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enter image description here

The image is a body free-falling with two forces acting on it. If we take moments about A, we clearly get anticlockwise moments, and that is what will happen if this were to take place in real life.

However, if you take moments about B, the resultant moments would be clockwise - this is clearly wrong, but why is it wrong? I can see some people saying we have to take moments about the centre of mass, but why is that the case?

Thanks

-----edit sorry the two forces should act from the same side

------edit diagram edited

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  • $\begingroup$ I guess the arrow from A to B is not meant to be an arrow especially related to A and B ? :-) $\endgroup$
    – Fabrice NEYRET
    Oct 26, 2015 at 21:30
  • $\begingroup$ That's the weight and the points A and B are where the weight crosses the forces $\endgroup$
    – RelativisticDolphin
    Oct 26, 2015 at 22:08
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    $\begingroup$ @VincentL, if the forces displayed on the graph are not right, please update the graph. It's pretty confusing otherwise. $\endgroup$
    – BowlOfRed
    Oct 26, 2015 at 22:13

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I'm not sure your graphic shows it well. By my take, both the weight force and the one at the bottom go through B, so there is zero torque from them about that axis. So only the upper force from the right creates a torque, and that is still counter-clockwise.

But the general reason you can't do this is that your choice of axis is accelerating, and is therefore in a non-inertial frame. An additional, fictitious force appears in this frame that opposes the acceleration. Because this force exactly opposes the gravitational force, you get no torque contribution from the weight.

When you analyze the torques about the COM axis, you are already ignoring the weight. So this step can be skipped in that case without causing problems.

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  • $\begingroup$ if the force on the right were pointing to the right, that is both forces on the same direction, would your argument still hold? (you will get a clockwise torque if calculated on B and a anticlockwise if calculated on A $\endgroup$
    – user83548
    Oct 26, 2015 at 22:04
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    $\begingroup$ Because then you would have a net force to the right, so the block would accelerate to the right. That would lead to a fictitious force to the left (through the COM) appearing. Total torque would then be CCW for analysis about both A and B. $\endgroup$
    – BowlOfRed
    Oct 26, 2015 at 22:10
  • $\begingroup$ Yep thanks very much - I've got another question. For a free-falling body with only weight acting on it, if we take moments about a point which isn't on the body or the line of weight, we would get a moment about that point. However, the body isn't rotating - why can't I take moments about that point? I think the point is not on the body so it's still an inertial frame so it's not related to frame of refernce $\endgroup$
    – RelativisticDolphin
    Oct 26, 2015 at 22:13
  • $\begingroup$ That should be a separate question, not a comment. But let's just say that you can have angular momentum without rotating. $\endgroup$
    – BowlOfRed
    Oct 26, 2015 at 22:24

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