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Work out force transmitted to sensor

  1. Dec 19, 2014 #1
    1. The problem statement, all variables and given/known data
    There is a wind tunnel force balance allowing the
    measurement of the drag force resultant at point A of a test object in the air
    stream from an arm and lever system mounted on a pivoting point O and
    pressing onto a force sensor at point B.

    1) What will be magnitude of the force transmitted into the sensor at B if the air
    flow results in creating a lift force of 1.75 N and a drag force of 1.20 N?

    I couldn't paste in the picture it wouldn't let me. but there is a 90 degree lever with the first part of the lever being vertical in the direction of lift and is 1.2m long. There is then a pivot, O, and to the right of that at 90 degrees is a horizontal part of the lever and is 0.5m long. sensor B is underneath this part so that when drag occurs it will push on the lever and push down onto B.

    2) Calculate the horizontal component Ox and vertical component, Oy of the transmitted force in pivoting point O.
    (Consider the coordinate system used to be positive in the direction against the airflow for the x component, and positive in the upward direction for y)
    Give your answer in Newtons (N)


    2. Relevant equations
    Pythagoras?

    3. The attempt at a solution
    question 1 - The sensor is used to measure lift so is the answer simply 1.20N?
    I tried resolving the force and then did some moments but then realised I only need an answer in Newtons and I would get an answer in N/m.

    question 2 - is it -1.20 since the force transmitted is in the opposite direction to airflow? and vertical component 0? I'm confused because the lengths of the lever arms seem to be irrelevant.

    This is my first post, thank you for reading and helping if you can!
    Thanks.
     
    Last edited by a moderator: Dec 19, 2014
  2. jcsd
  3. Dec 19, 2014 #2
  4. Dec 20, 2014 #3

    CWatters

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    Is it? The problem statement says..

    The answer isn't 1.2N. There are levers of different lengths involved.
     
  5. Dec 20, 2014 #4
    Sorry I meant to say drag,
    So how do you work it out then? Because why would you have a sensor that measures drag that doesn't read 1.2N?
    Thank you!
     
  6. Dec 21, 2014 #5

    CWatters

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    (For Q1) This diagram is effectively the same problem. Any easier to understand?
    Equivalent.jpg
     
  7. Dec 21, 2014 #6

    CWatters

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    The question doesn't ask what the sensor displays it asks for the "force transmitted into the sensor"
     
  8. Dec 21, 2014 #7
    Ah that is a bit easier, is it 0.5N then?
     
  9. Dec 21, 2014 #8

    CWatters

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    Try again. Have a look at the length of the levers. Will the force be greater or less than 1.2N ?
     
  10. Dec 21, 2014 #9
    Actually it would be more, but how do you do it without getting the answer in Newton Meter's? Is it 2.88N?
    Thanks
     
  11. Dec 21, 2014 #10

    CWatters

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    Correct.

    Torque = force * distance
    = 1.2N * 1.2M
    = 1.44 NM

    It's the same torque on the other arm so using..

    Force = Torque/Distance

    = 1.44NM / 0.5M
    = 2.88N

    Have a go at Q2. May help to add the 2.88N force to the diagram.
     
  12. Dec 21, 2014 #11
    Great thank you!
    How about the x and y components? How would I go about solving that? Trig?
     
  13. Dec 21, 2014 #12

    CWatters

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    Start with the original drawing in the link you posted and produce a free body diagram for the cranked beam.

    eg Add..

    a) the 2.88N force that the sensor puts on the beam.
    b) arrows representing the x and y forces at the pivot.

    The beam isn't accelerating (in any direction) so what does that mean for the net force on the beam (in any direction)?

    PS: Remember the question says...

     
    Last edited: Dec 21, 2014
  14. Dec 21, 2014 #13
    It is 0?
    So you'll have the reaction force in the y direction?
     
  15. Dec 21, 2014 #14

    CWatters

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    The net force in the horizontal (or vertical) direction will be zero because F=ma so if a = 0 then F=0.

    But show your working. The recommended procedure for any similar problem is to:

    Make the drawing
    Define which direction(s) are positive
    Label the forces
    Write the equations
    Solve any simultaneous equations (if necessary)
    Substitute values
    Check answer makes sense.

    PS In most exams I've sat you get marks for getting the procedure right even if you make an arithmetic somewhere near the end. If you only give the answer and it's wrong due to a trivial error you get zip.
     
  16. Dec 21, 2014 #15
    Thank you so much! That has helped lots thank you! Yeah we get that in the UK too, could get most the marks with all the wrong answers
    So in the x direction it'll be -2.88N since it'll be in the opposite direction to air flow
     
  17. Dec 21, 2014 #16

    CWatters

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    I'm also in the uk.

    The sign is correct but how did you get the 2.88N part?
     
  18. Dec 21, 2014 #17

    CWatters

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    Correction the sign is also wrong...

     
    Last edited: Dec 21, 2014
  19. Dec 21, 2014 #18
    Because it is transferring 2.88N of force to the sensor, so if y is 0N, then the x component must be the remaining 2.88N?
    Damn. All wrong
     
  20. Dec 21, 2014 #19

    CWatters

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    Err that sounds all mixed up.

    Back in a moment.
     
  21. Dec 21, 2014 #20

    CWatters

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    Here is my Free Body Diagram.

    FD = Drag Force = -1.2N
    FL = Lift Force = 1.75N
    FS = Sensor force = 2.88N
    FY = Y component of force at pivot
    FX = X component of force at pivot

    The drag force is to the left so that is also the direction of the airflow.

    The problem says to assume:
    X is "positive in the direction against the airflow" so +ve is to the right.
    Y is "positive in the upward direction"

    Have a look and then write an equation that sums all the horizontal forces to zero.

    Wind tunnel balance.jpg
     
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