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Angle of a force when towing a suitcase

  1. Mar 19, 2005 #1
    hi guys
    ive got a problem where a woman is towing a 20kg suitcase at constant speed. She is pulling on a strap of unknown degrees from the horizontal. She pulls on the strap with a 35N force and the frictional force on the suitcase is 20N.
    It asks for the angle that strap makes to the horizontal as well as the normal force the ground exerts on the suitcase.
    Any clues
    Thanx
     
  2. jcsd
  3. Mar 20, 2005 #2
    ok guys, ive worked it out
    ill post my solutions just incase anyone was interested or you have a similar question

    for part a
    It says its moving at constant velocity, therefore to overcome the 20N friction there must be a 20N force in the horizontal direction, therefore this allows us to use reverse cos(20/35). This gives an angle of 55.2 degrees.

    part b

    use component method in vectors to find the vertical force of the strap, which is 35sin(55.2). minus this from mg( which is equal to N) and this gets 167N
     
  4. Mar 20, 2005 #3
    Let the unknown angle=@

    H: Tcos@-F=0

    =>Tcos@=F

    =>35cos@=20

    =>cos@=(20/35)=(4/7)

    =>@=arccos(4/7)=55.2deg

    V: Tsin@+N-mg=0

    =>N=mg-Tsin@

    =>N=20g-35sin55.2=167.3N

    Dirac.
     
  5. Mar 20, 2005 #4
    Did you not read his second post? He worked out the solutions already..
     
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