# Relating pressure and height in a container

by Linus Pauling
Tags: container, height, pressure, relating
 P: 190 1. A container of uniform cross-sectional area A is filled with liquid of uniform density rho. Consider a thin horizontal layer of liquid (thickness dy) at a height y as measured from the bottom of the container. Let the pressure exerted upward on the bottom of the layer be p and the pressure exerted downward on the top be p+dp. Assume throughout the problem that the system is in equilibrium (the container has not been recently shaken or moved, etc.). Since the liquid is in equilibrium, the net force on the thin layer of liquid is zero. Complete the force equation for the sum of the vertical forces acting on the liquid layer described in the problem introduction. 2. 0 = sum of forces in y direction 3. Ok, I know that F_up = pA F_down = A(p + dp) weight of the thin layer = pAg dy So I did: pA - A(p+dp) - pAgdy = -Ap(d + dyg) = 0 And it's telling me that the answer does not depend on d.
Mentor
P: 41,440
 Quote by Linus Pauling 3. Ok, I know that F_up = pA F_down = A(p + dp)
OK.
 weight of the thin layer = pAg dy
Do you mean: rho*Ag dy ?

When you set up your equation, you won't have a 'd' by itself.
P: 190
 Quote by Doc Al OK. Do you mean: rho*Ag dy ? When you set up your equation, you won't have a 'd' by itself.

My bad, yes it's rho*Ag dy

Thus:

F = F_up - F_down - rho*Agdy
= pA - A(p + dp) - g*rho*A dy
= -dp - g*rho*A dy

which is incorrect

Mentor
P: 41,440
Relating pressure and height in a container

 Quote by Linus Pauling F = F_up - F_down - rho*Agdy = pA - A(p + dp) - g*rho*A dy = -dp - g*rho*A dy
Redo the last step--you dropped an A.

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