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I came across the following example. I have a mass [itex]m'[/itex] that is suspended from a massless rope which is wound around a drum of mass [itex]m[/itex]. The radius of the drum is [itex]R[/itex]. The drum is fixed on a shaft with bearings of radius [itex]r[/itex]. I am supposed to determine coefficient of friction between the bearings and the drum but for now I am stuck with the forces diagram.

I created something like that:

In my solution I add a pair of forces of the same magnitude and opposite direction [itex]F^{T''}[/itex] and [itex]F^{T'}[/itex]. [itex]F^{T''}[/itex] and [itex]F^{T}[/itex] make a force couple which generates some momentum about the axis of the drum (which contributes to rotation motion). However, [itex]F^{T'}[/itex] acts on the center of mass of the drum and adds up with gravitation force. Thus, the resulting force exerted by the drum on the bearing is bigger than if there was no [itex]m'[/itex]. Consequently the friction is bigger too.

My question is whether this is correct. The solution I have only calculates the torque due to [itex]m'[/itex] and does not introduce the force couple therefore it does not have the additional force on the center of mass and consequentially no additional item in the normal force and friction respectively.

I am also curious, the friction itself produces a torque, if my reasoning above is correct, it would create another force couple, giving rise yet to another force acting on the center of mass of the drum which could go on and on. This is confusing for me at the moment.

I created something like that:

In my solution I add a pair of forces of the same magnitude and opposite direction [itex]F^{T''}[/itex] and [itex]F^{T'}[/itex]. [itex]F^{T''}[/itex] and [itex]F^{T}[/itex] make a force couple which generates some momentum about the axis of the drum (which contributes to rotation motion). However, [itex]F^{T'}[/itex] acts on the center of mass of the drum and adds up with gravitation force. Thus, the resulting force exerted by the drum on the bearing is bigger than if there was no [itex]m'[/itex]. Consequently the friction is bigger too.

My question is whether this is correct. The solution I have only calculates the torque due to [itex]m'[/itex] and does not introduce the force couple therefore it does not have the additional force on the center of mass and consequentially no additional item in the normal force and friction respectively.

I am also curious, the friction itself produces a torque, if my reasoning above is correct, it would create another force couple, giving rise yet to another force acting on the center of mass of the drum which could go on and on. This is confusing for me at the moment.

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