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Torque and distance problem

  1. Jan 17, 2010 #1
    1. The problem statement, all variables and given/known data

    http://img195.imageshack.us/img195/3310/exerciciospdfadobereade.png [Broken]

    The uniform object AB mesures 4 meters and weighs 500N.
    There is a fixed point C along which the object can rotate. The object rests on point A. A person of mass 75 kg walks over the object, starting in A.
    40.1 Calculate the maximum distance x the person can walk keeping the object in equilibrium

    2. Relevant equations


    [tex]\tau = F \times \Delta r[/tex]


    [tex]F = ma[/tex]

    3. The attempt at a solution

    I must confess beforehand that the concept of torque isnt yet confortable for me, so I'm not really sure of the following process.

    first, the torque at A:
    using 1)

    [tex]\tau = 500 \times \Delta 2,5
    = 1250[/tex] (N.m-1)

    and the torque of B:

    [tex]\tau = 500 \times \Delta 1,5
    = 750[/tex] (N.m-1)

    Now, concerning the person:
    The Force applied by gravity =
    [tex] 735.75 = 75 \times 9,81[/tex]

    the torque of the person, then :

    [tex]\tau = 735.75 \times \Delta (x-2.5)[/tex]

    My guess is, when the torque of the person + torque of B = torque of A, the bar will lose equilibrium.

    so :

    [tex]1250 = 735.75 \times (x-2.5) + 750[/tex]

    This equals aproximately 3,18 meters. However, the solution claims it to be 2.84.

    What am I doing wrong?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 17, 2010 #2


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    Your image is not showing up, can you attach a file or otherwise explain the location of the supports A and C, the distance between A and C, and between C and the end of the object?
  4. Jan 17, 2010 #3
    sure. ive attached it to this post

    Attached Files:

  5. Jan 17, 2010 #4


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    OK, I got your picture now. When you sum torques and set their sum equal to zero for the equilibrum condition, you can choose any point about which to sum torques, but you must use the same point when detremining the torques from each force. HINT: choose point C as the point about which to sum torques, and note that the weight of the beam acts at its center of gravity, and that the reaction force at A will be zero when the object just starts to tip, because the beam will start to lift off of support A as it starts to tip..
  6. Jan 17, 2010 #5
    i think i considered C as the point of reference for Torque
    (could you please point out where I didnt?).

    And why do I need the center of gravity? should I consider it to weight differently in positions other than the center of gravity?
  7. Jan 17, 2010 #6


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    First, the resultant force of the beam weight, 500 N, acts at the center of gravity of the object, which since it's 4 m long, is located 2 m from A, or 0.5 m from C . When you sum torques of that beam weight about C, what's the torque? Your torque equation for the torque from the person about C is correct, so set them equal to solve for x.
  8. Jan 18, 2010 #7
    I'll be sure to try that, thanks.
  9. Jan 18, 2010 #8
    So, I'm sorry to keep bothering about the same problem, but I'm still having trouble with torques, and I would need some help
    So, first of all, to be in equilibrium the sum of torques must be 0, right?

    So, ill just submit my thinking, please correct me wherever i'm mistaken.


    [tex]\Sigma \tau = 0[/tex]


    So, if I choose to calculate the torque about C:

    [tex]\tau_{A} =F_{beam} \times 2.5 [/tex]

    [tex]\tau_{B} = F_{beam} \times 1.5[/tex]

    [tex]\tau_{person} = F_{person} \times (x-2.5) [/tex]

    Fbeam is applied on the center of gravity, 0.5 meters away from C.

    I think the confusion comes from having a supporting point in A, that is really making it hard for me to understand: Otherwise it would be just a matter of finding a place for the person where the sum of torques would equal 0.


    Now, subsituting with the values :

    [tex]\tau_{A} =500 \times 2.5 = 1250[/tex]

    [tex]\tau_{B} = 500 \times 1.5 = 750[/tex]

    [tex]\tau_{C} = 500 \times 0 = 0[/tex]


    the torques of B and C are negative, since they make the beam move clockwise, and the one of A is positive, since it makes the beam move counterclockwise.
    However, we must also consider the reaction force on A and C

    [tex]\Sigma \tau = \tau_{A} + R_{A} - \tau_{B} - \tau_{C} + R_{C}[/tex]

    [tex]\Sigma \tau = 1250 + R_{A} - 750 = 0[/tex]

    [tex]R_{A} = -500 N m[/tex]


    So, that would mean that in order for the person to take the beam off equilibrium it would have to equal a torque of 500 N m

    [tex]\tau_{person} = 735.15 \times (x-2.5) = 735.15x - 1835,75[/tex]

    [tex]500 = 735.15x - 1835,75[/tex]

    And solving for x:

    [tex]x = 3,18[/tex].

    when the correct answer would be

    [tex]x = 2,84[/tex].
  10. Jan 19, 2010 #9


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    Since the problem is asking you to find the farthest distance the peson can walk before the beam goes from its equilibrium stage to 'tipping over', this occurs when the end reaction at A is 0 (the beam lifts off that point at the start of rotation). So forget any torque from A, just sum moments about C ffrom the beam weight (500 times 0.5, ccw), and the persons weight (75g times (x-2.5), clockwise), set their algebraic sum equal to 0 for equilibrium (one is a plus, the other a minus value), and solve for x. (Note there is also no torque from the reaction force at C about C, since there is no moment arm)
  11. Jan 19, 2010 #10
    it's correct, thanks.
    so there's no need for torque at B or A?
  12. Jan 19, 2010 #11
    oh, i see. The torque must be calculated only when forces are applied (center of gravity and the person).
    There is no forces applied elsewhere so, no torque.
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