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Friction in wedges.

  1. Feb 19, 2010 #1
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    1. The problem statement, all variables and given/known data
    What horizontal force P must be applied to the weightless wedges to start them under the 450kg block,(the angle of friction for all contact surfaces is 10 deg.).


    2. Relevant equations



    3. The attempt at a solution

    I tried out the sum in the method that came to my mind first...this is illustrated in the diagrams...
    (coeff. of friction at all suraces (mu)=tan10=,176)
    the weight of the 450kg block on either wedge is 225kg(each share the weight equally)...but the friction along the top surface of the wedge(in green) is calcuated by mu*(component of the block's weight perpendicular to the wedge's upper surface i.e 225gcos15,where g=acc. due to gravity).
    Also,to calculate the friction at the base,we calculate the net downward force....
    N-225g+fsin15=0
    => N=225g-fsin15
    and F(in pink)= mu*N
    Thus,we can easily calculate the net horizontal force P to overcome these two frictional forces.....the problem is,my answer does not match the answer in the book(1417.31N)
     

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    Last edited: Feb 19, 2010
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  3. Feb 20, 2010 #2

    tiny-tim

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    Hi Urmi Roy! :smile:

    Won't P increase the normal force at the top?
     
  4. Feb 20, 2010 #3
    How?
     
  5. Feb 21, 2010 #4

    tiny-tim

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    Well, imagine it's all flat on a table (ie horizontal instead of vertical, so that there's no weight involved) …

    the P forces will create a normal force, won't they? :wink:
     
  6. Feb 22, 2010 #5
    Which one should I imagine as a flat surface? (I still haven't been able to figure this out.)
     
  7. Feb 23, 2010 #6
    I've been thinking,the P forces are along the horizontal itself,so how can they have any vertical component at all?
     
  8. Feb 23, 2010 #7

    tiny-tim

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    You're right, they can't.

    But they can have a component along any other direction than vertical …

    draw any pair of perpendicular axes, and P will have components along both of them, Pcosθ and Psinθ …

    the only exception is when θ = 90º (or 0), when the components are P and 0. :smile:
     
  9. Feb 24, 2010 #8

    nvn

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    Urmi Roy: Your normal force on the slanted interface is currently incorrect. Hint 1: You know the direction of the weight of the upper block (m*g). You know the direction of the normal force, N1, on the upper block. You know the direction of the frictional force on the upper block. Therefore, draw a free-body diagram of the left half of only the upper block showing these three forces, and perform a summation of vertical forces on the upper block, and then solve for the normal force, N1, on the upper block. Show your work, and show your numeric answer for N1; and then someone might check your math.
     
  10. Mar 1, 2010 #9
    The procedure nvn suggested is just what I did...and I checked and rechecked the calculation(even my teacher did)...I think I made a comceptual error......Thiugh I didn't understand which force tiny-tim was talking about,....perhaps that's where there's something wrong with my working?
     
  11. Mar 1, 2010 #10

    nvn

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    Urmi Roy: Notice the last sentence of post 8. Show your work, and then someone will be able to check your math.
     
  12. Mar 1, 2010 #11

    tiny-tim

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    Urmi Roy, I'm talking about the same normal force that nvn :smile: is.

    As he says, show us your calculations, and then we'll see what went wrong, and we'll be able to help.
     
  13. Mar 1, 2010 #12
    Okay,
    If N is the normal reaction at the bottom of the wedge(in pink),and f is the friction on the top surface of the wedge,we have

    f=mu(225gcos15)

    F=mu(225g-fsin15)

    Also,summing forces along the x axis,
    sigma Fx=(225gsin15)cos15 -fcos15 +P-F=0

    Since we know all the quantities in the third equation except P,we can find out P.

    (mu is the coefficient of friction at all surfaces)
    (PLease refer to the diagram to check the components of the forces)
     
  14. Mar 1, 2010 #13

    nvn

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    Urmi Roy: That is currently incorrect. It currently seems you did not understand post 8 yet. Let's call the pink arrow N2. And let's call the normal force on the slanted surface N1. Reread and perform post 8.

    Also, when you show your work, list not only the equations, but also show the numeric values you plug into the equations, and the numeric answer you calculate from each equation. Try again.
     
  15. Mar 11, 2010 #14
    I couldn't get back to this sum,since I had exams in the last one and a half weeks...anyway,you're right,I didn't really understand post 8...even now.....it seems what you are talking about is what I showed you in my working out.:((

    I just can't see where my current working is wrong.
     
  16. Mar 12, 2010 #15

    nvn

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    Urmi Roy: Pick apart post 8, sentence by sentence, and explain which words are causing you trouble. If you give post 8 a try (whenever you have time), I think you might find it works. Give it a try, and show your work.
     
    Last edited: Mar 12, 2010
  17. Mar 8, 2012 #16
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