# Principle of transmissibility

1. Sep 21, 2007

### DaNiEl!

Hello. Sorry if it is a bother, but i'm starting Applied Mechanics (statics) and my books states this "principle of transmissibility" with no proof (except for the one given in the dynamics volume which i don't have). could anyone provide me an explanation or point me in the direction of a website which explains where this comes from? my main problem with it is that, if you slide a force vector along a line which doesn't intersect the center of mass, you should get a different angular momentum...
i probably didn't understand it. either way, it would be useful to see the proof.

2. Sep 21, 2007

### Staff: Mentor

A force applied to a rigid body will have the same effect if applied anywhere along its line of action. That's a statement of the principle of transmissibility.
Note that sliding a force along its line of action does not affect the torque it creates. (You move it along its own line of action, not sideways.)

3. Sep 21, 2007

### DaNiEl!

oh, i see. the explanation should lie in the fact that the arm length and the projection of the force on a perpendicular to r vary in the same proportion so the torque is always the same. it's good to see it intuitively, though a proof would be better (to confirm this hypothesys).
another thing which i have trouble with when i visualize it is that if a force perpendicular to r has no translational component, when u slide it along the line of action it should have some, or am i seeing it wrong?

Last edited: Sep 21, 2007
4. Sep 21, 2007

### Staff: Mentor

Realize that the lever arm doesn't change as the force slides along its line of action. Equivalently, $\vec{r}\times\vec{F} = Fr\sin\theta$ is constant.

If I understand what your saying, what counts is $r\sin\theta$, which remains constant.

5. Sep 22, 2007

### DaNiEl!

let's say you have a body at the origin and a force (1,0) applied at the point (0,1) (r=(0,1)). the external product is not null, but the internal product is. now if the same force is applied at the point (1,1)(r=(1,1)), the external product of the force with r (fxr) is still 1, but the internal product is not null this time. i need to know what i am doing wrong so i can understand this principle, because it seems to be used a lot throughout the book.

6. Sep 22, 2007

### Staff: Mentor

You're not doing anything wrong. What counts in determining the effect of a force on a rigid body is the external (or cross) product, which gives you the torque it produces. The internal product has no particular significance.

7. Sep 22, 2007

### DaNiEl!

the reason i refered the internal product is that, "apart" from a multiplication of a constant (the inverse of the distance (||r||)), it should give the component of the force paralel to r. if i understood the principle of transmissibility, a force should have the same effect if placed on any point of it's line of action. unless i miss-understood this, the principle should account for the torque and the translational force. what you are telling me is that the principle refers only to the torque?

8. Sep 22, 2007

### Staff: Mentor

I assume that by "translational force" you mean the translational effect of the force (as opposed to the rotational effect, which is captured by the torque). That will certainly not change by just sliding the force, as long as you don't change its direction.

Realize that as the forces slides, the direction of $\vec{r}$ changes, so the component of the force along that direction will surely change. But that doesn't mean anything.

9. Sep 23, 2007

### DaNiEl!

I don't understand what you are saying. you mean, the force component changes but the effect doesn't? this is too puzzling for me. i don't understand how the translational component change yet this having no importance. doesn't that change the movement of the body?

10. Sep 23, 2007

### Staff: Mentor

Imagine a fixed set of coordinate axes. As long as you just slide a force vector parallel to its line of action, its orientation with respect to those fixed coordinates does not change.

You are treating $\vec{r}$, the position vector describing the location of the point of application of the force with respect to the origin, as if it were a fixed direction in space. It's not!

Take the example you gave in post #5. You have a force (1,0) applied at point (0,1). The force points in the +x direction. Now if you apply the same force at point (1,1), it still points in the +x direction! No change in its translational components. (Beyond that, the torque it produces doesn't change--that's what the principle of transmissibility says.)

IMHO, "transmissibility" is a hilariously overblown term used only by engineers. (Just teasing! )

11. Sep 23, 2007

### DaNiEl!

then, my missunderstanding seems to be at the basic level of forces and torque. what you are telling me is that when, as in my example, a force (1,0) acts on the point (0,1), it not only produces a rotation (torque=(-1) x 1 N.m) but a translation according to a force of 1N in the x direction?

12. Sep 23, 2007

### Staff: Mentor

That's correct. The rotational effect of a force depends on the torque it produces about some reference point, while the translational effect is always just parallel to the direction of the force (no matter where the force is applied.)

13. Sep 23, 2007

### DaNiEl!

i was just reading about reducing a force to a binary-force system and this made sense. i always thought it was only the component paralel to the line from the center of mass to the force that counted but i guess i was wrong. makes me wonder if it has energy implications.
thanks for the help!