1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Force exerted by water on a tube

  1. Feb 12, 2012 #1
    1. The problem statement, all variables and given/known data

    A tube of uniform cross-section is bent to form a circular arc of radius R, forming three quarters of a circle. A liquid of density ρ is forced through the tube with a linear speed 'v'. Find the net force exerted by the liquid on the tube.


    2. Relevant equations

    F = ρav2 ?

    3. The attempt at a solution

    Since the arc is three quarters of a circle, the water travels through 3∏/2 radians (Is this relevant?). Since it says net force, then the force exerted by the liquid must be greater than that exerted by the tube on the liquid. I am not sure what to do next.
     
  2. jcsd
  3. Feb 12, 2012 #2

    ojs

    User Avatar

    Think of centripetal acceleration, the tube is forcing the liquid to turn so the tube must be applying some force on the liquid but do you know how to connect radius, angle, arc, velocity and centripetal acceleration?
     
  4. Feb 12, 2012 #3

    gneill

    User Avatar

    Staff: Mentor

    There should be action/reaction force pairs at the entrance and exit of the tube. Think of it in terms of rocket-like thrusts at the input and output.

    attachment.php?attachmentid=43812&stc=1&d=1329063237.gif

    I think your Relevant Equation should yield the reaction force at each end of the tube... what then is the net force?
     

    Attached Files:

  5. Feb 12, 2012 #4
    No.

    But I think centripetal acceleration should be R3∏/2.
     
  6. Feb 12, 2012 #5
    OK, so force = pressure in this scenario. Since the cross sectional area is not given, I can assume it as A. Velocity is given as 'v'. So the force exerted by the liquid on the tube is ρAv2. But what is the force exerted by the tube on the liquid? Is it -ρAv2? But this cannot be right, as there is a net resultant force exerted by the liquid which enables it to come out of the tube. It is all hazy.
     
  7. Feb 12, 2012 #6

    gneill

    User Avatar

    Staff: Mentor

    It's an action/reaction scenario -- equal and opposite forces are at work (assuming that the arrangement is being held stationary in some way). No matter what the details are regarding the forces that occur between tube and fluid within the system, the fact is that the system is 'catching' the moving fluid that enters at one port and 'expelling' fluid at the other port. Both scenarios demand action/reaction pairs that result in a net force on the system. Unless the system is accelerating as a result, all the forces must net out to zero. That means there will be some tension on the supply hoses that sum to be equal to the force that the fluid imparts to the tube during its travels through it.
     
  8. Feb 12, 2012 #7
    So this means that ρAv2 = - ρAv2? What other forces act on the tube? I still don't get it.
     
  9. Feb 12, 2012 #8

    gneill

    User Avatar

    Staff: Mentor

    The only forces you need to worry about are those that end up at the input and output ports (ends of the tube). No matter how any other forces between the fluid and tube play out within the system, the NET RESULT that ends up transmitted to the tube ends will be the sum of all those forces.

    It's a bit like weighing a chair. You don't care about the particular forces that the seat back places where on the seat, or if there's tension between the legs stretchers, or any other incidental minutia. Just bung the thing onto a scale and measure the total force with which the legs are pressing down.

    The fluid flow into the input engenders a force ρAv2. The flow out of the other end of the tube engenders another force ρAv2. These forces are in different directions. Add accordingly. Somewhere within the tube system there will be suitable interactions that exactly balance the net of these forces -- but you really don't need the details!
     
  10. Feb 13, 2012 #9
    OK, so the vectors for the input and the output forces are at right angles. On adding, they give a vector which bisects the angle at 45°? In other words, it is the diagonal of a square. In that case, the resultant vector would be sqrt of 2.ρAv2 (Sorry, but LaTEX is a pain, I am unable to use it properly). Is this correct?
     
  11. Feb 13, 2012 #10

    gneill

    User Avatar

    Staff: Mentor

    Yes, that's the idea.

    LaTEX is not so bad after a bit of practice :smile:
     
  12. Feb 13, 2012 #11
    All that talk about centripetal accelaration, radius, arc and action_reaction etc threw me off track. Thanks for your patience.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook