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What is the minimum pushing force required

  1. Nov 7, 2008 #1
    1. The problem statement, all variables and given/known data

    So there are 2 boxes: 1 has a mass of 16kg and the other is 88kg. There is friction between 2 boxes, the surface which box 88kg is on is frictionless. Static friction coefficient is 0.38. Box 16kg is not on the surface, it's is push against the 88kg box. What is the minimum pushing force required in order for box 16kg not drop

    2. Relevant equations


    3. The attempt at a solution

    So I do F=ma in the y direction which is Fs - mg=0 to find Fs. Then I do in the x direction F-Fn= 0. From Fs I can get Fn by dividing Fs by 0.38.

    Did I solve the problem right ?
  2. jcsd
  3. Nov 7, 2008 #2
    Re: Friction


    It's in the middle of the night for me and I've got high fever but I think you made one single mistake. You should not divide Fs by 0.38 but rather multiply them.

    Think of it this way. If you divide Fs by the friction coefficient and the friction coefficient --> 0 that would imply the friction force would --> infinity, and that dos not seem right... :wink:

    Hope this helps!

  4. Nov 7, 2008 #3
    Re: Friction

    But Fs = Fn * coefficient of static friction right ? so Fn =Fs[/SUB / 0.38 right ?
  5. Nov 8, 2008 #4
    Re: Friction

    I might be confused by your nomenclature. I think you might have mixed up Fs and Fn, though it might be me reading the problem or your solution wrong as well. Anyway, here's how I would solve the problem:

    Mass m1 = 16 kg
    COF u = 0.38
    acc. g = 9.82 m/s2

    Find the friction force FF

    F = mg [1]
    u = FF/FN [2]

    My calculations:
    Force normal to the plane, FN, is found using [1] (FN=m1*g) [3]

    [2] is rewritten as FF=u*FN [4]

    Using [3] in [4] yields FF=u*m1*g

    This gives a numerical value of roughly 60 N

    Questions, comments?

  6. Nov 8, 2008 #5
    Re: Friction

    probably I confused you guys.

    The problem is not like that.

    The 16 kg is on the left side of the 88kg and kinda pushed against the 88kg but in the air, 16kg box is not on the frictionless surface. We need to find the min F pushing the 16kg box against the other in order for the 16kg not drop. Get it ??
  7. Nov 8, 2008 #6
    Re: Friction

    I see...
    Now I understand the problem and your solution. And now I agree with you and your solution. Sorry for my confusion.

    Just for clarification:
    The friction force, FF, is linearly dependent on the applied force,FP (P for Push), on the 16kg-box:

    FF= u*FP [1]

    FN=m16*g [2]

    To balance the box you need top satisfy FN = FF [3]

    [1] and [2] in [3] gives: m16*g = u*FP [4]

    And [4] can be rewritten as FP = (m16*g)/u

  8. Nov 8, 2008 #7
    Re: Friction

    So I did it right ?

    Peace !!! It was on my test. I screwed up a lot so at least I did something right.
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