Calculating Frictional Force in a Water Eroded Tube: A Case Study

[Meis_113]

This one is a doozy . . .

A fresh water pond that is 15.7 m deep is contained on one side by a cliff. The water has eroded a nearly horizontal "tube" through a bed of limestone, which allows the water to emerge on the other side of the cliff. If the "tube" has a diameter of 4.05 cm, and is located 5.6 m below the surface of the pond, what is the frictional force between the "tube" wall and a rock that is blocking the exit?

I've spent a bit too long on this one, and I am getting nowhere . . . any assistane would be greatly appreciated. The sooner the better as well lol. Thanks for your help in advance.

 

[Arctangent]

actually I had the same question as well.

I'm wondering if I am missing a Force in my force diagram.

I used F = PA
And then the equal and opposite reaction to the force of the water pushing on the rock would be the friction between the rock and the wall going in a direction opposite to the direction of motion.
and where P = Patm + Dgh, where D= density of the fluid and Patm is the atmospheric pressure pushing down on the water.
Is that what you did?

 

[mezarashi]

Use the summation of forces Fnet = 0.

Your rock is acted on by
1. friction
2. air pressure - side 1
3. water pressure - side 2

Since, the hole is relatively small, we can make the calculations simple with not statics of bodies involved. The pressure of the water on side 2 of the rock is equal to:

[tex] p_w = \rho g h + p_{atm}[/tex]

Using F = pA, the answer shouldn't be far off.