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Friction Problem

  1. Oct 6, 2004 #1
    I am having trouble with this problem. I found the answer to a to be 14 N, but the online quiz says it is wrong. Fx-f=0. Fcos(theta)-f=0. f=18cos40=14 N. What am I doing wrong?

    A 3.5 kg block is pushed along a horizontal floor by a force F of magnitude 18 N at an angle = 40° with the horizontal (Figure 6-20). The coefficient of kinetic friction between the block and floor is 0.25.
    (a) Calculate the magnitude of the frictional force on the block from the floor.
    (b) Calculate the magnitude of the block's acceleration acceleration.

    [​IMG]
     
    Last edited: Oct 6, 2004
  2. jcsd
  3. Oct 6, 2004 #2

    arildno

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    Welcome to PF!
    In order to delete in the other forum, press the "Edit button". On top of that, there's a "Delete" option.
     
  4. Oct 6, 2004 #3

    Pyrrhus

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    [tex] F_{x} - F_{f} [/tex] is not 0, It is moving on the x-axis.

    Remember

    [tex] F_{f} = \mu N [/tex]
     
  5. Oct 6, 2004 #4

    arildno

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    For your problem:
    Note that the NORMAL force acting on the block must be GREATER than the weight, due the vertical component of F.
    Hence, the frictional force is also greater..
     
  6. Oct 6, 2004 #5
    So if [tex] F_{f} = \mu N,[/tex] then [tex] F_{f} = \mu * (mg+18sin40) = .25 * ((3.5*9.8)+18sin40) = 12 N? [/tex]
     
  7. Oct 6, 2004 #6

    arildno

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    That looks correct, yes.
    (Do you understand, to your own satisfaction, why you need that addition to the weight?)
    I haven't checked your numbers, though..
     
  8. Oct 6, 2004 #7
    Oops. I mistyped. The correct answer is 11 N I'm hoping. Yes I do now see why the normal force is greater than the weight. Thanks for all of the help.
     
  9. Oct 6, 2004 #8

    arildno

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    Perhaps you shouldn't round down to an integer answer.
     
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