Is force a a bound vector or a free vector?

[fog37]

TL;DR Summary

Force, free or bound vector?

Hello Everyone,

A small dilemma: is force, which is a vector, a free vector, since it can be slid along its along of application, thus changing its point of application (principle of transmissibility) or a bound vector, since the point of application of the force is crucial for the effect the force produces?

A bound vector, by my definition, is one whose point of application cannot be changed without changing the effect of the vector...

Thank you!

 

[KingOfDirewolves]

Force is a Bound Vector, it's effect is different for different points of application.

 

[DrStupid]

Force is a Bound Vector

Not by definition. Force is defined to be the reason for changes in the state of motion and to be proportional to the change of momentum. That is independent from the point of application.

Yes, torque for example depends on the point of application of a force. But does that make force a bound vector?

 

[Mister T]

A bound vector, by my definition, is one whose point of application cannot be changed without changing the effect of the vector...

Can you give us an example of a vector that doesn't fit your definition of a bound vector?

 

[fog37]

For instance, a couple (two equal magnitude, parallel and displaced forces) is a free vector. The point about which we describe the couple does not need to be specified making it a free vector...

 

[Mister T]

For instance, a couple (two equal magnitude, parallel and displaced forces) is a free vector. The point about which we describe the couple does not need to be specified making it a free vector...

But that's just a pair of vectors. I was looking for an example of a vector (not a pair of vectors) that doesn't fit your definition of a bound vector.

 

[fog37]

mmm... I know there is a distinction between bound and free(sliding) vectors. For example, I was reading, the translation vector is a free vector since what is important is not its point of application but its direction and magnitude...

 

[Mister T]

For example, I was reading, the translation vector is a free vector since what is important is not its point of application but its direction and magnitude...

I'm familiar with the translation of vectors, but I'm not familiar with the translation vector. What units, for example, are used to measure the magnitude of this vector?

 

[sophiecentaur]

But that's just a pair of vectors.

Is that right? The vector, corresponding to a couple lies in the direction of the axis but can be achieved with many different pairs of force vectors. You can apply a couple on a floating body at any point on it and it will still have the same angular acceleration. I think that implies that it's unbound.

 

[vanhees71]

I'm not sure what's meant by "free" vs. "bound vector". I can only try to guess. In #1 you said a bound vector is one that only makes sense given some fixed point. I thus guess you are referring to cases, where position vectors are involved, like angular momentum or torque, and I also assume we are talking about Newtonian physics.

Indeed, the position vector needs an arbitrary chosen reference point in space. Only then you can describe each position precisely by a vector. Since the choice of the reference point is arbitrary, physics should not depend on its choice, and indeed the Galilei-Newtonian spacetime is translationally invariant, i.e., the choice of the reference point doesn't change the physics of any phenonon, just its description. Formally that's included in the Newtonian theory by translation invariance in space (i.e., it's based on the homogeneity of the affine Euclidean space used to describe space in Newtonian physics).

From Newton's 2nd Law we have
$$m \ddot{\vec{x}}=\vec{F}.$$
Now already the velocity $\dot{\vec{x}}$ does not depend on the choice of the reference point and thus is in your definition a free vector. The same consequently holds true for the acceleration $\ddot{\vec{x}}$. Also the mass of a point mass doesn't change with position within Newtonian mechanics. Thus it's a scalar independent of position, and thus also force is a "free vector".

That's not true for torque or angular momentum, because it's definition depends on the choice of the reference point, $\vec{\tau}=\vec{r} \times \vec{F}$ and $\vec{J}=\vec{r} \times \vec{p}=m \vec{r} \times \dot{\vec{r}}$.

 

[Mister T]

The vector, corresponding to a couple lies in the direction of the axis but can be achieved with many different pairs of force vectors.

That sounds like the description of a torque vector. It would not fit the OP's definition of a bound vector since it will have a different effect if it's applied in different places. For example, the same torque vector applied to a bridge girder will not have the same effect when applied to a tree.

The point I'm trying to make to the OP is that the vectors he's encountering have a magnitude and a direction and those properties are unaffected when the vector is translated. If the vector is applied in different locations it will have different effects, but changing the effect a vector has need not change the properties of the vector itself. Thus the distinction between bound and unbound vectors is superfluous.

 

[sophiecentaur]

For example, the same torque vector applied to a bridge girder will not have the same effect when applied to a tree.

But those two situations involve other forces. If you take a girder out in space and apply a torque vector, does it matter where you apply it? It's difficult to apply personal experience to this because it's pretty well impossible to apply just a torque by hand. Perhaps rotating a floating boat from a quayside could give a clue about what happens.

 

[Mister T]

If you take a girder out in space and apply a torque vector, does it matter where you apply it?

Put the tree in outer space, too. Apply the same torque to the tree as you do to the girder. They will, in general, have different effects. My point is that even though these vectors can be translated, it doesn't change the vector, but it can change the effect that the vector has on objects. Thus it seems silly to call them all bound vectors when according to the definition given by the OP, every vector is a bound vector. It's a distinction without a difference.

 

[sophiecentaur]

Apply the same torque to the tree as you do to the girder. They will, in general, have different effects

Why? Newton 2 still applies - as long as they have the same MI so why would the angular acceleration be different?
Both objects will rotate about their CMs.

 

[vanhees71]

Put the tree in outer space, too. Apply the same torque to the tree as you do to the girder. They will, in general, have different effects. My point is that even though these vectors can be translated, it doesn't change the vector, but it can change the effect that the vector has on objects. Thus it seems silly to call them all bound vectors when according to the definition given by the OP, every vector is a bound vector. It's a distinction without a difference.

I think it's the distinction between vectors that depend on a specific location in space (an affine manifold not simply a vector space!) like the position vector, torque, and angular momentum.

Take a closed two-body system interacting via a central force (i.e., an interaction force $\vec{F}_{12}=-\vec{F}_{21}$ that's always in direction of the line connecting the two points) as the most simple example. You start in an arbitrary inertial reference frame. A reference frame consists of the choice of an arbitrary reference point in space, the origin $O$. Then any other point $P$ is uniquely determined by the position vector $\vec{r}_P=\overrightarrow{OP}$ (just remember the definition of an affine (Euclidean) space).

The total angular momentum is given by
$$\vec{J}=m_1 \vec{r}_1 \times \dot{\vec{r}}_1+m_2 \vec{r}_2 \times \dot{\vec{r}}_2.$$
It depends on the choice of the arbitrary reference point.

Changing this reference point means to use $$\vec{r}'=\vec{r}+\vec{r}_0$$ for all position vectors, wehre $\vec{r}_0=\text{const}$. Then you have
$$\vec{J}'=m_1 \vec{r}_1' \times \dot{\vec{r}}_1'+m_2 \vec{r}_2' \times \dot{\vec{r}}_2' = \vec{r}_0 \times (m_1 \dot{\vec{r}}_1 + m_2 \dot{\vec{r}}_2) + \vec{J}.$$
Thus $\vec{J}$ is a "bound vector". The same holds true for the total torque $\vec{\tau}=\dot{\vec{J}}$, while velocities, accelerations, and thus also forces are "free vectors", i.e., independent of the choice of the origin of the reference frame.

 

[fog37]

From Wikipedia:
Being an arrow, a Euclidean vector possesses a definite initial point and terminal point. A vector with fixed initial and terminal point is called a bound vector.[9] 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. Thus two arrows

and

in space represent the same free vector if they have the same magnitude and direction: that is, they are equipollent if the quadrilateral ABB′A′ is a parallelogram. If the Euclidean space is equipped with a choice of origin, then a free vector is equivalent to the bound vector of the same magnitude and direction whose initial point is the origin.

The term vector also has generalizations to higher dimensions and to more formal approaches with much wider applications.