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

Click For Summary
SUMMARY

The discussion centers on calculating the force and torque required to balance a cylinder subjected to static fluid pressures. The Euler equation is utilized, leading to the conclusion that the pressure gradient is given by $$\nabla p = -\rho_0 g \hat{k}$$. The pressure function is derived as $$p(z) = -\rho_0 g z + \lambda$$, where $$\lambda$$ represents the pressure at the reference point. The participants seek clarification on the relationship between pressure at the reference height and the integration limits for calculating force and torque.

PREREQUISITES
  • Understanding of fluid dynamics principles, specifically static fluid pressure.
  • Familiarity with the Euler equation in fluid mechanics.
  • Knowledge of integral calculus for force and torque calculations.
  • Basic concepts of pressure gradients and their physical implications.
NEXT STEPS
  • Study the derivation and applications of the Euler equation in fluid dynamics.
  • Learn about pressure distribution in static fluids and its implications on forces acting on submerged objects.
  • Explore integral calculus techniques for calculating forces over variable geometries.
  • Investigate the relationship between pressure, height, and density in hydrostatics.
USEFUL FOR

Students and professionals in engineering, particularly those specializing in fluid mechanics, as well as anyone involved in the design and analysis of submerged structures or systems.

mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
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)
 

Attachments

  • cylindrical.png
    cylindrical.png
    5.4 KB · Views: 103
  • z_h.png
    z_h.png
    1.2 KB · Views: 113
  • P(theta).png
    P(theta).png
    4.7 KB · Views: 123
  • d_theta.png
    d_theta.png
    452 bytes · Views: 129
Physics news on Phys.org
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)
 

Similar threads

Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
669
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
6K