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I think I'm missing something about torques. Do supports exert torques? Consider these two situations I'm confused about.

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A rod is attached in to a wall so that it cannot rotate. Since weight exerts a torque, the wall must do it too. But my doubt is: if I take A as point to measure torque, then it is necessarily zero and $\sum M =mg \frac{l}{2}\neq 0$. Nevertheless there is static equilibrium, how can that be?

Another situation is the one with a rod that can rotate about a point B, with another support in point C. Suppose there is also a weight on the rod in a particular position. Taking point B to measure torques, does the support in C exerts torque? In some exercises I noticed that the torque of the support in C is not considered for equilibrium. Furthermore, suppose to take a point to measure torque that differs from B, then does B exert torque?

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Torques are always calculated about some specified reference point, if you change the reference point, you change the individual torque which each force correlates to (``produces''?). If the sum of the torques about a certain point is non-zero, then changing the point may change the total torque, but if the sum of torques about one point is zero (and the sum of the forces is zero), the sum of torques about any point will be zero. Notice, that it's the sum of torques which is zero.

Now to address your first diagram, if you calculate the torques about point A, then you must consider the interaction of the walls of the hole with the rod. Those interactions produce the torques which make the sum equal to zero. If the rod is in static equilibrium, you can't ignore the effects of the walls of the hole. And you can't say that the entire hole is a point.

Edit: If the rod is secured to the wall by attaching it to the outside by sending a screw or nail through the rod, then the friction between the wall and rod, or the friction between the rod and screw (or nail) will exert a torque about A.

Regarding the second diagram, if support C is touching the rod, it exerts a force on the rod and therefore has a torque about point B. If a different point (away from B) is chosen and the rod is touching support B, then B exerts a torque. But if the system is in rotational equilibrium, the sum of the torques about a chosen point must be zero.

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    $\begingroup$ Nitpick: if the sum of torques about one point is zero and the sum of the forces is zero, then the sum of the torques about any point will be zero. Granted, this additional assumption is always satisfied in statics problems, but it's not generally the case when things are in motion. $\endgroup$ – Michael Seifert Mar 18 '16 at 18:06
  • $\begingroup$ Ah, yes! If the sum of the forces isn't zero, then the object will linearly accelerate, which could be considered as rotational acceleration (of the object as a whole) about other points. I'll edit. $\endgroup$ – Bill N Mar 18 '16 at 19:10
  • $\begingroup$ @BillN Great answer! Considering the second diagram, if I'm asked to determine how far from C the weight can be placed and still have equilibrium (considering also the weight of the lever), should I take into account also the torque of the support in C? (I'm asking because I saw in a similar exercise that such torque is not considered) $\endgroup$ – user104324 Mar 19 '16 at 6:55
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    $\begingroup$ You have to include all forces and all torques about which ever point you want to consider. If you have equilibrium with the weight at one point, then you move it, the force (and resulting torque) at each support will change until you get a physically unrealistic solution. For example, if C can only push up, but can't pull down, then investigate the arrangement for C to have zero force. Basically, don't ignore any force or torque, but you can let their values go to zero. $\endgroup$ – Bill N Mar 19 '16 at 15:11
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The answer to your question is simply "yes" , supports do exert torques.So as you mentioned , in your first example there is an opposite toque (-mgl/2) in support A , in order of static equilibrium. Similar situation in example 'B'

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This is mostly about definition of "exert".

A contact between two rigid bodies that "exerts a torque" can mean two things, though I think only the first one is really useful and makes sense:

  1. The contact is fixed in such a way that at the point of contact the contact can produce a non-zero torque. That is, it's a weld and not a hinge.
  2. The contact can exert a force, so that effectively creates a torque somewhere else. That is, it's just about anything, a weld and hinge, etc, and this definition is therefore fairly useless.

Also, useful here is "support a torque", which is similar to exert a torque, but is passive rather than active, eg, welds can support torques, hinges can't.

In your diagrams: A would typically refer to a contact that could exert a torque, and B and C could not exert a torque (ie, like a fulcrum).

That is, in the second diagram, if the bar isn't going to move, the net torques about B and C must both be zero, since these points can't exert torques. In the first diagram, A can support a torque, so the bar won't move in general.

These are not cut-in-stone definitions though, and the context must be considered. For example, it's also fair say "F exerts a torque about B" where B is away from F, or that a "string exerts a torque on a pully" (where it's assumed we're talking about the center of the pully and not the contact point of the string -- since the string can't exert a torque about this point).

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