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!

Fluid Mechanics, simple task.

  1. Jan 22, 2014 #1
    1. The problem statement, all variables and given/known data
    Task number 2.
    The problem is that my solution is wrong but the lecturer didnt want to explain what is wrong, she insists on doing it by myself. But to be honest I'm not really good at it and I just need someone either explain it or solve or at least check it and tell me what is wrong there.

    SkNglAel.jpg



    2. Relevant equations



    3. The attempt at a solution

    TkYMKC4l.jpg
     
  2. jcsd
  3. Jan 22, 2014 #2
    What is the pressure at depth z below the surface? What is the direction of the pressure acting on the flat ends of the cylinder, and what is the direction of the pressure acting on the curved surface of the cylinder?

    Chet
     
  4. Jan 22, 2014 #3
    I would really like to respond to that, but to be honest I dont know how. I didnt know I have to include anything about pressure there :/
     
  5. Jan 22, 2014 #4
    Well the forces on these surfaces are the result of the water pressure acting on them. Are you familiar with the equation p = po+ρgz, where po is the pressure at the air interface, ρ is the water density, and z is the depth?
     
  6. Jan 22, 2014 #5
    I am not, sooo...
    P1=ρgz=1000*10*0.9=9000 kg/m3
    P2=ρgz=1000*10*2.1=21000 kg/m3

    I just dont know from where will I get po.
     
  7. Jan 22, 2014 #6
    Ummm I have problem with units now. po is F/s where F = P*s (s is the area of base)
    F1= p1*s=qgh1*s ;
    F2= p2*s=qgh2*s

    F1 = 10179 kg/(m^2 * s^2)
    F2 = 23751 kg/(m^2 * s^2)

    V = (h2-h1)*s=(2.1-0.9)*1.131=1.3572

    F = qgV= 1000kg/m^3 * 10 kg/s^2 * 1.3572m^2 = 13572

    I made somewhere a mistake again. Units are totally wrong.
     
    Last edited: Jan 22, 2014
  8. Jan 22, 2014 #7
    These are the pressures at the axis of the cylinder at the two ends faces (over and above atmospheric pressure). But the pressure is not uniform on the two end faces. It is lower on parts of the faces that are shallower in depth, and higher on the parts of the faces that are deeper. Fortunately for you, the variations with depth cancel out, and you can use the pressure at the axis as the average pressure on the end faces. Knowing the diameter of the cylinder, what is the area of each end face? What is the direction of the pressure force on each end face?
    (a) parallel to the axis of the cylinder
    (b)perpendicular to the axis of the cylinder
    (c) in the vertical direction or
    (d) in some other direction?
    What is the magnitude and direction of the force on each end face?
    In terms of the length of the cylinder, what is the vector sum (resultant) of the forces on the end faces?
     
  9. Jan 22, 2014 #8
    Area is equal to 1.131 m^2.
    Direction of pressure force on each face will be perpendicular to the axis of cylinder. And the force can be split into two forces each like I did in the 2nd picture.
    Oh bloody hell.
    Average pressure is 15000 kg/m^3.
    I cant really imagine how the forces vectors should look like. Force on the bottom of cylinder will be directed upwards and the one from top face will be directed opposite. Vector sum will be F= (p+ ρgH - p) = ρHAg

    Is that finally it?
     
  10. Jan 22, 2014 #9
    No. The pressure force on any surface is perpendicular to that surface. So the pressure force on each end face is parallel to the axis of the cylinder.

    You got the average pressure on each of the two faces correct: 21000 Pa (check your previous units) and 9000 Pa. You got the areas of the two faces correct, so what are the magnitudes of each of the forces on the two faces? What is the net of these two forces, and which way is it pointing:
    (a) up and to the left
    (b) down and to the right
    (c) some other direction
    [/QUOTE]
     
  11. Jan 22, 2014 #10
    Magnitudes at each face is:
    F1 = 10179 kg/(m^2 * s^2)
    F2 = 23751 kg/(m^2 * s^2)

    And the net of these two forces is... I think that the bottom Force is pointed up and left. And the top force is down and right.
     
  12. Jan 22, 2014 #11
    Those units should be Newtons. You are correct about the directions of the two forces. Which one wins out, and, numerically, by how much?
     
  13. Jan 22, 2014 #12
    The lower one has greater value, and the resultant will be 13 572 N.
     
  14. Jan 22, 2014 #13
    And how to calculate the force on the curved surface?
     
  15. Jan 22, 2014 #14
    Excellent. So the net resultant force from the two end faces is 13572 N, and it is up and to the left, parallel to the axis of the cylinder.

    Now let's turn attention to determination of the resultant pressure force on the curved face of the cylinder. There are two methods by which we can determine this force.
    1. Integrate the pressure vectorially over the area of the surface.
    2. Use Archimedes principle to determine the total force vector exerted by the water on the cylinder, and then vectorially subtract the forces on the end faces.

    Which method do you prefer? (One of these methods is easy, and the other is hard)

    Chet
     
  16. Jan 22, 2014 #15
    I think the 2nd one may seem more complicated but calculating integral is beyond my reach. We didnt have integrals yet.
     
    Last edited: Jan 22, 2014
  17. Jan 22, 2014 #16
    Method 2 is the simpler method. From Archimedes principle, what is the overall force exerted by the water on the cylinder, and what is its direction?

    Chet
     
  18. Jan 22, 2014 #17
    The force is directed upwards and its value is equal to the weight of our cylinder. Which is ummm, 13572 * sin30 degree?
     
    Last edited: Jan 22, 2014
  19. Jan 22, 2014 #18
    You're right about the direction, but not the magnitude. The magnitude of the force is equal to the weight of the displaced volume of water (which is equal to the volume of the cylinder). What is the volume of the cylinder? What is the density of water? What is the mass of the displaced water? What is the weight of the displaced water?

    Chet
     
  20. Jan 22, 2014 #19
    Volume of cylinder is V=3.14*(0.6^2) * 2.4 = 2.714m^3
    Density of water is 1000kg/m^3
    mass of water will be the the same as volume of cylinder so 2.714kg also.
    And to be honest my english isnt that good, mass and weight? Isnt that the same? :/


    So it should be: N= g*m*sin30= 2.714 kg * sin30 * 10 m/s^2 = 13.57 N?
     
    Last edited: Jan 22, 2014
  21. Jan 22, 2014 #20
    No. The mass of the water will be the volume times the density = 2714 kg
    No. The weight is a force (in Newtons). It is equal to the mass times g (i.e., mg). (I think you may have learned that in your course). So, again, what is the weight of the displaced water? This will then be equal to the total upward force of the water on the cylinder.

    Chet
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted