Statics Problem Submerged Object

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SUMMARY

The discussion focuses on calculating the reactions at points A and B for a submerged gate AB measuring 0.5 x 0.8m, hinged at A and resting on a frictionless stop at B. The user attempts to determine the pressure exerted by the water and the weight of the gate, using the equations of static equilibrium (Fx=0, Fy=0). The calculations for pressure yield 2119.005 N, while the weight of the gate is calculated as 4327.68 N. The user expresses confusion regarding the conversion of mass to force and the subsequent calculations needed to find the reactions at points A and B.

PREREQUISITES
  • Understanding of hydrostatic pressure calculations
  • Knowledge of static equilibrium principles
  • Familiarity with force conversion from mass (kg) to weight (N)
  • Basic geometry for area calculations of shapes (triangles and rectangles)
NEXT STEPS
  • Review hydrostatic pressure calculations in fluid mechanics
  • Study static equilibrium problems involving submerged objects
  • Learn about converting mass to weight using gravitational acceleration (9.81 m/s²)
  • Explore detailed examples of calculating forces on submerged surfaces
USEFUL FOR

Students in engineering or physics courses, particularly those studying fluid mechanics and statics, as well as professionals involved in structural analysis of submerged structures.

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Homework Statement


A 0.5 x 0.8m gate AB is located at the bottomof a ank filled with water. The gate is hinged along its top edge A and rests on a frictionless stop at B. Determine the reactions at A and B when cable BCD is slack.
Statics4.jpg



Homework Equations


Fx=0
Fy=0


The Attempt at a Solution


I thought I would have to find the pressure of the water would be the x component, and then the weght of the water would be for the y component...

Since all of the widths are the same I figured they would null each other out because if they weren't the same water would either over flow under B and A.

Pressure=(1/2)bh*density=(1/2)*.93*.64*1000=297.6 kg (not a force)
Weight=Area of triangle+area of rectangle
=(1/2)bh+bh=((1/2)*.64*.48*1000)+(.64*.93*1000)=748.8 kg(not a force)

I don't understand excatly what the next step is what I tried doing was making Fx and Fy equations and solve for one...but i don't know how to turn kg into N, because when I multiply by gravity I get a huge number which gives me the wrong answer...

Here is my work drawn out on my picture...
Statics4-WOrk.jpg

Orange triangle is the pressure
The purple square and blue triangle is the weight. Does this make sense? Or am I in the wrong direction
 
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I have a better understanding now I believe is this the correct value for P and W...

P=width*specfic weight*hieght*height*(1/2)
=.5m*(9800N/m^3)*.93 m* .93m *.5m=2119.005 N

W=(area of triangle+area of rectangle)*9800
=(.5*.64*.48)+(.64*.45)=.4416*9800=4327.68 N
 

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