- #1
Dr.Brain
- 538
- 2
Ok there is a dam of width "b" and water is fille don one side of it to a height h . Now i need to calculate the force on the wall of the dam on the side where water is filled .
This is what I did:
I considered a small layer of water of dx (vertical) at a distance x from top of water surface .
Then:
[itex] \int dF= \int \delta xg (bdx)[/itex]
Limits 0--->h on RHS
[itex] F=\frac {b \delta gh^2}{2} [/itex]
Therefore the force on the wall depends on :
Width of the dam
The height "h" to which water is filled
Density of water
and g
But one thing i noticed is that the force on the wall doesn't depend on the length of the river..!
2) Now suppose i want to calculate force due to water ina cuboid lying horizontally on floor ..Its length is l breadth is b and height h ... now if it lies on floor with breadth as its height...then it means as per previous conclusion that length of the cuboid lying horizontally wouldn't make any difference...So this means i keep on increasing its length without any effect on force due to water on the front face?
This is what I did:
I considered a small layer of water of dx (vertical) at a distance x from top of water surface .
Then:
[itex] \int dF= \int \delta xg (bdx)[/itex]
Limits 0--->h on RHS
[itex] F=\frac {b \delta gh^2}{2} [/itex]
Therefore the force on the wall depends on :
Width of the dam
The height "h" to which water is filled
Density of water
and g
But one thing i noticed is that the force on the wall doesn't depend on the length of the river..!
2) Now suppose i want to calculate force due to water ina cuboid lying horizontally on floor ..Its length is l breadth is b and height h ... now if it lies on floor with breadth as its height...then it means as per previous conclusion that length of the cuboid lying horizontally wouldn't make any difference...So this means i keep on increasing its length without any effect on force due to water on the front face?