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Block pushed up against a wall

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data
    A horizontal force of 12 N pushes a 0.5-kg block against a vertical wall. The block is initially at rest. If us = 0.6 and uk = 0.8, which of the following is true?
    (a) The frictional force is 4.9N
    (b) The frictional force is 7.2N
    (c) The normal force is 4.9
    (d) The block will start moving and accelerate
    (e) If started moving downward, the block will decelerate

    2. Relevant equations
    fk = UkN
    N = mg

    3. The attempt at a solution
    N = 0.5kg * 9.8m/s2 = 4.9

    I marked down the answer as C, but I was wrong...
  2. jcsd
  3. Nov 10, 2008 #2
    I think the friction is related only to the 12N and not to the mg component.
    So the reaction force at the wall is equal to 12N.
    12*us=7.2N (b)
    and we can disprove (d) and (e) by stating 0.5*9.81=4.9N
    The mass of the block is not great enough to overcome friction.

    Short answer: (b)
  4. Nov 11, 2008 #3
    I'm still having a hard time understanding how you knew it was B.
  5. Nov 12, 2008 #4


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    1. The normal force is the force of the wall perpendicular to the block.
    2. The friction force between the wall and block, which is u_k(N) if the block moves, or less than or equal to u_s(N) if it doesn't move, acts parallel to the wall , upward.
    3. The weight of the block acts parallel to the wall, downward.

    So the correct answer is________________??
  6. Nov 12, 2008 #5
    ahh so the Normal force is actually 12 N (horizontal).
    The frictional force is then 12*.6 = 7.2

    And like craigman proved, the block is not moving, so it can't be D or E.

    Makes sense. So the answer is B.

  7. Nov 12, 2008 #6


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    But what's the weight of the block? If the block is not moving, what must the friction force be?
  8. Nov 12, 2008 #7
    The weight is 0.5kg, no?

    If the block is not moving, the frictional force (7.2 N) must be than F = m * a = .5 * 9.8 = 4.9... Which it appears to be.
  9. Nov 12, 2008 #8


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    that's the mass, the weight is 4.9N
    I'm not sure if you answered the question. The weight down is 4.9N, and the block is not moving. Thus the friction force acting up, in accord with Newton 1, must be equal to ________________??
  10. Nov 12, 2008 #9
    Well then the friction force must be 4.9 as well. So A.
  11. Nov 13, 2008 #10


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    Yes. A common oversight is to assume that F_f(static) = u_s(N), but that is its maximum value, only true when the block is just on the verge of moving with respect to the surface it is in contact with. Otherwise , in general, F_f is less than or equal to u_s(N). For dynamic or kinetic friction, F_f(kinetic) = u_k(N), always.
  12. Nov 13, 2008 #11
    If the block were in contact with the ground and not the wall, the friction force would be calculated by 9.81*0.5*u=friction force.
    In this case gravity does not actually push the block into the wall and the force of m*g is acting perpandicular to the wall, which means only the horizontal force of 12N is acting to be a multiplier in the friction equation.
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