Force and moment on submerged surface

Click For Summary
SUMMARY

The discussion focuses on calculating the force and moment on a submerged vertical plate acting as a dam in a muddy lake, where the density variation is defined as ρ = ρ₀ + kh. Given parameters include ρ₀ = 1000 kg/m³, k = 100, and a water depth of 10m. The force per meter width is calculated using the integral F = ∫(ρg*y) dy, yielding a result of 815.5 x 10³ N. For the moment calculation, the formula M = ∫(r*P) dA is applied, where P is defined as ρ₀ + k(h-y), and the integration is performed from y=0 to y=h.

PREREQUISITES
  • Understanding of hydrostatic pressure and fluid mechanics
  • Familiarity with integration techniques in calculus
  • Knowledge of the concepts of force and moment in structural analysis
  • Basic principles of density variation in fluids
NEXT STEPS
  • Study the derivation of hydrostatic pressure formulas in fluid mechanics
  • Learn about the application of integration in calculating moments in structural engineering
  • Explore the effects of varying density on submerged surfaces
  • Investigate the principles of torque and reference axes in mechanics
USEFUL FOR

Students and professionals in civil engineering, mechanical engineering, and fluid dynamics who are involved in analyzing forces and moments on submerged surfaces.

boileroo
Messages
11
Reaction score
0

Homework Statement



I am having trouble understanding problems involving the force and bending moment on a submerged surface.

The problem in question is as follows:

variation in density of a muddy lake is given as rho = rhoo + kh, where rhoo = 1000, k = 100 and h = depth of lake.

A vertical plate is used as a dam, and the depth of the water is 10m.

Find the force/meter width and moment/meter width.

My problem lies in part b, calculating the moment

The Attempt at a Solution



F = int(rho*g*y) dy = int((rhoo +kh)*g*y) dy = (rhoo*g*h2)/2 + (k*g*h3)/3 = 815.5x103 N

It is this next part with the moment I am struggling on. The formula i have been taught to use is:

M = int(r*P) dA

I think that this goes to M = int((rhoo +kh)(y-h) dy but am unsure why, if it is infact it is the case.
Any help would be appreciated!
 
Physics news on Phys.org
To calculate torque, you need to select a reference axis, because you get different answers depending on where you define "r" to be 0. Suppose you choose the bottom of the dam as your reference.

The moment exerted on a small section of the dam, of length dy and width w, would then be y*P*wdy: the distance from the reference axis, times the force exerted on the infinitesimal section. P is just rho_0+k(h-y), and integrating from y=0 to y=h would give you the moment.
 
thank you, I think that helps!
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
6K
Replies
7
Views
1K
Replies
6
Views
4K
  • · Replies 22 ·
Replies
22
Views
1K
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
14
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K