Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion

Discussion Forum : Hydraulics and Fluid Mechanics - Section 7 (Q.No. 38)
38.
A vertical wall is subjected to a pressure due to one kind of liquid, on one of its sides. The total pressure on the wall acts at a distance __________ from the liquid surface.
H/3
H/2
2H/3
3H/4
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
7 comments Page 1 of 1.

Abhi said:   5 years ago
We know the formula for Centre of pressure:

C.P = IG/Ax + x.
Where, IG= MOI about an axis passing through CG.
A = Area of plane surface submerged in liquid.
X= distance of CG from the free surface of the liquid.
If we consider the wall as rectangular then,
IG= bh^3/12 ; A= b * h ; x= h/2.
Putting all these values in C.P formula we get an answer as - 2h/3.

Gopi said:   8 years ago
Actually, the intensity of pressure is directly proportional to the depth of the liquid level so that the total pressure also acts at the bottom of the liquid but in the above question the total pressure acts on the wall is equals to 2H/3 distance from the liquid level.

So please clarify my doubt.

Shu said:   6 years ago
Let the area of vertical wall be,A= h* b and moi be I= (b*h^3)/12. This problem related to hydrostatic and condition is of vertical case , so in this case pressure act at h/3 from bottom side , and 2h/3 from top side.( just like uvl case)

Ishrat said:   7 years ago
Total pressure will act at the centre of pressure which is equal to;
CP= h/2 + moi/ area * h/2,
= 2h/3.

Suresh said:   4 years ago
Total pressure act at h/3 distance from the bottom.

So h-h/3= 2h/3 from the liquid surface or free surface.

ASHIR said:   5 years ago
I=bh^3/12 and A=bh while x=h/2.
using h=I/Ax +x.
we get;
2h/3.

Shakti said:   3 years ago
How we get x=h/2? Please explain.

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