MHB Find the force and the torque so that the cylinder is in balance

AI Thread Summary
To determine the force and torque needed for the cylinder to be in balance, the discussion references the Euler equation and the relationship between pressure and gravitational force. It establishes that the pressure gradient is equal to the product of fluid density and gravitational acceleration, leading to the pressure function p(z) = -ρ₀gz + λ. The user seeks clarification on why the pressure at z=0 equals atmospheric pressure (p_a) and the reasoning behind substituting z with h. Additionally, they inquire about the integration process for calculating the total force and the limits for the angle θ when evaluating forces from point A to B. The conversation emphasizes understanding the calculations for both force and torque as outlined in the notes.
mathmari
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Hey! :o

At the cylinder of the picture there are static pressures from environment fluid of density $\rho$. If we neglect the atmospheric pressure, calculate how much force and how much torque is needed so that the cylinder balance.

View attachment 4436

In my notes there is the solution, but I haven't really understood it...

The Euler equation is $$\rho \frac{D \overrightarrow{u}}{D T}=-\nabla p+\rho \overrightarrow{b}$$

Since the cylinder should be in balance we have that $\overrightarrow{u}=0$.

We also have that $\overrightarrow{b}=\overrightarrow{g}$

That means that $$\nabla p=\rho \overrightarrow{b}=\rho \overrightarrow{g}$$

Since $\overrightarrow{g}=-g\hat{k}$ we have that $$\nabla p=-\rho_0 g \hat{k} \Rightarrow \partial_xp=0 \ \ , \ \ \partial_yp=0 \ \ , \ \ \partial_zp=-\rho_0 g$$

$$\Rightarrow \frac{dp}{dz}=-\rho_0 g \Rightarrow p(z)=-\rho_0 g z+\lambda$$

Is it correct so far?? (Wondering) After that in my notes there is the following which I don't understand:

$$p(z)=-\rho_0 g z+\lambda$$

View attachment 4437

$$z=0 \ \ \ \ \ p(z=0)=\lambda =p_a \\ p(z)-p_a=-\rho g z=-\rho_0 g h$$

View attachment 4438

$$P(\theta)=\rho_0 g h(\theta)=\rho_0 g\frac{D}{2}(1-\cos \theta)$$

$$P(\theta) dA=P(\theta)\left (1 \cdot \frac{D}{2}d\theta\right )$$

View attachment 4439

$$F_{AB}=\int_{\theta=0}^{\theta=\pi} p(\theta)\frac{D}{2}d \theta \\ F_{B \Gamma}= \dots $$

Could you explain to me the part I don't understand?? (Wondering)
 

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I have the following questions:

Why is for $z=0$, $p=p_a$ ?? And also why do we replace $z$ with $h$ ??

At the end to compute the force do we take the integral to add the force at each part at which we divide the cylinder??

Since $\theta$ is the angle from $A$ and we want to calculate the force from A to B we take the limits $\theta=0$ to $\theta=\pi$, right?? So, to find the force $F_{B \Gamma}$ we have to calculate the integral $\int_{\phi=0}^{\phi=\frac{\pi}{2}} p(\phi)\frac{D}{2}d \phi$, right??

Which is the force and which is the torque that we are looking for at the solution of my notes??

(Wondering)
 
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