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Incline Plane

  1. Nov 19, 2005 #1
    Knowing that the incline plane has an angle of 30 degrees and the coefficient of friction between a 60 lb block and the incline is 0.25, determine the smallest force P for which motion of the block up the incline is impending and the corresponding angle the force makes with the incline plane.

    Is P= 0.25*cos(30)*60+sin(30)*60? But the book gives the answer of 41.7 while my answer is 42.99...any ideas?
     
  2. jcsd
  3. Nov 19, 2005 #2
    First, where is your gravitaional constant g ?

    Isn't lb just pounds ?

    Secondly, the gravity component (ie the sine-part) is bigger then the friction component (the cosine part). This means that the block is going down with a force equal to the difference between the two parts in your equation. You need to apply a force , equal to that difference but opposite in direction, to make sure that the block does not go down anymore

    marlon
     
  4. Nov 19, 2005 #3
    Doesn't the sine part points at the same direction as the friction part...?
     
  5. Nov 19, 2005 #4
    And pound is a force.
     
  6. Nov 19, 2005 #5

    Fermat

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    P is at an angle to the incline. If you keep P acting along the plane then you will get P = 42.99...
    But if you angle P upwards from the plane. then it will have a lifting effect on the block which will reduce the friction, hence reduce the slope component required from P.

    Edit: looks like a minimisation problem, perhaps ???
     
  7. Nov 19, 2005 #6
    Can you get me started?
     
  8. Nov 19, 2005 #7

    Fermat

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    Hold on a moment.
     
  9. Nov 19, 2005 #8

    Fermat

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    Use the sketch below.
    Do the same as before, but adjust the friction value and the force normal to the plane to take the new position of P into account.
    You will get a function involving beta, the angle of P.

    Now differentiate P wrt beta in order to minimise P.
    http://img508.imageshack.us/img508/1593/physicss0fb.th.jpg

    Edit: changed maximise to minimise.
     

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    Last edited: Nov 19, 2005
  10. Nov 19, 2005 #9

    Fermat

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    I'll post my attachment onto image shack.
     
  11. Nov 19, 2005 #10
    Is this right?
    sum of forces alont the incline:
    P*cos(B)-0.25*cos(30)*60-sin(30)*60=0

    P*sin(B)+cos(30)*60-cos(30)*60=0
     
  12. Nov 19, 2005 #11

    Fermat

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    Just worked through it myself. I got 41.7 deg :biggrin:
     
  13. Nov 19, 2005 #12

    Fermat

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    P.sinB reduces the normal force on the plane and so reduces the friction component. You haven't included that in your calculations yet.
     
  14. Nov 19, 2005 #13
    I don't quite get it...
     
  15. Nov 19, 2005 #14

    Fermat

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    The normal force on the plane is mg.cos@ less the normal component of P, which is P.sinB.
    So, the normal force on the plane is NR = mg.cos@ - P.sinB.
     
  16. Nov 19, 2005 #15
    P*cos(B)-0.25( cos(30)*60-P*sin(B) ) - sin(30)(60)=0?
     
    Last edited: Nov 19, 2005
  17. Nov 19, 2005 #16

    Fermat

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    Yas!

    Now get P as a function of B and "minimise" it.
     
  18. Nov 19, 2005 #17
    Thank you!
     
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