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Liquid Pressure at Depth

  1. Jun 9, 2012 #1
    The pressure of a liquid of constant density at depth 'h' is given by the equation:

    [tex] P = hρg [/tex]

    I am trying to understand the derivation of this equation. I understand it perfectly for liquid columns with rectangular liquid columns but I can't see why it works if the shape of the fluid column is irregular.

    [tex] P = \frac{F}{A} = \frac{mg}{\frac{V}{h}} = hρg [/tex]

    The problem is that [itex] \frac{V}{h} = A [/itex] only if it is rectangular shaped, right?
    So why should we allowed to generalize it for all fluid columns?

  2. jcsd
  3. Jun 9, 2012 #2
    No. For example, the area of a circle with radius r multiplied by a height h gives the volume of a cylinder with radius r and height h.
  4. Jun 9, 2012 #3
    Ok but what if the area of the cross-section changed with the height? Then would it still apply?

  5. Jun 9, 2012 #4


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    Yes, it always applies. Suppose you had liquid in a container of irregular shape and you inserted a tube vertically to some depth. If the pressure at the base of the tube resulting from the liquid in the tube were different from that elsewhere at the same level then there would be a flow into or out of the tube. The level at the top of the tube would now be different from that in the rest of the container.
  6. Jun 10, 2012 #5
    The last equation holds for any vertical column with constant cross-section (prism). If the liquid has any other shape, you can still derive the same equation for imaginary vertical column in the liquid. "vertical" means directed in direction of gravitational acceleration. This way the pressure forces of surrounding liquid on the sides of the column don't affect the force balance in vertical direction (force is a vector and F in the first equation is its vertical component).
  7. Jun 10, 2012 #6
    It is important to remember that force is a vector and that pressure force in liquid is perpendicular to the surface. If the liquid container does not have the shape of a vertical prism, then the force on the bottom surface is not equal to the weight of the liquid, because weight of the liquid is also supported by the walls of the container.

    Before I understood that I had troubles understanding the liquid pressure on the bottom of a V shaped vessel: I thought it should be infinite, because "the weight of the water is supported by a single point (zero area)". This assumption is wrong, because V shaped vessel has inclined sides and pressure force on them will have a nonzero vertical component.

    Only in case of a vertical column we can assume that the pressure force on the bottom equals the weight of the liquid. Of course we can always define an imaginary vertical column in any liquid container. So the formula

    [tex] P = \frac{F}{A} = \frac{mg}{\frac{V}{h}} = hρg [/tex]

    is valid for all cases where the density and gravitational acceleration are constant.
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