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My textbook states this principle as follows

" The conditions of equilibrium or of motion of a rigid body will remain unchanged if a force acting at a given point of the rigid body is replaced by a force of the same magnitude and same direction, but acting at a different point, provided that the two forces have the same line of action"

My question is as follows:

Case 1: Assume an unconstrained rigid disc lying on a smooth horizontal surface such that its centre is at the origin of the coordinate system and z axis perpendicular to the horizontal plane. Now, apply two equal and unlike forces at (0,-1) and (0,1) parallel to x-axis so that they form a couple which will induce rotation about origin.

Case 2: Now, move the two forces along their lines of action such that the two equal and unlike forces now act at (1,1) and (1,-1) respectively. Clearly the forces now induce a rotation about the point (1,0).

How can the above two cases be equivalent. The motion of rigid body does change in the above scenario. But the principle of transmissibility of forces states that the motion of rigid body remains unchanged.

I know that I am missing something obvious. I would be happy if someone could point it out.

Reference book: Vector Mechanics for Engineers by Beer and Johnston

Sai Teja
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  • I must admit I don't understand what "The conditions of equilibrium or of motion of a rigid body will remain unchanged" means in this context. I think your teachers are on far safer ground if they stick to Newton's laws including the third law. – James Arathoon Sep 02 '18 at 14:47

1 Answers1

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The above two cases are equivalent, we know that due to the couple, the disc would be at equilibrium. The torque about the center, would also be same because it is given by $\tau = F \times R = FRsin(\theta)$ and if $R$ has changed, so has the $sin(\theta)$, you will take components of the force perpendicular to the radius and the torque would remain as $R$ will become $\sqrt{2}$ and $sin(\theta)$ will become $\frac{1}{\sqrt{2}}$

Harshit Joshi
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  • I understand that the net torque about any point in the entire plane is the same. So, If the disc is fixed about some point in the plane, I would agree with your assessment that the body behaves the same in both the cases. But clearly we can observe that the actual motion of the body when no point is fixed is different in the two cases. How can we say that the motion of the rigid body remains unchanged when no point of ghe disc is fixed. – Sai Teja Sep 02 '18 at 14:08
  • If the net torque is same and the couple CANCELS out to give NO MOTION of the center of mass then it does not matter whether the body is fixed or not it will not move and will only rotate. – Harshit Joshi Sep 02 '18 at 14:23
  • In case 2, the center of mass of the body moves. Also, when the author says motion, he is considering both translation and rotation. I understand that in both the cases the body only rotates and there is no translation. But, in case 1 it rotates about the origin and in case 2 it rotates about (1,0). We cannot say that both the motions are equivalent because, for example, in case 1 the centre of disc is at rest and in case 2 the centre of disc is in motion. – Sai Teja Sep 02 '18 at 14:40
  • No, the center of mass only moves if there is a net non zero force acting on the body. Try this yourself using a circular plate or something. Couple forces never cause motion of Center of Mass. – Harshit Joshi Sep 02 '18 at 14:46
  • The centre of rotation moves, not the centre of mass. If the centre of rotation moves, about which the angular acceleration will occur in this case, and this does not always coincide with the centre of mass of the object, then continuously changing unbalanced accelerations will occur in part due to the changing moment of inertia. If the word "equilibrium" is interpreted as internal forces to the rigid body being in an inertial frame of reference, then this seems not to hold in this case. – James Arathoon Sep 02 '18 at 15:14
  • @SaiTeja Reply to the comment. – Harshit Joshi Sep 02 '18 at 16:29
  • @JamesArathoon I don't know about the center of rotation. I just know the fact that forces that balance out do not cause translation of the body. – Harshit Joshi Sep 02 '18 at 16:30
  • @Harshit Joshi: You obviously have to distinguish between static balance and dynamic balance. What about the case of the gyroscope? Apply a perpendicular balanced couple force to a dynamically balanced spinning object and precession will occur. – James Arathoon Sep 02 '18 at 17:13
  • @JamesArathoon Sorry sir, i am just a pre university student and answered the question correctly as far as my knowledge goes. I think that OP does not even care now. – Harshit Joshi Sep 02 '18 at 17:15
  • @James Arathoon and Harshit Joshi: If I understand correctly, you mean to say that the centre of mass stays at rest in case 2 as well as case 1. Now, please consider a different case: Assume a force F acting at centre of mass of the disc in positive X direction. Also assume another force of same magnitude parallel to this force acting in negative X direction at a point other than centre of mass. Do you mean to say that even in this case the centre of mass wouldn't move as the net external force is zero. It just seems practically impossible. – Sai Teja Sep 02 '18 at 17:50
  • @SaiTeja Yes it should not move at all. – Harshit Joshi Sep 02 '18 at 17:52
  • I understand that what you are saying is corect. Because, ultimately net external force must equal to mass times acceleration of centre of mass. It's just that I am unable to visualize it in this example and it kept bugging me. So, I guess in all these cases that I have described, the body purely rotates about its centre of mass with no translation. Thanks Harshit and James for your help – Sai Teja Sep 02 '18 at 18:15
  • @SaiTeja Please upvote and mark the final answer. – Harshit Joshi Sep 02 '18 at 18:23