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Friction and Wedges

  1. Nov 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Provided the following diagram determine the reversed force of P needed to pull out the wedge, A.
    6a9b6b6b3b4780d93b80dd6917a408fc.png

    2. Relevant equations
    Given the coefficient of static friction between A and C and between B and D is 0.2 and between A and B static friction is 0.1. Weights of wedges are neglected.
    Force of friction = static coefficient * normal (When impending motion is occurring)

    3. The attempt at a solution
    Let NC be the normal at C
    ND be the normal at D
    NB be the normal between wedge A and wedge B
    FC be the friction force between wedge A and wall C
    FB be the friction force between wedge A and wedge B
    FD be the friction force between wedge B and wall D

    For FBD of wedge A:

    X-direction:
    0 = 0.2 NC - NBsin15 - 0.1NBcos15 - P
    simplifying 0= 0.2 NC - 0.3554NB - P (Equation 1)
    Y-direction:
    Nc - NBcos15 + 0.1NBcos15=0
    simplifying Nc=0.94 NB (Equation 2)

    For FBD of wedge B:


    X-direction:
    NB= ND/(sin15) (Equation 3)

    Y-direction:

    NBcos15 - 3000 + 0.2ND = 0 (Equation 4)

    substituting equation 3 into 4 and solving ND = 763lb


    From equation 3 then NB = 2947.9 lb

    From Equation 2 NC = 2771lb

    From Equation 1 P = 493.5 lb

    My answer is considerably off the correct answer which is 106 lb
     
    Last edited: Nov 30, 2015
  2. jcsd
  3. Nov 30, 2015 #2

    SammyS

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    (Try to not use Bold type face except where necessary.)

    Isn't P supposed to be pulling the wedge out? That would mean that friction forces acting on wedge A are exerted toward the right.[/B]
     
  4. Nov 30, 2015 #3
    This is why the question says the reverse force of P so -P. I accommodated for this in my equations. Sorry about the bold my bad
     
  5. Nov 30, 2015 #4

    SammyS

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    Equation 1 is not an equation. There's no " = " in it.

    In the line before that there is no P, and shouldn't 0.1NBcos15° be to the right, so have " + " ?
     
  6. Nov 30, 2015 #5
    Sorry I forgot to add that variable when typing the corrections have been made,
    0 = 0.2 NC - NBsin15 - 0.1NBcos15 - P
    simplifying 0= 0.2 NC - 0.3554NB - P (Equation 1)
     
  7. Nov 30, 2015 #6

    SammyS

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    That doesn't address the sign on 0.1NBcos15° .
     
  8. Nov 30, 2015 #7
    Sorry you are absolutely correct let me work through this with this change, thank you!
     
  9. Nov 30, 2015 #8
    Excellent, Thank you so much for your time it worked out! Sorry for wasting your time on such a trivial mistake. I was looking at this for hours and didn't notice it.
     
  10. Nov 30, 2015 #9

    SammyS

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    No problem.

    Lately I've been helping some people who understand almost nothing and at times seem to act clueless on purpose. It's great to be able to make a few suggestions and have them recognized for what they are,
     
  11. Nov 30, 2015 #10
    Yeah I really appreciate that I think it is important to work from ones mistakes and learn from them. I appreciate you not just giving me the answer but rather guiding me.
     
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