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Calculating work when 2 forces are present

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data
    This should be an easy one. Please tell me what im missing.

    A man is pushing a 75 kg crate at constant
    velocity a distance of 12 m across a warehouse.
    He is pushing with a force of 225 N at an angle
    of 15° down from the horizontal. The coefficient
    of friction between the crate and the floor is
    0.24. How much work did the man do on the
    crate?

    Given - m=75kg, Δd=12m, Fa=225N [15° down from hor], μ=.24

    2. Relevant equations

    Ffriction = μ*Fn
    W = F*Δd*cosθ

    3. The attempt at a solution
    First attempt
    Ffriction = 176.4N
    So
    W = F*Δd*cosθ
    W = (225 - 176.4)*12*cos15
    W = 563.33N/m (wrong answer)

    Second attempt
    W = F*Δd*cosθ
    W = 225*12*cos15
    W = 2608N/m

    W = F*Δd*cosθ
    W = 176.4*12*cos180
    W = -2116.8N/m

    W = 2608 - 2116.8
    W = 491.2N/m (wrong answer)

    I've tried so many combinations I cant seem to figure it out. The answer is 320N/m btw
    Please tell me what im doing wrong
     
  2. jcsd
  3. Oct 3, 2011 #2
    The thing is, it's the work the man is doing on the crate. So why is friction relevant?

    It should just be

    [itex]W_{man}=\int \mathbf{F}^{man} \cdot d\mathbf{r} = \int (F^{man}_{x},F^{man}_{y}) \cdot (dx, dy) =\int F^{man}_{x} dx=F^{man}_{x} \Delta x= F^{man} \cos \theta \Delta x[/itex]

    I computed that and got 2608 N/m. Maybe they meant the work done by both friction and the man. I also computed that but got 2525.08 N/m. Anyone else have any ideas? Am I just being an idiot? xD

    Perhaps the answer in the back of the back is wrong. Maybe you should check with your prof.
     
  4. Oct 3, 2011 #3
    The answer is 320N/m btw

    Would you mean N-m for work.
     
  5. Oct 3, 2011 #4
    ya n-m

    does anyone have an answer for this though?
    I found a question from the text book that had a wrong answer in the back so I dont care about that answer anymore. Im starting to think my first attempt is the correct way. Can anyone confirm that for me please?
     
  6. Oct 4, 2011 #5
    ^Thing is, it's the work the MAN does on the crate. You just take the force from him into account, nothing else. So I think the second one is correct.
     
  7. Oct 4, 2011 #6

    SammyS

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Total work done by the man minus the work he does in overcoming friction. That is 321.3 N-m . (Using g =9.91 m/s2)

    Don't forget, when computing normal force that there is a component of the 225 force that's downward in addition to the weight of the box, mg. Therefore, the normal force must be equal but opposite to the sum of those downward forces.
     
  8. Oct 4, 2011 #7

    gneill

    User Avatar

    Staff: Mentor

    Hmm. If the man is pushing the crate at a constant velocity, how is it that the horizontal component of the force he applies is greater than the friction force? Something got by the editors!
     
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