- #1
AdkinsJr
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This is a problem right out of the text of my calculus book. So I actually have the work, but I'm confused about something the did.
Problem: Find the hydrostatic force on one end of a cylindrical drum with radius 3 ft. if the drum is submerged in water 10 ft deep.
I attached a diagram
Pressure can be defined:
p=density, d=distance below the surface
[tex]P=pgd=\delta d[/tex]
The area of the ith strip is:
[tex]A_i=2\sqrt{9-(y^*_i)^2}\Delta y[/tex]
The average distance below the surface is:
[tex]d_i=7-y^*_i[/tex]
Then when they write the pressure, they give:
[tex]\delta_i=62.5(7-y^*_i)[/tex]
The implies that [tex]pg=62.5[/tex].
The density of water is [tex]p=62.4 lb/ft^3[/tex]
http://en.wikipedia.org/wiki/Density_of_water
Gravitational acceleration is [tex]32.174 ft/s^2[/tex]
So how does [tex]\delta d_i=pgd_i=62.5(7-y^*_i)[/tex] ? Is this an error? It seems that they accidently neglected to account for gravitational acceleration.
Problem: Find the hydrostatic force on one end of a cylindrical drum with radius 3 ft. if the drum is submerged in water 10 ft deep.
I attached a diagram
Pressure can be defined:
p=density, d=distance below the surface
[tex]P=pgd=\delta d[/tex]
The area of the ith strip is:
[tex]A_i=2\sqrt{9-(y^*_i)^2}\Delta y[/tex]
The average distance below the surface is:
[tex]d_i=7-y^*_i[/tex]
Then when they write the pressure, they give:
[tex]\delta_i=62.5(7-y^*_i)[/tex]
The implies that [tex]pg=62.5[/tex].
The density of water is [tex]p=62.4 lb/ft^3[/tex]
http://en.wikipedia.org/wiki/Density_of_water
Gravitational acceleration is [tex]32.174 ft/s^2[/tex]
So how does [tex]\delta d_i=pgd_i=62.5(7-y^*_i)[/tex] ? Is this an error? It seems that they accidently neglected to account for gravitational acceleration.