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Torque: Why F*r?

  1. Jan 11, 2004 #1
    We all know that torque = force x lever arm, through education and experience, but I can't recall ever being taught an explanation for why this is. It's always just been explained to me as a definition without underlying reasons.

    Does anyone have a good causal explanation for why you can cause a greater change in angular velocity by applying the force farther away from the axis or rotation (specifically, why this varies in a linear manner)?
  2. jcsd
  3. Jan 11, 2004 #2
    See topic "explanation of torque arm?" that you started earlier.
    When you have a force at a certain distance then we need a momentum to have equilibrium, it's as easy as that. Without torque no equilibrium.
    Suppose we have bar with a fixed point. At the free end of the bar there is a force. Action=Reaction but we can't have a reaction force in the free point because there's nothing that can cause a reactionforce so the force will try to press the bar down. With the force being perpendicular on the bar and the bar being indeformable, it will cause a rotation with the radius being the distance from the free point to the fixed point on the bar. Thus the force causes a rotation, thus in the fixed point we need to write a momentum in the opposite direction to have equilibrium.

    So in short: The reason for torque is to write equilibrium.
  4. Jan 11, 2004 #3
    Excuse me. I had forgotten that I started the other thread.

    Why do you state that "we can't have a reaction force in the free point because there's nothing that can cause a reactionforce"? Wouldn't the mass at the free point (along with all the rest of the mass in the object) provide the reaction force?

    What do you mean by "pressing the bar down"?

    Actually, I can't follow what you said to see how it results in the torque being proportional to the arm. Perhaps some equations would be helpful for me.
  5. Jan 11, 2004 #4
    The first (upper) drawing is wrong as you can see.
    You can only have reactionforces in points that are connected to something, wich is obviously not the case with the free point at the end of the bar (that point is part of the body you're making free and not connected to anything but the free body).
    If you have a force at a point,that point will have the tendency to move downwards unless an equal force (a reactionforce for example) stabilizes the point. But this is not the case so the point will have the tendency to move down. But the point is a part of a bar that's connected to a fixed point, so this bar prevents the point from moving. If the bar doesn't move,there has to be equilibrium, thus there has to be a momentum M that equals the rotation.

    An equation wich leads to M=F.r I can't give you, but it's a necessity to have equilibrium. But remember the momentum is nothing else then a kind of a 'force' to opposite the rotation caused by a force.

    P.S.: I just see a mistake in the second drawing. The arrow with the M is in the wrong direction ;).

    Attached Files:

  6. Jan 12, 2004 #5
    I thank you for your attempts to educate me, but I am still not understanding your explanations.

    However, I've been thinking about it quite often, and I have come to an explanation which I think is the truth.

    It's funny, because we are normally taught these basic laws as they apply to rigid bodies. However, I think that torque, (and rotation itself) depends on bodies being deformable.

    When you apply a force at a point, it causes a momentum change in the particle that you apply the force to. This momentum change propogates down any particles that the first particle is connected to. This occurs because the particle changes its position relative to the other particles that it is connectd to. This repeats again and again.

    It takes a time for this momentum to propogate to the pivot axis. This time should be rougly proportional to the distance from the pivot axis at which the force was applied. During this time, the momentum is being transferred to the other particles on the same side of the pivot axis as the point at which the force was applied. This means more mass for the momentum to transfer to/through before arriving at the pivot axis. This mass then has less velocity relative to the pivot axis than if it had been applied closer, meaning that less momentum will be transferred to the pivot point, which would either transfer the momentum to a stationary object, if nailed down, and also to the side at which the momentum would oppose the rotation. This momentum being more and more lopsided as the force application point moves away from the center of rotation results in the rotation being greater.

    What is really interesting about all this is that if everything was perfectly rigid, then there would be no rotation! I, for one, am glad that I can turn around.
  7. Jan 13, 2004 #6
    Anyone have any comments on my explanation?
  8. Jan 19, 2004 #7
    Dan--you've got the right idea, but what matter is how far you can displace the point of action, not the time it takes for the force to propogate.
    Consider a simple lever arm (with a mass at one point). You provide a push at one point. The rest of the lever is carried along by forces inside the lever. Each piece of the lever provides a little bit of force proportional to how much it is bending at that point (yep, it's not really rigid--this is important).
    Basically, the farther out you are, the more the internal forces tend to pull the mass along in the direction you want to go.The closer in you are, the more they tend to resist you. If you push exactly at the point of mass, the lever arm doesn't resist you or help you.
    If the lever arm cannot generate forces as large as it's supposed to, the relevant point will accelerate according to the sum of internal and external forces, i.e., the lever breaks. :)
    If you actually want to do the calculation remember that the pivot point is exerting exactly the right force so that it stays in place. A simpler version is the distribution of forces on a plank supported at both ends with a weight in the middle. What is the force at each end?
  9. Jan 19, 2004 #8


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    Why all this talk of momentum? Its much easier to consider static forces.
  10. Jan 19, 2004 #9
    Suppose that we begin by assuming these two things

    (1) Newton's Second Law:

    [tex]F = \frac{dp}{dt}[/tex]

    where p = mv = linear mechanical momentum

    (2) Newton's Third Law: Both Weak and Strong Form

    Weak form: [tex]F_{ij} = -F_{ji}[/tex]
    Strong form: [tex](r_{i} - r_{j})\times F_{ij} = 0[/tex]

    respectively. Fij is the force on particle j due to particle i.

    Define the angular momentum, L, of particle i about a particular origin

    [tex]L_{i} = r_{i} \times p_{i}[/tex]

    Define the torque, N, on particle i as

    [tex]N_{i} = r_{i} \times F_{i}[/tex]

    Then it can be shown that the total angular momentum of a system of particles equals the total external torque on the system. If the total external torque is zero then the total angular momentum of the system is zero.

    The rest follows from that. Thus it can be shown from the above that when a see-saw has zero angular momentum then nothing is moving (static equilibrium) and F1d1 = F2d2. This is the lever law.

    For proof of above statements please see --
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