A Circular area of diameter ‘d’ is vertical and submerged in a liquid. Its upper edge is at liquid surface. Its center of pressure is at a depth of
5d/8
d/2
d/4
3d/8
Answers
Answer:
B. FORCES ON SUBMERGED PLANE AREAS
8.
If a triangle of height d and base b is vertical and
submerged in liquid with its vertex at the liquid surface,
derive an expression for the depth to its center of pressure
(c.p.); c.g. – center of gravity.
Answer:
4
3d
hcp =
9.
If a triangle of height d and base b is vertical and submerged in
liquid with its vertex a distance a below the liquid surface,
derive an expression for the depth to its center of pressure (c.p.);
c.g. – center of gravity:
Answer:
(6 2 )3/
6 8 3
2 2
a d
a ad d
hcp
+
+ +
=
10.
If a triangle of height d and base b is vertical and submerged in
liquid with its base at the liquid surface, derive an expression
for the depth to its center of pressure (c.p.); c.g. – center of
gravity:
Answer:
2
d
hcp =
11.
A circular area of diameter d is vertical and submerged in a
liquid. Its upper edge is coincident with the liquid surface.
Derive an expression for the depth to its center of pressure