Respuesta :
Solution:
Given data:
A rectangular tank with dimensions 5 m as length, wide of 2 m and height of 8 m is filled with water tot a depth of 5 m.
A). If we select an elemental part of the tank of thickness dy at a depth y from the top, its mass will be :
m = p.v
⇒ m = p.(Ady)
⇒ m = p.( 5 x 2 dy)
⇒ m = 10 p dy
Now work done in moving mass from a depth y to the top will,
dw = m.g.y
dw = (10 p dy).g.y
or dw = 10 p.g.y.dy
Total work done is :
[tex]W= \int_{2}^{8} 10 p.g.y.dy[/tex]
[tex]$ W=10 pg[\frac{8^2-2^2}{2}]$[/tex]
[tex]$ W=10 pg[\frac{60}{2}]$[/tex]
W = 300 p.g
B). Origin at the bottom and consider an elemental part of the tank at a height y from the bottom, then its mass is :
⇒ m = p.( 5 x 2 dy)
⇒ m = 10 p dy
Now, work done in moving this mass from a height y above origin to a height 8 m above the origin is,
dw = mg(8 - y)
dw = (10 p dy).g.(8 - y)
dw = 10 pg.(8 - y). dy
The total work done is
[tex]W= \int_{0}^{5} 10 p.g.(8-y).dy[/tex]
[tex]W=10 p.g \int_{0}^{5} (8-y).dy[/tex]
[tex]W=10 p.g \left [ 8y-\frac{y^2}{2} \right ]_0^5[/tex]
[tex]W=10 p.g \left [ 8(5)-\frac{(5)^2}{2} \right ][/tex]
W = 10 pg (40-12.5)
W = 275 pg
C). Origin at the water level and consider y axis positive in the downward direction.
The work done to move and elemental ,as from a depth y below the water level to the top of the tank is,
dw = 10 pg (y+2) dy
The total work done is :
[tex]W=10 p.g \int_{0}^{5}(y+2) dy[/tex]
[tex]W=10 p.g \left[ \frac{y^2}{2}+2y \right ]_0^5[/tex]
[tex]W=10 p.g \left[ \frac{5^2}{2}+2(5) \right ][/tex]
W = 10 p g (12.5 + 10)
W = 225 pg
