Civil Engineering - Hydraulics - Discussion
Discussion Forum : Hydraulics - Section 5 (Q.No. 44)
44.
The depth of the centre of pressure on a vertical rectangular gate (4 m wide, 3 m high) with water upto top surface, is
Discussion:
19 comments Page 1 of 2.
Niranjan said:
2 years ago
The depth of the center of pressure= 2H/3.
(2*3)/3 =2m.
(2*3)/3 =2m.
(1)
Jero M said:
2 years ago
We know that, the centre of pressure for a plane uniform vertical surface lies at the depth of two-thirds of the height of immersed portion.
So, here we get 2/3 of 3m which is equal to 2m.
The given answer is correct.
So, here we get 2/3 of 3m which is equal to 2m.
The given answer is correct.
(1)
Nikitha said:
2 years ago
2h/3 = (2*3)/3 = 2.
Muhammad Hashim said:
4 years ago
CP = (Ig/Ax)+x.
CP = ((bd3/12)/bd*d/2)+1.5,
CP = (d/6)+1.5,
CP = 1/2+1.5,
CP = 2.
C is the right one.
CP = ((bd3/12)/bd*d/2)+1.5,
CP = (d/6)+1.5,
CP = 1/2+1.5,
CP = 2.
C is the right one.
(1)
Dipunku said:
5 years ago
It should be 2.5.
Sanku said:
5 years ago
It is 2.5.
Imran said:
5 years ago
Area = b*d.
CG (y)= d/2.
Ig = bd3/12.
Center of pressure = y + Ig/Ay.
CG (y)= d/2.
Ig = bd3/12.
Center of pressure = y + Ig/Ay.
Santhosh said:
6 years ago
Depth of center of gravity,
H = 2h/3
= 2*3 / 3
= 2 m.
H = 2h/3
= 2*3 / 3
= 2 m.
Salahuddin said:
6 years ago
CG equals CP when water surface coincides the body surface. Therefore cg=cp=3/2=1.5.
Am I right?
Am I right?
M.Atiq said:
6 years ago
Center of pressure is always below the centre of gravity and CG is located at point H/2 = 1.5m so simply we can make an idea that it will be at 2m.
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