Answers to Chapter 9 Homework

1. We choose the coordinate system shown, with positive torques clockwise.

        We write St = 0 about the support point A from the force diagram for the board

        and people.

                      StA = -m1g(L-d) + m2gd =0;

                      - (30kg)(10m - d) + (70kg)d = 0,

        which gives d =  3.0m from the adult.
2. We choose the coordinate system shown, with positive torques clockwise.

        We write StA = 0 about the support point A from the force diagram for the seesaw and boys:

                       StA = +m2g1/2L + m3gx -m1g1/2L = 0;

                      + (35kg)1/2(3.6m) + (25kg)x - (50kg)1/2(3.6m) = 0,

        which gives x = 1.1m.

        The third boy should be 1.1m from pivot on side of lighter boy.
3. We choose the coordinate system shown, with positive torques clockwise.

        We write St = 0 about the point A from the force diagram for the pole and light:

                       StA = -FTH + MgLcos q +mg1/2Lcos q = 0;

                      -FT(3.80m) + (12.0kg)(9.80m/s2)(7.5m)cos37° +

                           (8.0kg)(9.80m/s)1/2)(7.5m)cos37° = 0,

        which gives FT = 2.5 x 102 N

        We write SF = 0 from the force diagram for the pole and light:

                       SFx = FAH - FT = 0;

                                FAH - 2.5 x 102N = 0, which gives FAH = 2.5 x 102N.

                      SFy = FAV - Mg -mg = 0;

                                FAV - (12.0kg)(9.80m/s2) - (8.0kg)(9.80m/s2) = 0

         which gives FAV = 2.0 x 102N.

 

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