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Statics: pushing a block up a wall w/ friction

  1. Jun 27, 2017 #1
    1. The problem statement, all variables and given/known data
    They ask for the smallest force P to apply if neither block is to slip. Coeff. of friction given as 0.25.
    The 20KG block is 196N; the 10KG block is 98.1N



    2. Relevant equations


    3. The attempt at a solution


    If we start at the 10KG block:

    Sum of forces x: N + cos26.6P – (cos55.33)(N of 20KG block)

    Sum of forces y: -98.1 –sin26.6P + 0.25N + (sin55.3)(N of 20KG block)

    So, P pushes the 10KG block away from the wall, and N pushes the block away from the wall; and therefore we need a corresponding force in the x direction to pull it back toward the wall. That force must come from the 20KG block, which is a function of its normal force (multiplied by mu, which we know).
    How do we find that?
    Well, for the bigger block:
    Sum of forces in y: -196.2 + N - (sin55.3)(98.1)-cos16.7P
    We have to add in the downward pull of the smaller block (as a function of its geometry) here, right?

    Stuck here. Too many variables: Two values of N (one for each block) plus P.

    Any help would be appreciated. Thank you :^)
     

    Attached Files:

  2. jcsd
  3. Jun 28, 2017 #2

    CWatters

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    You should start by drawing some FBD diagrams.
     
  4. Jun 28, 2017 #3
    I would find the force of tension applied by the rods. The force in the vertical direction could be found, I believe, by finding the force of gravity and friction on the 10 kg block. Knowing that force should help you with the problem. I solved the problem using that method and it seemed to work out.
     
  5. Jun 28, 2017 #4
    OK, here's my FBDs.
    I wasn't sure how to handle the influence of the 10KG block on the 20KG block. From the point of view of the 20KG block, the 10KG block is pulling it down and to the right. But aren't both these forces also a function of the friction the smaller block feels, since it is dragging against the wall?
     

    Attached Files:

  6. Jun 28, 2017 #5

    CWatters

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    I think you will find it easier if you make THREE free body diagrams where the third one is point B.
     
  7. Jul 6, 2017 #6
    OK, here's my FBD about point B.
    From B's point of view, the 10KG is pulling it to the right and down slightly. The downward pull is offset by mu times its N.
    From B's point of view, the 20KG block is pulling it up and to the left slightly. The leftward pull is offset by mu times its N.
    And we have force P pulling straight down.

    Is my logic correct? prob822FBDaboutB.jpg
     
  8. Jul 7, 2017 #7
    I wouldn't approach the problem this way. You should make an FBD of the system as a whole, that way you won't have to deal with the internal forces of each of the 4 rigid bodies.

    This FBD would consist of: the P force, the weights of both blocks, the 2 normal forces and the 2 friction forces (mu * Corresponding normal force)
    Now all you have to do is: sum of forces in the x and y directions and sum of torques at points A or C.
     
  9. Jul 9, 2017 #8

    OK, that makes sense. I am embarrassed to admit I overlooked torques. So here we go:

    Sum of forces in Y: -P -196.2 - 98.1 + Ntop + .25Nright = 0
    notation: Ntop is normal force of top block
    Sum of forces in X: -.25Ntop + Nright =0
    Cool! We immediately know a relationship between the 2 normal forces
    Sum of torques about A: -.075P - (0.225)(98.1) + (0.225)(.25)(Nright) + (0.325)(Nright)=0

    counterclockwise is positive

    For torques, P, is counterclockwise, the 10KG wt. is counterclockwise; these are balanced by clockwise friction and Nright

    Is my logic correct? I wasn't sure if I should include the Nright in the torque calculation.
    Three equations, 3 unknowns; should be solvable, right?
     
  10. Jul 10, 2017 #9
    Your logic is correct, now just do the math and solve for P.
     
  11. Jul 11, 2017 #10
    OK, here comes the math.

    -P -294.3 + Ntop + 0.25Nright=0
    We know that Ntop=4Nright, so we can immediately substitute, giving us:
    -P -294.3 + 4.25Nright Let's call this eqn 1

    The torque eqn is:
    -0.075P-22.07+0.056Nright+.325Nright=0
    -0.075P-22.07+0.381Nright=0 Let's call this eqn 2

    Let's multiply eqn 2 by (-11.15), giving us:
    0.83P + 246 -4.25Nright

    We add this to eqn 1; the Nright values cancel, leaving us:
    -0.17P -48.3=0
    P= -284

    Hmmmm.....a negative value of P?
    The book's answer is 371, although it does not specify direction.
    Did I make a math or logic error here somewhere?
     
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