0
$\begingroup$

I have to admit that I do not understand why the principle of transmissibility of forces works. The textbook states:

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

Ok, let's assume that I have some almost two-dimensional (Its hight is small) rectangular object. It has vertices A, B, C and D. In the first case, I attach a string to C and apply a constant force F (at a time t=0) parallel to the edge CD. After some time, t, passes, the line which connects point C with the center of mass of my rectangle will be parallel to the direction of the force F ( at first the rectangle rotates counterclockwise and then just slides in the direction parallel to force F).

In the second case, I attach my string to point D and apply the same constant force F parallel to CD ( so the line of action is the same as in the first case). Again, after some time t, the line which connects point D with the center of mass of my rectangle will be parallel to the direction of the force F and my rectangle will just slide in the direction parallel to the applied force.

However, in the second case, point C will be located to the left of the line which connects the center of mass and the point to which force F is applied (please, see attached image). This means that my rectangular object undergoes, in the second case, rotation on a larger angle. So, in the end, I get different results. I would be grateful if someone explained to me where I am making a mistake.enter image description here

$\endgroup$
0
$\begingroup$

The forces in each case are not acting through the same line of action after $t=0$, so the part you quote no longer applies. Note that the quote does not say that the same line of action at one instant in time is sufficient for the motion / equilibrium to be the same.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.