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Homework Help: Static equilibrium involving toque

  1. Sep 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Sir Lost-a-Lot dons his armor and sets out from the
    castle on his trusty steed (Fig. P12.19). Usually, the
    drawbridge is lowered to a horizontal position so that
    the end of the bridge rests on the stone ledge. Unfortunately,
    Lost-a-Lot’s squire didn’t lower the drawbridge
    far enough and stopped it at 20 degree above
    the horizontal. The knight and his horse stop when
    their combined center of mass is 1.00 m from the
    end of the bridge. The uniform bridge is 8.00 m
    long and has mass 2 000 kg. The lift cable is attached
    to the bridge 5.00 m from the hinge at the castle end
    and to a point on the castle wall 12.0 m above the
    bridge. Lost-a-Lot’s mass combined with his armor
    and steed is 1 000 kg. Determine the tension in the

    2. Relevant equations

    3. The attempt at a solution
    Actually I have the answer, but I just don't understand why you have to take the horizontal component of tension and the force from the hinge because it is perpendicular to surface of the bridge

    The angle between the wall and the cable is 24.5 degrees

    Take torques about the hinge end of the bridge:
    Hx (0)+ Hy (0)− Mg(4.00 m)cos20.0°
    (T sin 24.5°)(1.71 m)+ (T cos24.5°)(4.70 m)
    − mg(7.00 m)cos20.0° = 0
    T = 27.7 kN

  2. jcsd
  3. Sep 11, 2015 #2


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    What makes you conclude that ?
    Could you explain some of the symbols and the funny numbers, or perhaps even all of them, instead of letting us find out the hard way that M is the bridge and m is man + horse ?
    MAke a drawing (you probably did that already). Did you discover the error in the problem statement (or rather the unmentioned assumption) ?
  4. Sep 11, 2015 #3
    That's the diagram according to the answer's calculation. Any better?

    I don't understand how you can get the answer because you can see Tx is not perpendicular to the bridge. So how can you calculate torque of Tx by using (T sin 24.5°)(1.71 m)?
    Last edited: Sep 11, 2015
  5. Sep 11, 2015 #4


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    The picture at least explains where the 24.5 degrees comes from.
    Doesn't look as nice as most of the official solutions (such as in books) I've seen, but it looks OK.

    This way the problem solver avoids having to work out the angle between T and the bridge (85.5 degrees, I would say :smile:). T is the sum of Tx and Ty and the torque of T about the hinge is the sum of the torques about the hinge from Tx and from Ty.

    If you know about cross products and their linearity: $$
    \vec \tau = \vec r \times \vec T = \vec r \times (\vec T_x + \vec T_y) = \vec r \times \vec T_x + \vec r \times \vec T_y$$​
    Something seems to go wrong, though: one has a minus sign and the other a plus sign, but both torques are in the same direction. Brackets missing ?

    My ironic criticism on the problem statement was that the center of mass for man+horse is more than 1 m away from the end of the bridge (unless very small horse). What the problem composer (is that also the solver? hmmm...) means is that the center of mass is straight above a point on the bridge that is 1 m from the end. This isn't clear from the problem statement at all. Apart from my corny height, it could well be meant that its 1 m / cos 20 degrees from the end....
    Now what's this about the force from the hinge ?
  6. Sep 11, 2015 #5


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    The components of the tension which you have drawn in your diagram are oriented so that they are parallel and vertical to the ground. You can just as easily find different components which are parallel and vertical w.r.t. the bridge.
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