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- Thread starter phyeinstein_c
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tiny-tim

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i think you're confusing a bound vector with a free vector (i've forgotten the correct names, but it's

force is a bound vector, and so its line of application has to be included when specifying it

velocity is a free vector, and you can shove it around so as to put it on the end of another vector

so free vectors

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tiny-tim

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hmm, let me think …

yes, a force vector certainly generates torque (and a momentum vector generates angular momentum ,

perhaps i should have said that

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This statement is nonsense.

First, if one part of a rigid body possesses a certain velocity (vector) then all parts posess the same vector. That is a definition of a rigid body.

So velocity vectors cannot move around a rigid body willy nilly.

Second, velcocity cannot produce torque under any circumstances. That is the province of forces or couples.

Tinytim has already said that force is a bound vector, which means you cannot move it about. What he did not say is that bound vectors also have an origin or point of application as part of their specification.

So if you change the point of application of a force, yes you will induce different moments in a rigid body.

Finally for the record couples are free vectors which may be moved about.

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What do you mean?

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What system are you describing?

Is your rigid body translating or rotating about a pivot?

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mathwonk

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In applying the concept you may wish to distinguish between an individual arrow, and the class it represents. maybe this is what you call free and bound vectors, but i have never heard of them.

I chime in here because i thought maybe your teacher is a mathematician and does not know what you are talking about either.

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In applying the concept you may wish to distinguish between an individual arrow, and the class it represents. maybe this is what you call free and bound vectors, but i have never heard of them.

Yup that's about it (I'm well known for my precise mathematical statements ).

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tiny-tim

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hi mathwonk!

i'm sorry to disillusion you, but*wikipedia* has heard of them, see http://en.wikipedia.org/wiki/Euclidean_vector" [Broken] …

(though it then seems to get lost, see http://en.wikipedia.org/wiki/Euclidean_vector#In_Cartesian_space" )

In applying the concept you may wish to distinguish between an individual arrow, and the class it represents. maybe this is what you call free and bound vectors, but i have never heard of them.

i'm sorry to disillusion you, but

As an arrow in Euclidean space, a vector possesses a definite initial point and terminal point. Such a vector is called a bound vector. When only the magnitude and direction of the vector matter, then the particular initial point is of no importance, and the vector is called a free vector.

(though it then seems to get lost, see http://en.wikipedia.org/wiki/Euclidean_vector#In_Cartesian_space" )

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What system are you describing?

Is your rigid body translating or rotating about a pivot?

its a rigid body translating...

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neither did i ever hear of such bounded vectors.... in any VECTOR by definition its only REPRESENTATION of magnitude and direction..... hw does it matter if it is bounded or unbounded or free....

In applying the concept you may wish to distinguish between an individual arrow, and the class it represents. maybe this is what you call free and bound vectors, but i have never heard of them.

I chime in here because i thought maybe your teacher is a mathematician and does not know what you are talking about either.

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can u discuss some cases (examples) of the 2 cases u metioned..hi mathwonk!

i'm sorry to disillusion you, butwikipediahas heard of them, see http://en.wikipedia.org/wiki/Euclidean_vector" [Broken] …

As an arrow in Euclidean space, a vector possesses a definite initial point and terminal point. Such a vector is called a bound vector. When only the magnitude and direction of the vector matter, then the particular initial point is of no importance, and the vector is called a free vector.

(though it then seems to get lost, see http://en.wikipedia.org/wiki/Euclidean_vector#In_Cartesian_space" )

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in any VECTOR by definition its only REPRESENTATION of magnitude and direction

No there is more to it than just magnitude and direction.

Suppose I started in London and travelled 300 mile due north, turned and then travelled 300 miles due west.

Now suppose I did it the other way around ie 300 miles west then 300 miles north.

Would I be in the same place the second time around?

It is an important, and not often stated, requirement that

vector(a) + vector(b) = vector(b) + vector(a)

This is another way of saying that a+b is given by the parallelogram law.

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Since the English was so mangled I have no idea what you mean.

Was this an answer to my question,

"Would you end up in the same place either way?"

If, so was your answer yes or no?

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