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In The Feynman Lectures, vol.I, chapter 18, Feynman discusses torques on a rigid body in two dimensions and says:

Now we pause briefly to note that our foregoing introduction of torque, through the idea of work, gives us a most important result for an object in equilibrium: if all the forces on an object are in balance both for translation and rotation, not only is the net force zero, but the total of all the torques is also zero, because if an object is in equilibrium, no work is done by the forces for a small displacement. Therefore, since ΔW=τΔθ=0, the sum of all the torques must be zero. So there are two conditions for equilibrium: that the sum of the forces is zero, and that the sum of the torques is zero. Prove that it suffices to be sure that the sum of torques about any one axis (in two dimensions) is zero.

First, what does Feynman mean by "the forces on an object are in balance ... for rotation"? If I do not multiply axis-tangential forces by their levers, then what does it mean for them to balance? on the other hand, if I do multiply them by their levers, then we are looking at balanced torques, and the assertion becomes circular and vacuous; am I wrong? What does Feynman mean here?

Second, can you help me prove Feynman's assertion?

Prove that it suffices to be sure that the sum of torques about any one axis (in two dimensions) is zero.

(Note, this is not homework)

nir
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  • See relavent answers to related questions http://physics.stackexchange.com/a/111348/392 and http://physics.stackexchange.com/a/93742/392 – John Alexiou Apr 10 '15 at 14:25
  • I am surprised to see Feunman talk about net torque without discussing the point at which torques are measured about. For bodies in balanace, yes the point is irrelevent, but in general it makes a world of difference. – John Alexiou Apr 10 '15 at 14:28
  • The word "forces" in "all the forces on an object" is not really "force" but, as you said, includes torques. The preceding words are explaining this assertion. – Zheng Liu Apr 11 '15 at 03:18

2 Answers2

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The definition of being "in balance" for a rigid body is the absent of acceleration: both laterally and angularly.

For a rigid body, the condition of to be "in balance" is the sum of all torque has to be zero with respect to any axis, not just to any particular one. If this condition is not held i.e. there exists an axis that the net torque is not zero, then there would be an angular acceleration around that particular axis, contrary to the definition of being "in balance".

Tran Quyet Thang
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  • I found this definition in the wikipedia, which seems to cover rotations: "The static equilibrium of a mechanical system rigid bodies is defined by the condition that the virtual work of the applied forces is zero for any virtual displacement of the system. This is known as the principle of virtual work" - http://en.wikipedia.org/wiki/Rigid_body_dynamics#Static_equilibrium – nir Apr 11 '15 at 08:02
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What he means is:

  • Sum of forces is $\sum \limits_i \vec{F}_i = \sum \limits_i [F_x,F_y]$
  • Sum of torques is $\sum \limits_i \vec{r}_i\times\vec{F}_i = \sum \limits_i (F_y r_x-F_x r_y)$

So it is not a circular argument because you can have net forces zero but net torque not zero (two equal and oppsite forces a distance apart, or a force couple).

In fact, as long as the net force is not zero there is a point on the plane about which the net torque is zero. Extend the two lines of actions of the forces and where they meet is this point.

The second question is answered in https://physics.stackexchange.com/a/111348/392 and you can follow that the sum of the torques (above) is the same regardless of the point about which $\vec{r}_i$ are defined when the sum of forces is zero.

Community
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John Alexiou
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  • Feynman says the forces for rotation are in balance, what does that mean other than torques? – nir Apr 11 '15 at 08:07
  • It must mean forces that result in torque. A force trhough the center of mass isn't a force for rotation. – John Alexiou Apr 11 '15 at 17:43
  • right, but what does it mean for them to be in balance, if they are not multiplied by the length of their lever arms (becoming torques)? – nir Apr 11 '15 at 17:58