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A force with and without couple Suppose a force $\vec{F}$ is applied to the free floating body in absence of any other forces(Fig. 2), and $\vec{r}$ is a vector between COM and the point of application. Now, according to all resources and similar questions, the torque is $\vec{r}\times\vec{F}$. According to wikipedia, "Torque is defined as the product of the magnitude of the perpendicular component of the force and the distance of the line of action of a force from the point around which it is being determined." It makes sense if we assume that "the point around which it is being determined" is the center of rotation - that is, it does not move linearly. It is the case when the body is pinned as in the gif on wikipedia page, and a lot of articles about torque assume that the center of rotation is just a pin joint. My understanding of how pin joint works is that it creates the opposite coupled force which works as if we had the same force applied in the opposite direction on the other side of the pin point, thus cancelling out the first force and giving us "pure torque" according $\vec{r}\times\vec{F}$. That is, having a pin joint in the COM is the same as having the opposite force on the other side of the COM - as shown if Fig.1. Is it true? If it is true, then it means that in absence of pinning and a coupled force(Fig 2), the torque around COM should be half of what it is with a pin joint(or a couple) - $\vec{r}\times\vec{F}\over 2$. If it is not true, please explain why - how can we get the same torque in both Fig 1 and Fig 2.

UPD: i think i got Fig 1 wrong and the couple produced by the pin joint should be in the middle(where the pin is), which will give $\vec{r}\times\vec{F}$ as by original formula. But how do we compute torque for the second figure, in absence of couple force? And how can it be the same as with couple?

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2 Answers 2

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Now, according to all resources and similar questions, the torque is $\vec{r}\times\vec{F}$.

That's the torque about the COM. But there is a torque about any point on the object (except the point of application of the force), not just the COM.

According to wikipedia, "Torque is defined as the product of the magnitude of the perpendicular component of the force and the distance of the line of action of a force from the point around which it is being determined."

The key phrase is "from the point around which it is being determined". The COM is only one such point.

It makes sense if we assume that "the point around which it is being determined" is the center of rotation - that is, it does not move linearly.

The point about which the torque is being determined is not necessarily the center of rotation (COR). Whether or not it moves linearly depends on whether or not it is constrained.

It is the case when the body is pinned as in the gif on wikipedia page, and a lot of articles about torque assume that the center of rotation is just a pin joint.

The COR will be about the pin point simply because it is constrained to do so.

That is, having a pin joint in the COM is the same as having the opposite force on the other side of the COM. Is it true?

Yes. A frictionless pin support provides no moment reaction, only a force reaction. But this is true whether or not the location of the pin is at the COM. The pin is simply a constraint that forces the center of rotation to be at the location of the pin, wherever that might be.

If it is true, then it means that in absence of pinning and a coupled force, the torque should be half of what it is with a pin joint(or a couple)

I'm not sure what you are saying, but in the absence of pinning or the application of a couple (equal and opposite parallel forces) the torque will depend on the point about which it is determined and will be the product of the force and perpendicular distance to that point.

FIG 1 below shows an object of length L with a force $F$ applied at one end and a frictionless pin support at the other end. We will assume no other forces (e.g., gravity) are acting on the object. The moment about the pin support is $FL$. That is also the value of the couple produced by the equal and opposite reaction force at the pin. The COR is constrained to occur about the pin. Due to the couple, none of the points on the object undergo translation, only rotation about the PIN.

FIG 2 shows the same object without pinning or a coupled force. The moment about the left end of the object is still $FL$. But now there is a net force on the object which will give it linear as well as angular acceleration.

If it is not true, please explain why - how can we get the same torque in both pinned and free-floating cases.

As shown in FIG 2 you get the same torque about the left end with or without the pinning. The difference is you have linear acceleration of the object due to a net force without the pinning.

Hope this helps.

enter image description here

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  • $\begingroup$ You're considering torque around the other end of the stick. I updated my question and added a picture to show two situations, and the question is about torque around COM(which we assume is the center of the stick here). I don't get how can we get the same torque on both fig.1 and fig.2 on my picture. If we agreed that fig 1 is equivalent to having a pin joint in the center, shouldn't we get half the torque around center in the fig 2? $\endgroup$
    – idontgetit
    Jul 9, 2023 at 11:49
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I think that the two diagrams below might answer your question.

enter image description here

In Fig. 1 the two forces constitute what is called a couple and the torque is $F\,d$ independent of position of any axis etc.

In Fig. 2 by adding two equal and opposite forces at the centre of mass you can say that the couple (red) is $F\,d/2$ and there is a force $F$ acting at the centre of mass which, if allowed, could produce a translational acceleration.

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  • $\begingroup$ So, you're confirming that in Fig 2 we get half the torque that we get on the Fig 1? I don't get how we can get torque about anything but center of rotation, and if we only consider the center of rotation(which in that case is assumed to be COM in the middle of the stick), we should get F * d/2 for Fig 1 and F * d/4 for Fig 1, right? $\endgroup$
    – idontgetit
    Jul 10, 2023 at 18:05
  • $\begingroup$ @idontgetit In Fig. 1 for example choose any point and you will find that the torque is always $Fd$. Couple $\endgroup$
    – Farcher
    Jul 10, 2023 at 21:13
  • $\begingroup$ thanks, i got it, but why isn't the first figure equivalent to having a pin in the middle and just one of the forces? Doesn't pin create a couple? If it is equivalent, then the torque should be Fd/2 because the lever is d/2 which contradicts the definition by couples. $\endgroup$
    – idontgetit
    Jul 11, 2023 at 22:40
  • $\begingroup$ @idontgetit For your figure 1 a pin in the middle would constrain the rod to rotate about the centre of mass and the torque/couple would still be $F\,d$. Why would there be a force $F$ acting at the pin? If one of two forces $F$ were not there as per Fig. 2 then indeed there would need to be a force $F$ at the pin and the couple would be halves but you have drawn two forces acting on the rod in Fig. 1. $\endgroup$
    – Farcher
    Jul 12, 2023 at 7:38
  • $\begingroup$ But the question is - does having a pin and a force produce 2x more torque than not having a pin and a force? Intuitively it should, because a pin creates a couple, is it correct? $\endgroup$
    – idontgetit
    Jul 14, 2023 at 14:42

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