Fluids - bringing together pressure, depth and friction

[Arctangent]

Fluids -- bringing together pressure, depth and friction

Hi Again!

More troubles with fluids.

I was pretty sure I had this one.

There is a
freshwater pond
density, D=1000 kg/m^3
depth = 15.1 m
one of the sides is blocked off by a cliff.
A nearly horizontal tube
diameter = 3.17 cm
depth below the pond's surface is 6.2 m
eroded to the other side of the cliff
a rock blocks cuts off the flow of the water.

The question asks for the frictional force between the tube's wall and the rock blocking the exit.

I figured it would be fairly easy, since
F = P*A
and we can find out the hydrostatic pressure.
so
P = Patm + D*g*h
P = 101300 + 1000*9.8*6.2
P = 162060 Pa

A = pi*r^2 = pi* (0.01585)^2 = 7.89e-4 m^2

and therefore,

F = P*A = 127.9 N is the force on the rock blocking the exit

because there is no acceleration, the system is at equilibrium

so Fnet = F(ontherock) - F(friction) = 0
so F(on the rock) = F(friction)
so then my answer would be 127.9 N

Could you point out where I've made an error, and push me in the right direction?

 

[Tide]

I think you didn't take into account the pressure of the air pushing back on the rock from the other side!

 

[Meis_113]

LOL i just posted a question EXACTLY like this. Anyways, I understand arctans argument, but I don't know where you would add the pressure pushing back from the other side . . .

 

[Arctangent]

I think you didn't take into account the pressure of the air pushing back on the rock from the other side!

ahh good point!



So It'd be:

Fnet = F(airpushingonrock) + F(friction) - F(water) = 0

Am I correct?

 

[Tide]

That looks fine - just be careful with the signs of the forces.