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Explanation of torque arm?

  1. Jan 6, 2004 #1
    Well, we all know that the farther you move the point of force application from the point of rotation, the greater the torque. But why?

    I know the equations, but I am lacking a physical explanation of why this is. The best explanation that I've gotten is something along the lines of a balancing act of work, whereby distance is traded off with force, but that explanation seems pretty half-arsed to me.

    Any better explanations out there?
     
  2. jcsd
  3. Jan 8, 2004 #2
    It just applies to the law of "Action is Reaction".
    When we got a force working on a body (undeformable), we make that body free by adding a reactionforce in the fixed point. Suppose the only thing we add as a reaction is that force. Than in the middle of those two forces we'd get a couple, so our body would begin to rotate. But we have to get equilibrium so we have to add a rotationforce or momentum to keep the body from rotating. Since the momentum, caused by the force, increases linear with the distance we can write M=F.d
     

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    Last edited: Jan 8, 2004
  4. Jul 19, 2007 #3
    Torque

    What nobody can explain is the Above question : "Why is it more difficult to push the door closer to the fixed axe . . . What force create this difficulty ?
     
  5. Jul 19, 2007 #4
    I like to start by picturing this:
    Imagine a bar that is held tight by strong forces on each end.
    A----------------B

    You push in the center and what happens? Your force is distributed equally between A and B. Push near B and your force is concentrated near B, and a little gets transfered to A. Same thing with pushing near A. A will feel most of the force.
    Now keep A held in place, but let the resistance force on B be halfed.
    Push in the center again. You will notice that even though your force distribution is the same, since there is less pushing back at B, it begins to move forward. This, due only to A being hinged, forces a rotation.
    So your force that tries to move the whole bar gets transfered into a rotational force (torque) because A can't move, but B can, and they're hooked together.
    Now same situation, but move your push nearer to A. You'll see that as your force was distributed more to A than B, B begins to move slower than it did when you pushed in the center, solely because B gets a smaller fraction of the force than A gets.
    Push more near to B, and since B gets a larger fraction, he accelerates more, thus he rotates faster.

    So you notice that depending on where you place the force, only a certain fraction goes to each A and B. This is why when pushing on free-moving B, it is easier to move. Because your force required to move it is less than the force required to move near A.

    In these examples, all Torque is, is changing a linear force into a rotational one by keeping one end fixed. get it?
     
  6. Jul 19, 2007 #5

    FredGarvin

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    The quickest, but may not suit your needs is that of a lever arm.

    IMO, work is the best explanation. In terms of the work balancing act, that is the way I like to think of it. You have a choice, either rotate a small arc length distance but have to provide a large twisting force or move a large arc length distance with a small twisting force. The work is the same no matter what.
     
  7. Jul 19, 2007 #6
    why do you multiply by radius when going from linear so and so to angular so and so, good question.. my book just said that "it has to do with a property of angles" or something like that.. so I just grudgingly accepted it and read on....
     
  8. Jul 19, 2007 #7
    Simplest answer:
    You multiple Force by Radius to get the Angular Force (Torque) because assuming a constant Force, the Torque scales linearly with the distance(radius) away. Double the distance, Double the torque.
     
  9. Jul 19, 2007 #8

    mgb_phys

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    I think the easiest explanation is to consider energy.
    If you push at the outer end you are doing less force over a greater distance, rememebr energy=force x distance.
    Near the axis you are doing more force over a shorter distance = same work.
    It is exactly the same as a pulley. Lower force over greater distance = same work.
     
  10. Jul 19, 2007 #9

    russ_watters

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    Grab a golf club and balance it horizontally on your finger....

    That's torque expressed as a center-of-gravity problem.
     
  11. Jul 19, 2007 #10
    "I think the easiest explanation is to consider energy.
    If you push at the outer end you are doing less force over a greater distance, rememebr energy=force x distance.
    Near the axis you are doing more force over a shorter distance = same work.
    It is exactly the same as a pulley. Lower force over greater distance = same work."
    yeah, I guess you could think about it that way, but the force isn't parallel to the length of the lever it's perpendicular, the work would be 0
     
  12. Jul 19, 2007 #11

    mgb_phys

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    If you push on the outer end of a lever you are moving through a longer arc. The lever is the radius of the curve you move through, the circumference is the distance the force is applied over = the work done.
    The longer the level - the larger the radius - the greater the length of a segment of the arc.
     
  13. Jul 20, 2007 #12
    Ok..but I have the above question...You are trying to explain what torque is and to prove the equation . To prove that the force (A------------B) asked to A is smaller that the force asked to B (when puhing neer B) you use torque equilibre to points A,B . So you are trying to explain torque equation by using torque equation.
     
  14. Jul 20, 2007 #13
    Hmm I see what your saying. But thats like asking to describe Torque without talking about Force. Its tough to do. Torque is just force about a point.
    Maybe energy WOULD be a better way of talking asbout it.
     
  15. Jul 20, 2007 #14
    Why can't we combine angular momentum and energy to form four vector angular momentum?... and since we can't, how can we be sure that energy is conserved when discussing torques (aside from experimentally that is)?
     
    Last edited: Jul 20, 2007
  16. Jul 20, 2007 #15
    Unfortunately, this is just one of those "why does something move if a net force is acted upon it?" questions. The answer is in the definitions:

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

    It's just the recorded observation, and therefore, it happens to "just work that way." Sorry, there's not really a proof for this, it's just what the experimental evidence shows will happen and therefore its the definition we use.
     
  17. Jul 21, 2007 #16

    russ_watters

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    I would say you can prove or derive the torque equation using conservation of energy and the work equation. Just like with pulleys, if the distance doubles the force must be cut in half to maintain conservation of energy.
     
  18. Jul 21, 2007 #17
    Half true. However, similarly as Newton's third law can be derived out of conservation of momentum (they are equivalent), also this torque stuff can be derived out of conservation of angular momentum (then you can wonder if conservation of angular momentum can be derived somehow, and these questions continue quite far!). In general setting the proof would be quite difficult, it can probably found in good books about mechanics, but I can outline a highly simplified proof.

    synergyguru, I have couple of questions. Do you know the definition of angular momentum

    [tex]
    \boldsymbol{L}=\sum_{k=1}^{N} \boldsymbol{x}_k\times\boldsymbol{p}_k,
    [/tex]

    and do you understand that for a particle to have angular momentum, it does not really need to be attached to some axis and be rotating, but for example a projectile flying by can have angular momentum. A thrown stone has angular momentum with respect to some fixed point that is not in its way.

    Anyway, here's the explanation:

    Since I don't know how to put pictures here, I'll explain the picture with coordinates (you might want to draw it on paper). There is a stick that has length R. Its left end is in location (0,0), and the right end in (R,0). For simplicity, assume that it is very light elsewhere, but in location (r,0) it has mass m. This means that its mass is consenrtrated in a one point, just to make calculations easier. Then somebody exerts a force F in the right end of the stick, pointing upwards. That means the force vector is (0,F). Then we ask, that what is the angular acceleration of the stick? Or equivalently, what is the acceleration of the mass m in point (r,0)?

    If the stick has angular acceleration [itex]\alpha[/itex], then the mass m in point (r,0) is accelerating with an acceleration [itex]a_m=r\alpha[/itex]. So if we know [itex]a_m[/itex], we can solve [itex]\alpha[/itex].

    The force F isn't coming from thin air, but there must be some object right below the location (R,0), that is causing this force. By the Newton's laws, the stick is then exerting a force (0,-F) on this object. If the mass of this object is M, then this object is instantaneously accelerating downwards with an acceleration

    [tex]
    a_M=-\frac{F}{M}
    [/tex]

    This means that its angular momentum with respect to the origo is changing with a time rate

    [tex]
    \dot{L}_M = -RMa_M = -RF
    [/tex]

    (Because
    [tex]
    L_M = RM v \quad\implies\quad \dot{L}_M = RM\dot{v}
    [/tex]
    the dot means derivative with respect to time.)
    At the same time the angular momentum of the mass m in point (r,0) is changing with a time rate

    [tex]
    \dot{L}_M = rm a_m
    [/tex]

    where [itex]a_m[/itex] is its acceleration. But since the total angular momentum must not be changing, this numbers must cancel, and we get

    [tex]
    rma_m = RF
    [/tex]

    which is the same thing as

    [tex]
    \alpha mr^2 = RF
    [/tex]

    Now, without rigorous explanations, here's how this result can be generalized. This time the moment of inertia of the stick was [itex]I=mr^2[/itex]. If the mass distribution had been something different, then the momentum of inertia would be different, but the result would still be

    [tex]
    \alpha I = RF
    [/tex]

    Also if we had had several locations [itex]R_1,R_2,\ldots,R_n[/itex] along the stick, and all experiencing corresponding forces [itex]F_1,F_2,\ldots F_n[/itex], then using similar arguments as were used now, we would have got

    [tex]
    \alpha I = \sum_{k=1}^{n} R_k F_k
    [/tex]

    From this you see, that if you want the stick to remain still, then must have [itex]\alpha=0[/itex] which is the same thing as

    [tex]
    \sum_{k=1}^{n} R_k F_k = 0
    [/tex]

    This is just saying, that the torques must cancel. If I understood correctly, this was what you were after.

    If you didn't understand from where some equations popped out, just mention. They can be explained in better detail too, but it would have made the post too long.
     
    Last edited: Jul 21, 2007
  19. Jul 21, 2007 #18
    That was not an intuitive explanation, but since synergyguru was getting strict about avoiding circular conclusions in the derivation, there's how the result comes so that you know for sure what the assumptions are in the proof. And the assumption was conservation of angular momentum.
     
  20. Jul 7, 2008 #19
    i think my question is more like this threads question.So any ideas?
     
  21. Jul 7, 2008 #20
    i mean does newtons laws deals with only linear motion and for rotational we empirically accept new equations like rotational equivelants of newtons laws?
     
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