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A boy holds a metal sign that weights 100N against a wall by pushing on it horizontally?

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  1. Oct 15, 2014 #1
    1. The problem statement, all variables and given/known data

    If the coefficient of static friction between the sighn and his hands is 0.6, and the coefficient of static friction between the sign and the wall is also 0.6, with what force must the boy push to keep the sign in place?

    2. Relevant equations

    0.6(Fn)=100 N

    Fn=166.7 N

    3. The attempt at a solution

    So my question is this: Why does the boy only need to exert half of the normal force in order for the sign to stay in place?

    My reasoning is this: he pushes the sign, which pushes the wall, the wall pushes against the sign with the same force that the boy is exerting. But this makes me want to think that there are two normal forces, one from the boy and the wall and one from the sign and the wall. Is this true, or is there another reason the boy pushes with only half the normal force? Thanks!
     
  2. jcsd
  3. Oct 15, 2014 #2

    Simon Bridge

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    What would happen if the friction coefficients were different?
     
  4. Oct 15, 2014 #3
    Hmm, so if one coefficient was 0.6 and the other 0.7, the net coefficient force would be 1.3.

    So does that mean that there's only one normal force, but the boy is dancing with two frictions: one with him and the sign and the other with the sign and the wall? Which causes the net coefficient to be 1.2?
     
  5. Oct 15, 2014 #4

    Simon Bridge

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    1.3 is not a force.
    It is important to be careful in what you write. You mean the sum of the friction coefficients is 1.3.

    Can you write out the general equation for the minimum applied force ##F## needed to hold the sign up in terms of the mass ##m## of the sign, the acceleration of gravity ##g##, and the two coefficients of static friction ##\mu_h## between hand and sign and ##\mu_w## between sign and wall?

    Where does the "normal force" come from - what is it?

    But the answer to your question is that when the coefficients of friction are the same, boy and wall contribute the same amount to the friction.
    Which seems a bit like what you were trying to say.
     
  6. Oct 15, 2014 #5
    Sure:

    ##F## = (##m## X ##g##)/(##\mu_h## + ##\mu_w##).

    So what I'm getting is that since there are two friction coefficients, they add together?

    My first thought was that the force required was equal to 100/0.6. But I'm not sure why it's half of that.
     
  7. Oct 16, 2014 #6

    Simon Bridge

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    Well done - note: you don't need the "times" sign.
    If ##\mu_h=\mu_w=\mu##, then $$F=\frac{1}{2}\frac{mg}{\mu}$$ ... this is the situation you had: see the 1/2?
    If ##\mu_h=2\mu_w=2\mu##, then $$F=\frac{1}{3}\frac{mg}{\mu}$$ ... now the force is 1/3mu of the weight.

    Oh wait - you mean: why does the kid have to supply half this force and the wall supplies half as well?
    What happens when the boy and the wall do not provide the same normal force?
     
  8. Oct 16, 2014 #7
    Maybe a free body diagram for the bloc will help you to understand better.
    How may forces act on that sign? How many are up?

    It just happens here that the coefficients add up. But this is not a general rule.
    What you add up, in general, is forces. Here you have the same normal force on both sides of the sign (you understand why?) and this makes the coefficients to "add up".
     
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