Register to reply 
Principle of transmissibility 
Share this thread: 
#1
Sep2107, 04:55 PM

P: 23

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. thank you in advance. 


#2
Sep2107, 05:49 PM

Mentor
P: 41,440




#3
Sep2107, 06:26 PM

P: 23

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? 


#4
Sep2107, 06:59 PM

Mentor
P: 41,440

Principle of transmissibility



#5
Sep2207, 04:39 PM

P: 23

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.
thanks in advance. 


#6
Sep2207, 06:15 PM

Mentor
P: 41,440

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
Sep2207, 06:27 PM

P: 23

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 missunderstood 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
Sep2207, 06:48 PM

Mentor
P: 41,440

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 [itex]\vec{r}[/itex] changes, so the component of the force along that direction will surely change. But that doesn't mean anything. 


#9
Sep2307, 09:34 AM

P: 23

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
Sep2307, 09:52 AM

Mentor
P: 41,440

You are treating [itex]\vec{r}[/itex], 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 changethat's what the principle of transmissibility says.) IMHO, "transmissibility" is a hilariously overblown term used only by engineers. (Just teasing! ) 


#11
Sep2307, 10:52 AM

P: 23

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
Sep2307, 03:11 PM

Mentor
P: 41,440




#13
Sep2307, 03:23 PM

P: 23

i was just reading about reducing a force to a binaryforce 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! 


Register to reply 
Related Discussions  
Is this possible, IN PRINCIPLE?  Astronomy & Astrophysics  20  
Principle Value  Calculus  2  
Equivalence of DAlembert's principle and Action Principle  Classical Physics  4  
Principle Value of 1/x  Calculus & Beyond Homework  3  
First principle?  General Discussion  1 