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Find minimum force to move block

  1. Oct 3, 2013 #1
    1. The problem statement, all variables and given/known data
    "Figure 1 shows two wedge blocks A and B, weighing 1000 N and 50 N, respectively, in contact. The coefficient of friction between all contacting surfaces is 0.1 . What would be the smallest value of P required to push the block A upwards? What would be its value if there were no friction?"

    Ans, with friction: 887.3 N (given)
    Ans, without friction: 577.4 N (given)
    The gray rectangle in the picture is part of a separate problem.

    2. Relevant equations

    [itex]
    ƩF = 0
    [/itex]

    [itex]
    F_f = μF_N
    [/itex]

    3. The attempt at a solution
    I have solved problems of this nature before. I have a good understanding of vector addition and statics. I solved the problem sans friction in the following manner:

    [itex]
    F_N = W_A /cos30
    [/itex]

    [itex]
    ƩF_x = 0 ∴ F_N \cdot cos60 = P
    [/itex]

    [itex]
    P = W_A \cdot cos60 /cos30 = 577.4 N
    [/itex]

    With friction, however, I run into a problem: The side walls in contact with A exert a frictional force downwards. This of course depends on the normal force exerted by those walls, which partly balance the horizontal components of the normal and frictional forces exerted by B on A. But as far as I can tell, the vertical component of the normal force is equal to the sum of the weight and these frictional forces. So I encounter what appears to be a catch-22. I have tried various methods employing systems of equations but I get nowhere.
     

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    Last edited: Oct 3, 2013
  2. jcsd
  3. Oct 3, 2013 #2

    Simon Bridge

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    I would try the work-energy relation and conservation of energy.
    Otherwise you'd have to draw the free body diagrams.

    The wrinkle I see is that block A also gets pressed against the side of the whatsits that constrains it to the vertical - in proportion to the applied force. (The applied force ends up creating a couple with the reaction forces in the constraint - I suspect that this is "overthinking" the problem at the level it is presented.)

    You should get a number of equations where it appears that things have a circular dependence - and you can cancel terms.
     
  4. Oct 4, 2013 #3
    Ok, I will try that. Thanks for the help.
     
  5. Oct 4, 2013 #4

    Simon Bridge

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    At least you have nice angles:

    sin(30)=1/2
    cos(30)=(√3)/2
    tan(30)=1/√3

    ... you can replace the trig functions with the fractions (keep the surd until the end).
    Most of them seem to come out to 1/2.
    In all your triangles for this problem, find the 30deg angle and do all your trig with that as theta
    ... you will get less confused.
     
  6. Oct 4, 2013 #5

    Simon Bridge

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    Conservation of energy means: ##E_{in} = E_{out} + E_{lost}##.

    Applied force P over distance x, energy input: Ein = Px.

    This energy goes to gravitational PE, ##U##, in A, and three lots of friction:
    E1: between B and the ground
    E2: between the blocks
    E3: between A and the restraining channel.

    So: ##E_{in}=U+E_1+E_2+E_3##

    ##U=w_A\Delta y:w_A=m_Ag=1000\text{N}##
    B moving x means A moves up ##\Delta y=x\tan(30^\circ)=x/\sqrt{3}##
    ##\Rightarrow U=\frac{1}{\sqrt{3}}w_Ax##

    Without friction: ##P=\frac{1}{\sqrt{3}}w_A##

    ... which is where you are up to ;)


    As you have observed, the friction is the tricky bit.

    The ith energy lost to friction is ##E_i=f_id## where d is the distance moved against friction and ##f_i=\mu N_i## ... so you are finding expressions for the "normal force", ##N_i## ... which will mean drawing free-body diagrams. Assuming flush surfaces will probably avoid having to think about couples.

    i.e. For ##i=1## (the first friction): ##d=x##, and ##N_1 = w_B+\text{<contributions from }w_A\text{>} ##

    Try to express each normal force in terms of P and x and constants, where applicable.
    You'll end up with something like ##Px=\big(p(P)+q\big)x## where you can eliminate x and solve for P.
     
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