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Torque Problem from MCAT Physics Review Book

  1. Aug 21, 2012 #1
    The problem is based off a passage. The passage along with the answer choices are uploaded.

    With a full carriage, the second ride suffers a power outage with the mechanical arm perpendicular to the horizontal. How much torque must the mechanical arm provide in order to prevent the passengers from swinging down? (Assume the mechanical arm itself does not require any torque support.)

    The mechanical arm has a length of 5 m and a full carriage has a mass of 600 kg (300 kg cart plus 300 kg so I thought the answer should be 600(10)(5). They say that the answer should be 120,000 N m.

    The book explains that that the mechanical arm must provide enough torque to cancel out the torque produced by the gravitational force. Hence, τ = rfsinθ = mgr = 600(10)(20).

    Where did the the 20 come from?
     

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  3. Aug 21, 2012 #2

    cepheid

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    I must be missing something, because if the arm is "perpendicular to the horizontal", then this means that the arm is actually vertical. If the arm is vertical, then gravity does not produce any torque around the pivot point.

    I have no idea where the 20 comes from. The maximum lever arm is 5 m, the full length of the arm, and this is the lever arm that is present when the mechanical arm is horizontal. At any other angle, the lever arm will be smaller.
     
  4. Aug 21, 2012 #3

    CWatters

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    That was also my reading of the problem. If the answer =mgr then that implies the arm has stopped in the horizontal position. Yet the problem says the power is lost with the "mechanical arm perpendicular to the horizontal" eg in the vertical position.

    So I looked at other options..

    If power is lost as the cart passes through the bottom of the arc then it might be possible to work out the velocity needed fo it to to coast upto the top and stop up there without swinging back down. However I can't see how you can calculate the torque from that because you don't know over what part of the swing it has to acts before the power is lost.

    I conclude it's a badly worded problem and/or the answer is wrong.
     
  5. Aug 22, 2012 #4
    Thanks for your inputs. I too thought "perpendicular to the horizontal" meant that the mechanical arm was vertical (with the cart and its passengers upside down). What I didn't realize was that the line of the gravitational force is acting at the pivot point, meaning that there is zero force. Like cepheid said then, there cannot be any torque. I know test prep books have their fair share of errata but I'm so puzzled by where they could have gotten 20 from that I'm hoping I'm not overlooking anything.
     
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