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Sliding/toppling of a block on a ramp

  1. May 15, 2009 #1
    1. A uniformly loaded crate of weight 1000N is held at rest on an inclined place. A gradually increasing force P is applied horizontally as shown. If coefficient of friction, u, = 0.2, determine whether the crate first slides or topples and the least force required 34fiedg.jpg



    I am unable to find the least force required. please help me solve this - the answer is 726 N



    3. Reaction R = 1000 cos 30 + P sin 30
    F = uR
    P cos 30 - F -mg sin 30 = 0
     
  2. jcsd
  3. May 15, 2009 #2

    tiny-tim

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    Welcome to PF!

    Welcome to PF! :smile:

    (have a mu: µ :wink:)
    Yes, that looks ok for the sliding equation …

    now you need a toppling equation …

    Hint: it will start to topple when the reaction force is at the corner, so take moments :smile:
     
  4. May 15, 2009 #3
    Hiya! Thanks for replying..

    According to the way i've worked it out... P comes out to be 697 N.. while the book's answer is 726 N.. this is stressing me out.. I can't figure out where's the mistake..

    For the toppling bit, i guess i'll have to do a moment balance about the corner.. and if P < P for the sliding bit, then the block topples.. am i right?
     
  5. May 15, 2009 #4

    tiny-tim

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    Yes. :smile:

    (but you can do a moment balance about any point … it makes no difference … just choose the point which you think will involve the least amount of arithmetic :wink:)
    Show us your full calculations, and then we can see what went wrong, and we'll know how to help! :smile:
     
  6. May 15, 2009 #5
    Hiya! Thanks for replying..
    I found P to be 697 N.. which does not coincide with that of the book.. its stressing me out :s
     
  7. May 15, 2009 #6
    Alright will do that...

    ∑V = 0
    R = 1000 cos 30 + P sin 30
    = 866 + 0.5 P

    F = µR
    = 0.2 ( 866 + 0.5P)
    = 173.2 + 0.1P

    ∑H = 0
    P cos 30 = F + 1000 sin 30
    = 173.2 + 0.1P + 500

    P = 673.1/(0.1 + cos 30)
    = 697N
     
    Last edited: May 15, 2009
  8. May 15, 2009 #7
    Hiya! Thanks for replying..
    I found P to be 697 N.. which does not coincide with that of the book.. its stressing me out :s
     
  9. May 15, 2009 #8

    tiny-tim

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    Nope … (-0.1 + cos 30) :wink:
     
  10. May 15, 2009 #9
    Thanks loads! You saved my day ;) OOps now P is coming to 878 N :s
     
  11. May 15, 2009 #10

    tiny-tim

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    Don't panic!

    it's probably the other P, then :smile:
     
  12. May 15, 2009 #11
    Ah i give up.. Thanks for your kind help sir :)
     
  13. May 15, 2009 #12

    tiny-tim

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    But you haven't done the moment equation yet …

    that'll probably give you a P that's less than 697N :smile:
     
  14. May 15, 2009 #13
    Attempting a moment balance about far right corner :p

    (P cos 30 x 1.6) = (1000 cos 30 x 0.7) + ( P sin 30 x 0.7) + ( 1000 sin 30 x 0.8)
    P = 971 N .. doesnt sound right
     
  15. May 15, 2009 #14
    I get P = 753 for the minimum tipping force. I don't think the answer in your book is right.
    Let x=1.4, y=1.6, w=1000N. Then the torque = (W/2) (x sin 30 + y cos 30) - P y cos 30.
    This gives P = (W/2) ( x/y tan 30 +1) = 753 for zero torque.
     
  16. May 15, 2009 #15

    tiny-tim

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    can't follow this :confused:

    you need the perpendicular distance from the corner to the line of the external force to equal the perpendicular distance from the corner to the vertical line through the centre of mass (because both the friction force and the tipping-normal force have zero moment).

    Try again. :smile:
     
  17. May 15, 2009 #16
    (P cos 30 x 1.6) = (1000 cos 30 x 0.7) + (1000 sin 30 x 0.8)
    P = 726 N!!! Wheee!! At last!.. .. thanks a lot for the walk through! Ah i'm not really a lumiere when it comes to mechanics.. You're an Ace!!
     
    Last edited: May 15, 2009
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