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What is the smallest vertical force which will move the crate

  1. Jan 5, 2009 #1

    Air

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    1. The problem statement, all variables and given/known data
    A crate with a weight of [itex]50N[/itex] rests on a horizontal surface. A person pulls horizontally on it with a force of [itex]10N[/itex] and it does not move. To start it moving a second person pulls vertically upwards on the crate. If the coefficient of static friction is [itex]0.4[/itex], what is the smallest vertical force which will move the crate.


    2. Relevant equations
    [itex]f_s=\mu_sn[/itex]


    3. The attempt at a solution
    [​IMG]
    [itex]R(\rightarrow): 10 - f_s = 0 \implies f_s = 10N[/itex]
    [itex]f_s=\mu_sn_{tot} = (0.4)(n_{tot}) = 10 \implies n_{tot} = 25[/itex]
    [itex]n_{tot} = F_y+n \implies F_y = 25-50 =-25N[/itex]

    4. The problem I encounter
    Can you check my method please. I got [itex]-25N[/itex] which is negative hence this is giving me a doubt. Thanks.
     
    Last edited: Jan 5, 2009
  2. jcsd
  3. Jan 5, 2009 #2
    Re: Forces

    You did everything correctly. The reason for the negative is that your calculations used a downward direction as positive. Thus, your -25 N answer means the 25 N force will be applied upwards.
     
  4. Jan 5, 2009 #3

    PhanthomJay

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    Re: Forces

    What is [tex] n_{tot}[/tex]? The normal force is just [tex] n[/tex] . The applied upward force is [tex]F_y[/tex]. The crates weight is 50N down. You've got a couple of 'n's' in your equation that need to be corrected.
     
  5. Jan 5, 2009 #4

    Air

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    Re: Forces

    How can you tell that I took downwards as positive? Even if I resolve upwards I would get: [itex]R(\uparrow) : n - 50 = 0 \implies n=50N[/itex]?

    My [itex]n_{tot}[/itex] is just a notation I gave for the total force in the upper direction which would be the [itex]F_y + n[/itex]. I should have given that another notation but whats wrong other than that, is my calculation correct?
     
  6. Jan 5, 2009 #5

    Doc Al

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    Staff: Mentor

    Re: Forces

    The static friction is μn, where n is the normal force (not ntot). So μn = 10, thus n = 25.
    The vertical forces must add to zero:
    Fy + n - 50 = 0 (Where Fy is the applied vertical force.)
    Solve for Fy.
     
    Last edited by a moderator: Jul 17, 2014
  7. Jan 5, 2009 #6

    Air

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    Re: Forces

    Ah yes, that gives a positive answer, ##F_y = 50 - 25 = 25N##.

    The problem was that I took the vertical force combined with the normal force to equal the normal force and used that in the equation.

    It makes sence. Thanks to everyone for the help.
     
    Last edited by a moderator: Jul 17, 2014
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