[CJ6 9.P.024.] A womanwho weighs 5.11 102 N islean...

Answers ( 3 )

  1. neea erkkila
    neea erkkila
    a) Taking the torques about the shoe
                F n *1.5Sin60 - mg * 1.10Cos60 = 0
                                 Fn = ( mg * 1.10Cos60 ) / 1.5*Sin60
                                    Fn= 2.2 x 102 N

    b)
  2. melike ta?l?
    melike ta?l?
    2
  3. esther byrd
    esther byrd
    Since the net forces acting on the lady =0, wehave
                    åFx =Fsx - Fn =0
                             0r   Fsx =   Fn----------------------------(1)
                      åFy= Fsy - mg = 0
                                      Fsy = mg
                                            = 5.11 x 102 N
    We know that Fs = FsySin60
    substituting Fs iscalculated.            Fs = 4.42* 102N
               TheNormal force Fn = Fsx
                                                = F Cos60
                                          Fn = 2.2 * 102 N
    An alternate way of determining the torque is asbelow.
    Taking the torques about the point B, we have
                 FN*1.5Sin60 - mgCos60*1.10 = 0
                                    FN = (1.10*  mgCos60 ) / 1.5Sin60
                                          = 2.2*102 N

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