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Homework Help: Torque/Rotational Motion

  1. Apr 16, 2010 #1
    1. The problem statement, all variables and given/known data
    A 45 kg, 5.0 m-long beam is supported, but not attached to, the two posts in the figure . A 25 kg boy starts walking along the beam. How far can he get to the end without falling over?

    2. Relevant equations

    Support on the left is leftEnd, support on the right is rightEnd
    Torque equations. Equilibrium.

    3. The attempt at a solution

    Sum of Forces and Torque must be 0 for static equilibrium.

    Forces = normal Force on leftEnd + normal Force on rightEnd - weight of the boy
    When the board tips, the normal force on the leftEnd is 0, so:

    normal Force on rightEnd= weight of the boy = 25 * g

    Sum of the torques (leftEnd is the axis):

    Gravitational Torque is at the center of mass = -2.5 * 45 * g
    Torque of the boy is -25 * g * (3 + d)
    Toruqe of rightEnd = 25 * 9.8 (from above)

    Sum them and solve for d:

    25 * g * (3+d) = -2.5(45)(g) + 25 * g
    25 *(3+d) = -2.5(45) + 25
    3 + d = -3.5

    Which can't be right. Where did I go wrong?
  2. jcsd
  3. Apr 16, 2010 #2


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    Hi Cfem! :smile:

    Sorry :redface:, but you've done three things wrong …

    i] you left out the distance for rightEnd

    ii] you left out the weight of the beam in calculating the magnitude of rightEnd

    iii] you made it a lot more complicated than necessary by a bad choice of the place to take moments about …

    since leftEnd is zero, taking moments about the right support will eliminate rightEnd also :wink:
  4. Apr 16, 2010 #3
    i] Okay. I understand that one. Due to not paying attention ._.
    ii] This one I'm not sure about. Did you mean I should take the weight of the beam into account when calculating the Normal force or when I'm calculating the torque>
    iii] I hate rotational motion. Didn't even occur to me. So then it would be d*Force of the boy = the gravitational torque at the center of mass?

    Thanks in advance. Out of all the Physics I've done, I haven't had much experience with torque/rotational motion...
  5. Apr 17, 2010 #4


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    Hi Cfem! :smile:

    (just got up :zzz: …)

    i] he he :biggrin:

    ii] the normal reaction force … it has to support the whole weight, doesn't it? :wink:

    iii] I don't like this expression "gravitational torque" … why use it for the beam but not for the boy? … you'll confuse yourself unless you always use the same formula …

    always use moment = distance times (or "dot") force … in the case of the beam, the distance is from the support to the centre of mass of the beam, and the force is the weight of the beam.
  6. Apr 17, 2010 #5
    Me too ._.

    So I think I've nailed down most of what I missed.

    the normal force on the right pivot would be the weight of the boy + the weight of the beam, yes?

    Gravitational torque is just the term our textbook uses for the torque acting on the center of mass.

    Put all in all, the torque on the left support would be zero (because the beam comes off the support), the torque on the right support is zero (because it's the axis point), and then the only two torques there would be to balance would be the torque on the center of mass and the torque of the boy? That's my understanding, but that gives me a huge number as well/

    I hate rotation .-.
  7. Apr 17, 2010 #6


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    That's right …

    what do you get? :smile:
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