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Finding the angle when given the coefficient of friction

  1. Feb 11, 2011 #1
    1. The problem statement, all variables and given/known data
    A mop is being used to clean a floor. The coefficient of friction between the mop and floor is 0.3. Find the angle between the handle and the horizontal when the mop head is moving at constant velocity.


    2. Relevant equations
    The coefficient of friction = Friction/ Normal Force
    Pythagoras Theorem


    3. The attempt at a solution
    I figured, the main force would be gravity at 9.8 as mass is negligible. I also tried to get the components of the force. If the downward force was 9.8m/s, the horizontal would be 0.3*9.8 - 2.94 (?). The normal force would also be 9.8, as there is not vertical acceleration or velocity.

    I thought could you say perhaps, 2.94/9.8 and the inverse sine of this? It does give the right answer but it feels like the working is incorrect..

    Please help, any pointers would be appreciated.
     
    Last edited: Feb 11, 2011
  2. jcsd
  3. Feb 11, 2011 #2

    Doc Al

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    Staff: Mentor

    If the mass is negligible, then so is the force of gravity.

    Do this: Assume that some force is being applied along the line of the mop handle. Call it F. Call the angle that the mop makes with the floor θ. Now figure out the horizontal and vertical forces on the mop.
     
  4. Feb 12, 2011 #3
    Hint: you are missing a major point. there is a force holding the mop. otherwise the mop will fall.
     
  5. Feb 12, 2011 #4
    no...the mass is not negligible. if that is so then gravity, normal and friction force would all be negligible. consider all forces and apply the concept of torque..
     
  6. Feb 12, 2011 #5

    Doc Al

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    That is incorrect.
     
  7. Feb 12, 2011 #6
    Well the horizontal force would be cos θ = 0.3/F

    I'm not too sure on the vertical component as you don't actually have the opposite side?
     
  8. Feb 12, 2011 #7

    Doc Al

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  9. Feb 12, 2011 #8
    Thanks, I remember now,
    so the horizontal component is Fcosθ
    the vertical component would be Fsinθ

    i still don't see how that will help, as you don't actually have F or θ
     
  10. Feb 12, 2011 #9

    Doc Al

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    Good.
    Those aren't the only forces acting on the mop. What about the normal force and friction?
     
  11. Feb 14, 2011 #10
    The normal force would be equal to friction, as the mop is moving at constant velocity,
    Fr=coefficient of friction x the normal force?
     
  12. Feb 14, 2011 #11

    Doc Al

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    Not exactly. Since, the velocity is constant the net force is zero. That means that the horizontal component of the applied force (F) must equal the friction force.
    Right!
     
  13. Feb 15, 2011 #12
    Thanks, so if

    Fcosθ = Friction

    And if friction = the coefficient of friction x the reaction force,

    and the reaction force (R) = Fsinθ

    would it be...

    Fcosθ = 0.3 x Fsinθ?
     
  14. Feb 15, 2011 #13

    Doc Al

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    Looks good. Now just solve for θ. (Rearrange a bit and reach for your calculator.)
     
  15. Feb 19, 2011 #14
    Well I tried that, but then I got stuck...

    The F's cancel out.

    Move the sinθ to the other side, so you get

    Cosθ/Sinθ = 0.3

    then I got stuck, I was thinking perhaps you could use trigonometric proofs?
     
  16. Feb 19, 2011 #15

    Doc Al

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    So far, so good. Can you replace that ratio of trig functions--or its inverse--with a single trig function? Then use your calculator.
     
  17. Feb 19, 2011 #16
    THANKS!! I finally did it! Thank you so much! :D
     
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