- #1

AdkinsJr

- 150

- 0

This is a problem right out of the

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.

__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.