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Need help on a max and min force question!

  1. Oct 7, 2007 #1
    A block of mass 2.00 kg is pushed up against a wall by a force P that makes a 50.0° angle with the horizontal as shown below. The coefficient of static friction between the block and the wall is 0.258. (The attached picture might help).

    A.) Determine the possible values for the magnitude of P that allow the block to remain stationary.
    P max =
    P min =

    Any help would be greatly appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Oct 7, 2007 #2

    Doc Al

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    What forces (vertical and horizontal) act on the block?
     
  4. Oct 7, 2007 #3
    The force in the y-direction(vertical) would be Py = P(sin(50)) and in the x-direction(horizontal), the force would be Px = P(cos(50)). The x-direction would also include the normal force, n, and the y-direction would include gravity*mass, mg.
     
  5. Oct 7, 2007 #4
    I forgot to add that the x-direction would also include the normal force, n, which would be located along the negative x-axis. The y-direction would include gravity*mass, mg, on the negative y-axis.
     
  6. Oct 7, 2007 #5

    Doc Al

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    What about friction? What's its magnitude and which way does it act?
     
  7. Oct 7, 2007 #6
    Friction is the normal*coefficient of friction. I found that the normal force is n = Pcos50. To find the max wouldn't you make friction negative because it would be acting downwards and vice versa for minimum? I just not sure how to put all these into an equation to solve for P. I also know that the acceleration would be zero.
     
  8. Oct 7, 2007 #7

    Doc Al

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    Exactly right.

    Just set vertical forces equal to zero and you can solve for P.
     
  9. Oct 7, 2007 #8
    I don't think i'm getting the right answers....So would the equation (for the max) be
    ( Psin50 - (-friction(Pcos50)) - mg = 0 )? And the Psin50 and mg would cancel out because they would equal zero? I don't think i'm doing this right...
     
  10. Oct 7, 2007 #9

    Doc Al

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    So far, so good--except for that double minus sign. I'd write it like this:
    [tex]P\sin{50} - \mu P\cos{50} - mg = 0[/tex]

    Now just solve for P.

    Why would they cancel?
     
  11. Oct 7, 2007 #10
    I thought that's what you meant by set the vertical forces to zero because Psin50 and mg are vertical forces. When i solve for Pmax i'm getting 154.282 N. If you made the friction positive to find the min wouldn't that make the force the same but just negative?
     
  12. Oct 7, 2007 #11

    Doc Al

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    But friction is a vertical force also! There are three vertical forces, which add to zero.
    I don't see how you got this result.

    No.
     
  13. Oct 7, 2007 #12
    If the vertical forces, Psin50, -friction, and -mg add to zero, i don't see how you can solve for P or how it could be a reasonable answer. I think i'm making this harder than it is :S
     
  14. Oct 7, 2007 #13

    Doc Al

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    Solve the equation I gave in post #9. The only unknown is P; everything else is just numbers.
     
  15. Oct 7, 2007 #14
    Psin50 + P(.258)(cos50) - 2kg(9.8) = 0

    Psin50 + P(.258)(cos50) = 19.6

    2P = 19.6 / (cos50)(sin50)(.258)

    P = 154.282 / 2
    (I forgot to divide by two last time i got 154.282)

    P = 77.141 N

    Or am i making some stupid basic math mistake?
     
  16. Oct 7, 2007 #15

    Doc Al

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    That first + sign should be a minus sign.

    That should be:
    Psin50 - P(.258)(cos50) = 19.6

    No good. Factor out the P.

    For example, note that Xa - Xb = X(a - b)
     
  17. Oct 7, 2007 #16
    I always seem to make those basic math mistakes! I entered my answers and they were correct. Thank you for all your help Doc Al!
     
  18. Oct 7, 2007 #17

    Doc Al

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    Excellent. Why don't you mark this one solved.
     
  19. Oct 7, 2007 #18
    Actually, the is a part b and it has the same numbers except for an angle of 13 degrees from the horizontal. This time I'm getting a negative number for some reason
     
  20. Oct 7, 2007 #19

    Doc Al

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    If the only difference is that the new angle is 13 degrees (instead of 50), then you can solve it in exactly the same way.
     
  21. Oct 7, 2007 #20
    When i do that i get P = 19.6 / -.02 which turned out to be -714 ish
     
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