarrow R_1=\frac{W-F_p}{2}$$

I know that force due to pressure is $\int_A pdA ; p=\rho g z ; dA = Wdl ; dl=\frac{dz}{\sin\theta}$ so the force due to pressure is $dF_p=\rho g z W \frac{dz}{\sin\theta}$

I know that the torque due to the pressure is $d\tau_p = ldF_p$, but since l is changing, I must relate l to z using a triangle. Doing this I yield $l = \frac{z}{\sin\theta}$

Doing some plugging and substitution for the $dF_p$ and the relationship of $l = \frac{z}{\sin\theta}$ I get,

$d\tau_p=\rho g W \frac{z^2 dz}{\sin^2\theta}$

Integration this with respect from z=0 to z=H, $d\tau_p=\rho g W \frac{1}{\sin^2\theta}\int_0^H z^2dz$

$\tau_p=\frac{\rho g W L^2 H}{3}$

Then computing the Torque due to the reaction force I get

$\tau_R = R\cos\theta$

And lastly the toque due to the weight of the door is

$\tau_w=W\cos\theta \frac{L}{2}$

Here are my questions

  • What is the reaction force? I think I understand that R2 is coming from the ground pushing up on the door, but I don't understand R1 or why it is vertical in the y direction. Is R1 due to the liquid, or something completely else? And why are R1 and R2 equal?

  • Second, I don't understand why Fp, force due to pressure, is not normal to the surface. In his FBD it is completely in the y direction.

  • Lastly, since this is a hydrostatic situation, would I be setting the $\tau_p+\tau_R-\tau_w=0$? I assume that Reaction 1 has no effect as it is right at the hinge..I think?

  • $\begingroup$ Not to sound rude, but I'm not sure what you are getting at. I have attached a picture of me attempting a solution to my question. I needed assistance understanding some concepts which I posted. If you could elaborate that would be great. $\endgroup$ – JuliusDariusBelosarius Sep 10 '16 at 17:02
  • 2
    $\begingroup$ You just put a picture of a sheet full of formulas. How are we supposed to understand what your thoughts were? We have a nice system here that can help you to structure your question with equations. $\endgroup$ – Bernhard Sep 10 '16 at 17:12
  • $\begingroup$ OK. I can convert it to Latex if that is what you mean. I posted the picture as I want. I apologize for that, but appreciate the feedback on proper question asking. I just figured it would be easier since I also have drawings that accompany the problem and I don't believe I can make those in latex. $\endgroup$ – JuliusDariusBelosarius Sep 10 '16 at 17:32
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    $\begingroup$ I think @Bernhard is (somewhat less than politely) trying to point out that posting images of math is generally not a good idea for several reasons: 1) We can't edit it if we see a typo, 2) It's not searchable, 3) It's harder to read. Please go through and type up your work using mathjax, being very careful to use clear and consistent notation. You will still have to include the diagrams as images, and that's fine. Also, when asking for help solving specific problems, always identify a specific conceptual issue that has you stuck. $\endgroup$ – DanielSank Sep 10 '16 at 18:41
  • $\begingroup$ @DanielSank I know understand! Thank you for that. Ill get right on that shortly. I just wanted to know why it is taboo to post images, and now I am aware. $\endgroup$ – JuliusDariusBelosarius Sep 10 '16 at 18:59

All you need to do is balance the torque due to the weight of the plate with the torque due to water pressure on the plate. Take moments about the hinge to eliminate any reaction there ($R_1$), and assume the reaction at the gasket is zero ($R_2=0$) for minimum weight of plate.

In answer to your questions :

  1. Ignore reaction forces (see above).

  2. Force due to pressure of water is normal to the plate. The FBD seems to consider only vertical components of force.

  3. $\tau_R=0$ because we are assuming $R_2=0$; that leaves $\tau_p=\tau_W$.

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