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The center of mass of a uniform ring lies at its geometrical center i.e. outside the body. But we generally define center of mass as a point in which if we apply a force, the whole body will move in the same way as if the whole of the mass is assumed to be concentrated there and how that point mass moves under that force. So my question is how can we apply a force to a point not on body (here the point is the center of mass) and still see the force's effect on that body?

For example consider this: suppose a uniform ring hanging in mid air in a gravity free environment in $y$-$z$ plane. Now you have to move the ring along x-axis so that whole ring remains in a plain. No torque is to be applied and you can not put any charge on the ring and distribute it uniformly and apply an electric field along $x$-axis. How can we do it?

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    $\begingroup$ You have already mentioned that the whole mass of the system is assumed to be concentrated at the cm. So force is assumed to be acting on this fictitious mass $\endgroup$ Commented Sep 7, 2017 at 15:24
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    $\begingroup$ Why are you asking us how to do something which is obviously impossible? Who has told you to apply a force at a point where nothing exists? $\endgroup$ Commented Sep 8, 2017 at 15:59
  • $\begingroup$ @sammygerbil What i am asking is this suppose a uniform ring is hanging mid air in a gravity free environment in y-z plane now what u have 2 do is to move the ring along x-axis so that whole ring remains in a plain. No torque is to be applied and you can not put any charge on the ring and distribute it uniformly and apply an electric field along x-axis. How will you do this if someone asks you to do so :/ $\endgroup$
    – user168487
    Commented Sep 8, 2017 at 16:49
  • $\begingroup$ Your title is : Effect of external force on centre of mass of a body lying outside it. The question you have posted is : how can we apply a force to a point not on body and still see the force's effect?? If you wish to change your question you must edit the question, not provide comments. $\endgroup$ Commented Sep 8, 2017 at 17:32
  • $\begingroup$ The center of mass is defined in terms of the net force, which means the combined loading line has to pass through the center of mass. $\endgroup$ Commented May 28, 2019 at 3:49

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In rigid body mechanics there the concept of the extened rigid body. The includes points in space that are not attached to the physical body, but move with the body as if they were. In addition, any forces applied to the extended body are felt through the center of mass, with the equipollent torque added.

Example: At some instant a rigid body with center of mass located at $\vec{r}_{\rm cm}$ has a force vector $\vec{F}$ applied at some location $\vec{r}$ on the extended rigid body. Find the equipollent loading on the center of mass.

Answer: The force $\vec{F}$ transfers to the center of mass as-is. But a torque $\vec{\tau}_{\rm cm} = \left( \vec{r} - \vec{r}_{\rm cm} \right) \times \vec{F}$ needs to be applied to the center of mass also.

Here $\times$ is the vector cross product.

More specifically and uniform ring can be loaded on the body, but the combined net loading to have zero torque about the center of mass. This is equivalent to having a pure force acting through the center of mass.

There are several scenarios that can make this happen.

  1. Any radial (outward) force on the ring, can be thought as going through the center of mass as a force can slide along its line of action without changing the problem.
  2. A force is applied on the ring as well as a torque (force-couple) which causes then net equipollent torque at the center of mass to be zero.
  3. The body is under the influence of gravity, or any other so-called body force. When a uniform load is acting equally on every particle of a physical body, this is equivalent to a combined load through the center of mass.
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If you apply a force to a point not on the body, there will be no effect, obviously.

Perhaps an important clarification on the whole center of mass thing: if for example you're dealing with gravity, you aren't saying "well it's as if there's a force $mg$ applied at the center of mass"! What you're saying instead is "it's as if the whole thing is just a 0-D point mass at the center of mass".

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It appears that you want to know how to move the ring by pushing on it, without applying a torque. The only possibility is to apply force on the ring at two or more points in such a way that the torques due to the forces add up to zero.

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