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Determine angle of force

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data

    A woman at an airport is towing her 20.0-kg suitcase at constant speed by pulling on a strap at an angle θ above the horizontal (Fig. P5.40). She pulls on the strap with a 35.0-N force. The friction force on the suitcase is 20.0 N. Draw a free-body diagram of the suitcase. (a) What angle does the strap make with the horizontal? (b) What normal force does the ground exert on the suitcase?


    2. Relevant equations

    F - fk = ma

    W = mg

    3. The attempt at a solution

    (a) F - fk = ma

    35.0 N – 20.0 N = (20.0 kg)(sinѲ)(9.8 m/s2)

    sinѲ = (15.0 N) / (20.0 kg)(9.8 m/s2)

    Ѳ = sin-1[(15.0 N) / (20.0 kg)(9.8 m/s2)]

    Ѳ = °

    (b) W = mg

    W = (20 kg)(9.8 m/s2)

    W = 196 N

    I don't know if what I attempted was even on the right path. Could someone point me in the right direction? Thanks!
     
  2. jcsd
  3. Sep 22, 2009 #2

    rock.freak667

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    the 35N at an angle of θ, has what x and y components?

    It says it pulls at constant velocity which means the resultant horizontal force is what?
     
  4. Sep 22, 2009 #3
    y = 35N sinθ
    x = 25N cosθ - 20N

    ?
     
  5. Sep 22, 2009 #4

    rock.freak667

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    So now

    in the x direction the resultant force is zero so what is the equation for this sum of the forces in the x direction? You can find θ now.
     
  6. Sep 22, 2009 #5
    sorry that was a typo

    so

    0 = 35N cosθ - 20N

    20N / 35N = cosθ

    θ = cos^-1 (20N / 35N) = 48.19

    I really over complicated that. I need to remember to make sure I see all the parts before I begin. Can you tell me if I did the second part correctly? The normal force is the opposite of W since it is not moving in the Y direction right?
     
  7. Sep 22, 2009 #6

    rock.freak667

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  8. Sep 22, 2009 #7
    Got it, thanks!
     
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