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Basic Question: Multiple Torques on same shaft

  1. Jan 13, 2005 #1
    Ok. assuming that Torque is basically force at an angle (yes that is a very very loose description, I know). would it be fair to assume that net torque would be equal to all torques applied to one rotating shaft? Or are there exceptions to this?
     
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  3. Jan 13, 2005 #2

    FredGarvin

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    The angular form of Newton's 2nd law is:

    [tex] \sum T = I \alpha [/tex]

    So yes, they are added to come up with the net torque acting on the shaft. Remember that torque is a twisting force and therefore a vector. Make sure you establish which direction is positive.
     
  4. Jan 22, 2005 #3
    alright. couple more quick ones, since u seem to be well suited to answer them. I read somewhere that force applied in a rotational direction (if rotational is a word..) is considered torque, and to find the torque you must multiply the force times the radius of the object. now is that even true?

    also. if you say had a steel bar, attached perpendicular to a shaft to a shaft. and there was a force of say 50 N on teh y axis, and 75 N on the x axis (consider this to be a 2 dimensional problem) how would you figure out the torque of that? just add the vectors together then multiply by radius (if thats true)?
     
  5. Jan 22, 2005 #4

    Chronos

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    It's usually a bad thing. They like to couple and encourage shafts to commit unnatural acts.
     
  6. Jan 23, 2005 #5

    Q_Goest

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    I'd say torque is a more generic term. In engineering, we talk about moments, though torque is perfectly acceptable as well. Still, it helps to start with the concept of a moment.

    A moment can be applied to anything, take a wrench for example. A wrench sticks out radially from a bolt, like the spoke on a wheel. If you want to turn the bolt, you don't push or pull the tip of the wrench towards or away from the bolt, it wouldn't turn. You pull on the end of the wrench in a direction that is perpendicular to this radius. If you're not pulling exactly perpendicular to the radius, then only that component of the force you're applying to the wrench that is perpendicular to the radius actually creates a moment and can be used to calculate the moment. The moment around the bolt is equal to a torque which is equal to a radius multiplied by the perpendicular component of the force. That's why torque has the value of pound*foot or Newton*meters (force times length).

    If you think of a torque on a shaft for example, a torque created by a motor, it's the same thing. Let's say you have a motor that puts out 100 lb*ft of torque. It may be spinning at thousands of RPM, but that doesn't matter. The torque it applies to the spinning shaft is equal to a person with a wrench 1 foot long pulling with 100 pounds of force. Power is just a bit different then, power is the ability to move that force through a distance. The faster it rotates, the more power it is putting out.
     
  7. Jan 23, 2005 #6
    so in my example, if the steel bar is flat (horizontally) at the start. and the 50N, in teh y direction, pushes down on it perpendicularily the bar will produce a torque yet the force (75 N) in teh x direction will have no real effect on teh torque, but then when the bar is straight vertically the force in the x direction will be the force creating the torque and the y force will have no real effect? and then some where in between those two points both forces are added together at different angles to produce the torque?
     
  8. Jan 23, 2005 #7
    Last edited: Jan 23, 2005
  9. Jan 23, 2005 #8

    FredGarvin

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    "I read somewhere that force applied in a rotational direction (if rotational is a word..) is considered torque, and to find the torque you must multiply the force times the radius of the object. now is that even true?"

    Yes that is true.


    "if you say had a steel bar, attached perpendicular to a shaft to a shaft. and there was a force of say 50 N on teh y axis, and 75 N on the x axis (consider this to be a 2 dimensional problem) how would you figure out the torque of that? just add the vectors together then multiply by radius (if thats true)?"

    You're starting to get into the basics of a statics class. Your loads are going to combine to give you both bending moments (torques as you are calling them) and compressive or tensile forces. In your example, the force that you think has no real effect creates compressive or tensile stresses assuming that you have sufficient restraining forces. These forces result in things like buckling and elongation.
     
  10. Jan 23, 2005 #9
    ...ok so i'd need some sort of really strong material i take it. but a torque would be created no less, right?
     
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