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Homework Help: Very simple Friction problem.

  1. Nov 11, 2009 #1
    A horizontal force F of 12 lb pushes a block weighing 5.0 lb against a vertical wall. The coefficient of static friction between the wall and block is 0.60 and the coefficient of kinetic friction is 0.40. assume the block is not moving initially. (a) Will the block start moving? (b) What is the force exerted on the block by the wall?

    F = ma, Ff=usN

    I don't know what isn't clicking, but apparently the object is NOT moving.

    My attempt is as follows;

    The object is about to break when usN = mg. Since N = mg, usmg = mg.

    However, (0.6)(12) [tex]\neq[/tex] 5.
  2. jcsd
  3. Nov 11, 2009 #2
    What dont you like? I cant see a mistake and it seems normal that it wont move.
  4. Nov 11, 2009 #3
    My problem is that I continuously find the force of static friction to be larger than the force due to gravity; which would seem to suggest that the object should be moving upward?

    Or is it that since the maximum static friction is 7.2 N (0.6x12N) the object cannot surpass the threshold? Because that would make sense if you don't consider the static friction capable of pulling an object backwards.

    I'm just very confused.
  5. Nov 11, 2009 #4
    AA I see.. The frictional force is a very special force. The value you find is the Maximum frictional force that can exist between the wall and the body. This force in this situation is equal to the weight. Think about a heavy book on your desk.. You start pushing it ( its not moving ). you exert a force that is constantly increasing as you are pushing harder. At a point the book starts moving.The force before it started moving was constantly less than the maximum friction but obviously the box didnt go backwards.

    Think also about the role of friction. It resists the motion. It cant produce motion.
  6. Nov 11, 2009 #5
    That makes perfect sense.

    So by finding the maximum frictional force I inadvertently found how heavy the object would have to be to begin moving.

    The actual frictional force is equal to the force applied by the object until a certain threshold is met. In this case the threshold is [tex]\approx[/tex]7.2. But what this means is that the object will be held in place until Ffmax can be overcome.

    I think that's right.
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