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Reaction force in wheels?

  1. Jul 28, 2013 #1
    1. The problem statement, all variables and given/known data

    In the diagram below, work out the reaction force in each of the wheels.
    Note there are two wheels that make up Ra.

    15dkmq0.png



    3. The attempt at a solution

    0 = 25,000*2 + 5,000*15 -5 Rb - 8,000*2
    0 = 50,000 + 75,000 - 5 Rb - 16,000
    0 = 109,000 - 5 Rb
    5 Rb = 109,000
    Rb = 109,000 / 5 = 21,800 (5,600)

    8,000 + 25,000 + 5,000 = 38,000 N
    38,000 - 21,800 = 16,200 N
    Ra = 16,200 N
    Ra*2 = 32,400 N
    38,000 - 32,400 = 5,600 N
    Rb therefore = 5,600 N

    Ra*2 = 32,400. Ra = 16,200 N
    Rb = 5,600 N

    This has been marked incorrect, the '5 Rb' reference especially.
    And Ra and Rb are wrong.
     
  2. jcsd
  3. Jul 28, 2013 #2

    Doc Al

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    Staff: Mentor

    Looks to me like you are using the A wheel as your axis. So what's the distance between the wheels?
     
  4. Jul 28, 2013 #3
    16 meters between the wheels (Ra to Rb).
     
  5. Jul 28, 2013 #4

    Doc Al

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    Right. So why did you write "5 Rb"? Correct that.
     
  6. Jul 28, 2013 #5
    Would that make it "16 Rb" instead then?

    I thought it was 5 Rb due to the 5 kN force being 1 meter away from Rb = 5*1 = 5?
     
    Last edited: Jul 28, 2013
  7. Jul 28, 2013 #6

    Doc Al

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    Right.

    What matters is the distance of each force from wheel A, which is your axis of rotation. You already accounted for the torque from the 5 kN force, which is 15 m from wheel A.
     
  8. Jul 28, 2013 #7
    So would it just be the case of completely removing "5 Rb" from the equation if I've already accounted for it via (5,000*15)?
     
  9. Jul 28, 2013 #8

    Doc Al

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    5,000*15 is the torque due to the 5 kN force, not the torque due to the unknown force Rb. You still need a torque term due to Rb.
     
  10. Jul 28, 2013 #9
    And that would be "16 Rb"?
     
  11. Jul 28, 2013 #10

    SteamKing

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    You must also be careful in finding the reaction at each of the front wheels once you have solved for RA.
    You should write a 'sum of the forces' equation to go along with your 'sum of the moments about A' equation.
     
  12. Jul 28, 2013 #11

    Doc Al

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    Exactly. (Making sure to give it the proper sign.)
     
  13. Jul 28, 2013 #12
    0 = 25,000*2 + 5,000*15 -16 Rb - 8,000*2
    0 = 50,000 + 75,000 - 16 Rb - 16,000
    0 = 109,000 - 16 Rb
    16 Rb = 109,000
    Rb = 109,000 / 16 = 6812.5 N (round it to 6800)

    8,000 + 25,000 + 5,000 = 38,000 N
    38,000 - 6800 = 31200 N
    Ra = 31200 / 2 = 15600 N
    38,000 - 31,200 = 6,800 N
    Rb therefore = 6,800 N

    Is this any better?
     
  14. Jul 28, 2013 #13

    Doc Al

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    Yes, looks good. (Assuming that Ra is the force on each front wheel.)
     
  15. Jul 28, 2013 #14
    Yep, 15,600 N on each front wheel (15,600*2 = 31,200)
     
  16. Jul 28, 2013 #15
    Right, still a bit unsure.
    Just dug up some old notes related to moments;

    Moment of a force (Nm) = force (N)*perpendicular distance from pivot to line of action of the force (M)
    M = F*X

    Perpendicular distances = moment = force*distance from pivot

    Balanced object = sum of clockwise moments = sum of anti-clockwise moments

    • Decide where the pivot is.
    • Decide whether the object is balanced or unbalanced.
    • For each force, decide whether the force is tending to turn the object clockwise or anti-clockwise.
    • Decide whether any distance given in the question is the perpendicular distance from the pivot to the line of action.


    Is the aircraft in the question balanced or unbalanced? (just want to clarify before I re-submit for marking)
     
  17. Jul 28, 2013 #16
    What do you think? What criteria should be used?
     
  18. Jul 28, 2013 #17
    I'd say unbalanced.

    The pivot is Ra.
    25 kN and 5 kN forces are tending to turn it clockwise and the 8 kN force anti-clockwise.
    So a total of 30 kN clockwise force and 8 kN anti-clockwise, would tend to make one assume it's unbalanced?
     
  19. Jul 28, 2013 #18

    Doc Al

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    Well is the plane sitting in equilibrium or not? Any reason to think it is rotating?

    Note that you had to assume equilibrium (and balanced torques and forces) in order to solve for the reaction forces.
     
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