Another Hydrostaic force Problem

[AdkinsJr]

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.

 

[ideasrule]

Yeah, it's an error.

 

[kerimek]

Hello,

Thank you for bringing this to my attention. It does seem that there may be an error in the given solution. The correct expression for pressure in this problem should be P = ρgd, where ρ is the density of water, g is the gravitational acceleration, and d is the distance below the surface. In this case, the correct value for ρ is indeed 62.4 lb/ft^3 and the correct value for g is 32.174 ft/s^2. Therefore, the correct expression for pressure should be P = (62.4 lb/ft^3)(32.174 ft/s^2)(7-y*_i).

I suggest double checking your calculations and possibly consulting with your instructor or a classmate to confirm the correct solution. It is important to make sure all units are consistent and accounted for in order to obtain an accurate result.

I hope this helps clarify the confusion. Let me know if you have any further questions or concerns. Keep up the good work in your calculus studies!