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HELP! problem involving force of friction and normal force

  1. Feb 6, 2007 #1
    A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force of magnitude F = 6.0 N and a vertical force of magnitude P are then applied to the block. The coefficients of friction for the block and surface are μ_s = 0.40 and μ_k = 0.25. Determine the magnitude of the frictional force acting on the block if ...
    a) ...the magnitude of P is 8.0 N. (Hint -- is the block moving?)

    b) ...the magnitude of P is 10 N. (Hint -- is the block moving?)

    c) ...the magnitude of P is 12 N. (Hint -- is the block moving?)


    I have been stumped with this problem for a long time now. I understand the force being applied is both horizontally and vertically and that I have to decompose each component but I have no clue on how to do it. I tried finding the normal force by multiplying the weight 2.5 kg by 9.8 m/s2 and then using the kinetic mu to find the force of friction but this was wrong. Please help! I am really stumped.
  2. jcsd
  3. Feb 6, 2007 #2
    This is not clear, what is the direction of P?
  4. Feb 6, 2007 #3
    Sorry. P is going straight up on the box. It is going vertically up.
  5. Feb 6, 2007 #4

    Tom Mattson

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    Staff Emeritus
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    Gold Member

    And for that matter, what's the direction of F? It's not enough to say "vertical" or "horizontal". We need to know "up or down" and "right or left".
  6. Feb 6, 2007 #5
    Direction of F is right. Direction of P is up.
  7. Feb 6, 2007 #6
    You're on the right track. Don't forget to calculate the maximum force of static friction (using static mu). What might help is to draw a free body diagram. In many problems, the normal force will be equal in magnitude (but opposite in direction) to the weight. This isn't the case if the surface the object rests isn't horizontal. It's also not the case if something else is lifting (at least partially) your object. If you stand on a bathroom scale placed on a level surface, the bathroom scale will give you your weight, which would be equal to the normal force (the amount that the scale is pushing back up on you.) Now, imagine someone hanging on the ceiling attaches a rope to you and pulls up slightly... what will happen to the reading on the scale? This is what's going on in your problem - there's a vertical force to consider also.
  8. Feb 8, 2007 #7
    Can I get some real help with the problem please. I still do not understand how to find acceleration or normal force. If anyone could help me that would be great. Thanks
  9. Feb 8, 2007 #8

    Doc Al

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    Staff: Mentor

    To figure out the normal force, do this: Identify the vertical components of all forces acting on the block. (I count three vertical forces.)

    What must these forces add up to? (Assume the block does not leave the surface.) The only unknown force is the normal force; solve for it in each case.
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