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Homework Help: Help with a Friction Question: Maximum mass a Man can Push

  1. Sep 26, 2010 #1
    1. The problem statement, all variables and given/known data

    A man of 72.0kg is pushing a heavy box along a flat floor. The coefficient of sliding friction between the floor and the box is 0.20, and the coefficient of static friction between the man's shoes and the floor is 0.65N.

    a) If the man pushes downward on the box at an angle of 30 degrees, what is the maximum mass of the box be can move?

    b) If the man pushes upward on the box at an angle of 30 degrees, what is the maximum mass of the box he can move?

    2. Relevant equations
    fk= ukN


    3. The attempt at a solution

    So, according to the free body diagram for the man:
    Fx man= -ukmmang-Pcos(60)
    For the box:
    Fx box= Pcos(30)-ukmboxg

    I don't know where to go from here. If I set the equations equal to one another I have two unknowns (P and mbox). If I add the equations, I get the same problem, plus acceleration thrown in there. If anyone can steer me in the right direction, that would be great!
     

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  2. jcsd
  3. Sep 26, 2010 #2

    fzero

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    Forget about the box for a second and imagine just a force acting on the man. How large does the force have to be before he begins to slide along the floor?
     
  4. Sep 26, 2010 #3
    The force should be equal to the force due to friction of the man?
     
  5. Sep 26, 2010 #4

    fzero

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    Right, it's the force due to static friction on the man. Can you see how this relates to the original question about pushing the box?
     
  6. Sep 26, 2010 #5
    Not exactly. If the force is equal to the static friction of the man, where does the angle of the force come into play? I've spent 3 days on this problem and I am a little lost and frustrated.
     
  7. Sep 26, 2010 #6

    fzero

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    As for the angle, in the equations you wrote down, you neglected to work out the vertical forces on the box. The normal forces involve more than just the force of gravity. Note that this also affects the force due to static friction on the man.
     
  8. Sep 26, 2010 #7
    Ok, so the vertical equations are:

    Fy man= N-mmang + Psin(60) =0
    Fy box= N-mbox - Psin(30) =0

    If I sub in the values of N, I'm still left with two variables, P and mbox
     
    Last edited: Sep 26, 2010
  9. Sep 26, 2010 #8

    fzero

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    You should really write Nman and Nbox. Now we consider the horizontal force on the man. There is static friction, so as long as P cos(60) does not exceed usNman, the man does not slide. The value Pmax at which he does begin to slide is the maximum value of the force he can exert on the box. The horizontal force equation on the box will let you solve for mmax.
     
  10. Sep 26, 2010 #9
    Ok-

    So Fx man= -us(mmang - Psin(60)) + Pcos(60)= 0 (man not moving) then P=usmmang/(cos(60)+ussin(60) and P=432N

    Can I just use this P value in the Fx box and assume the equation is zero?
     
  11. Sep 26, 2010 #10

    fzero

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    Yes, you can set Fx box=0 because we're looking for the point where the acceleration becomes non-zero. Just make sure you use the proper Nbox depending on whether the man is pushing up or down.
     
  12. Sep 26, 2010 #11
    Using P=-432N gives me an mbox of 213kg, which my homework says is wrong

    I'm using mbox= (ukPsin(30) + Pcos(30))/(-ukg)
     
  13. Sep 26, 2010 #12

    fzero

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    Sorry, I hadn't been following the algebra closely, I was mainly trying to give hints on the concepts. When the man is pushing down, I find

    [tex]F_{\text{man},y} = N_{\text{man}} + P \sin(30^\circ) - m_{\text{man}} g [/tex]

    [tex]F_{\text{man},x} = - P \cos(30^\circ) + u_s N_{\text{man}}[/tex]


    [tex]F_{\text{box},y} = N_{\text{box}} - P \sin(30^\circ) - m_{\text{box}} g [/tex]

    [tex]F_{\text{box},x} = P \cos(30^\circ) - u_k N_{\text{box}}[/tex]

    Then I find

    [tex]P_{\text{max}} = \frac{m_{\text{man}} g}{\frac{\cos(30^\circ)}{u_s} + \sin(30^\circ)}[/tex]

    and

    [tex]m_{\text{box, max}} = \frac{\frac{\cos(30^\circ)}{u_k} - \sin(30^\circ)}{\frac{\cos(30^\circ)}{u_s} + \sin(30^\circ)} m_{\text{man}}.[/tex]

    I can't guarantee that there are no sign errors or typos anywhere, but hopefully it helps.
     
  14. Sep 26, 2010 #13
    Yes! I got it. It ended up being a small geometry error in my free body diagram that was my downfall. THANK YOU for all your help, I completely understand where I went wrong and know how to fix it!
     
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