# Vector is a function of its position or not?

• Ebby
In summary, the conversation discusses a force vector that was initially thought to be a function of x and y positions, but is actually a force with a constant magnitude and direction. It is also mentioned that the force can be split up into its x and y components using integration. The conversation then shifts to discussing the relevance and significance of the mass and trajectory of the particle in relation to the work done by the force. The conversation concludes with the reminder that the work done is equal to the force multiplied by the distance and cosine of the angle between the force and the displacement.
Ebby
Homework Statement
Is this vector a function of x and y?
Relevant Equations
W = fd

At first I thought that this force vector ## \vec F = 3 \hat x + 2 \hat y ## is a function of ## x ## and ## y ##, which is to say that its magnitude and direction vary with the x and y positions, but this is not so, right? It's just a force with a constant magnitude and direction.

And I can split it up using ## \int \vec F \cdot \, d \vec r = \int F_{x} \, dx + \int F_{y} \, dy ##.

Ebby said:
Homework Statement: Is this vector a function of x and y?
Relevant Equations: W = fd

View attachment 326770

At first I thought that this force vector ## \vec F = 3 \hat x + 2 \hat y ## is a function of ## x ## and ## y ##, which is to say that its magnitude and direction vary with the x and y positions, but this is not so, right? It's just a force with a constant magnitude and direction.

And I can split it up using ## \int \vec F \cdot \, d \vec r = \int F_{x} \, dx + \int F_{y} \, dy ##.
yes

Ebby said:
Homework Statement: Is this vector a function of x and y?
Relevant Equations: W = fd

View attachment 326770

At first I thought that this force vector ## \vec F = 3 \hat x + 2 \hat y ## is a function of ## x ## and ## y ##, which is to say that its magnitude and direction vary with the x and y positions, but this is not so, right? It's just a force with a constant magnitude and direction.

And I can split it up using ## \int \vec F \cdot \, d \vec r = \int F_{x} \, dx + \int F_{y} \, dy ##.
##\hat{x}## is a unit vector in the x direction. As such, it is a constant vector, with nothing at all due to the value of x. A similar comment holds for the y direction.

-Dan

The standard notation is that x "hat" and y "hat" are unit vectors (constants not variables) in the x and y directions respectively.

The vector values depend on the "units" and "direction" which are defined by the "unit vectors" and what is assumed but not stated are the units. (kg or N or ?)

Not only does the force vector have a direction, but it might also have a vector position on some object or point relative to some datum. In this question, the force position is over all xy space on a particle, but in an xy direction.

This problem begs the question why the work done does not depend on the mass or specific trajectory of the particle?

It also begs the question to what extent does random mathematical input relate to something physically meaningful?

MatinSAR
In this question we are told the particle must travel from 0 to x,y with a Vector Force applied to it.
The mass does not matter, as that only affects the time to reach a certain kinetic energy rise at destination. That kinetic energy is the same as the work done which is W=Fd, but the only useful force is in the vector direction of the final destination as we must assume for now the particle is not free to move in any other direction as we are told it must arrive at XY. If it does, then the lossless work done is "Fd" times the ratio of angle of the force, where cos 0 deg if along the same axis = 1 and = 0 if the force is at right angles.

Edit
After feedback, of course lossless conservation of energy in motion, we need only assume the particle arrives at some XY destination then the work done is ##W=F\cdot d \cdot cos \theta=\vec F.\vec d##.

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TonyStewart said:
In this question we are told the particle must travel from 0 to x,y with a Vector Force applied to it.
The mass does not matter, as that only affects the time to reach a certain kinetic energy rise at destination. That kinetic energy is the same as the work done which is W=Fd, but the only useful force is in the vector direction of the final destination as we must assume for now the particle is not free to move in any other direction as we are told it must arrive at XY. If it does, then the lossless work done is "Fd" times the ratio of angle of the force, where cos 0 deg if along the same axis = 1 and = 0 if the force is at right angles.
You can't and shouldn't assume that because the question setter did not specify something, it does not matter. It's a poor question, IMO, because it actively encourages the student simply to plug numbers into equations without any throught of either a) whether those equations are valid in this case; or, b) the physics of the scenario. Perhaps the particle is constrained and perhaps it is not. In fact, a better question would be to explain what is going on and what sort of physical experiment could lead to this data.

TonyStewart
TonyStewart said:
as we must assume for now the particle is not free to move in any other direction as we are told it must arrive at XY
There is no need to assume any such thing. The vector field across which we are integrating is conservative. A constant vector field is always conservative.

[Yes, this means that we are free to assume the direct path -- it will yield the same result as any more complicated path]

In more basic terms, the work done on a particle by a constant force in a fixed direction does not depend on the path the particle follows from point A to point B. It only depends on point A and point B (and the fixed force vector).

This is essentially the same thing as saying that the potential energy of a body at the top of a hill is well defined. No matter what frictionless path it follows to the bottom, it will arrive at the bottom with the same kinetic energy increment due to gravity.

I am not as disturbed as @PeroK by the lack of a physical justification. This seems to be a typical toy problem intended purely to exercise (or test) the student's skills at performing mathematical manipulations. Maybe we have a charged pith ball in the passenger seat of a toy car mounted on non-conductive rails and positioned between a pair of oppositely charged parallel plates.

Or maybe we have balls of a standard mass being rolled down a groove that is carved into the face of an inclined plane that has been cross-hatched with coordinates at an angle to the slope.

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nasu, TonyStewart and MatinSAR
PeroK said:
This problem begs the question why the work done does not depend on the mass or specific trajectory of the particle?
Why should it depend on mass? We want to calculate work done by force ##\vec F## on the object. And ##\vec F.\vec d## does not depend on mass.
We can prove that ##\vec F## is a conservative force and we know that work done by a conservative force doesn't depend on path.

MatinSAR said:
Why should it depend on mass? We want to calculate work done by force ##\vec F## on the object. And ##\vec F.\vec d## does not depend on mass.
We can prove that ##\vec F## is a conservative force and we know that work done by a conservative force doesn't depend on path.
That's the justification. That the force is conservative. That needs to be said.

MatinSAR
jbriggs444 said:
I am not as disturbed as @PeroK by the lack of a physical justification. This seems to be a typical toy problem intended purely to exercise (or test) the student's skills at performing mathematical manipulations.
A nice problem might be additionally to give the measured speed of the particle at the two points and ask for the particle's mass. That would get the student thinking more about the physics and not just the maths. It's also plausible that you might know the charge of a particle and know the electric field (hence the force) and be able to measure the particle's speed. That gives you a way to calculate the particle's mass. At least that bears a passing resemblance to real physics!

jbriggs444
PeroK said:
That's the justification. That the force is conservative. That needs to be said.
How would the work depend on mass for a given, non-conservative force?

The force required to overcome friction or air resistance typically increases with mass. Therefore, if the mass of an object increases, more force is needed to overcome these non-conservative forces, resulting in more work being done.

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But the force is given. You know the force and you want to calculate the work. The question is not how a force can depend on mass but how the work of a given, known force may depend on mass. In the OP the force is completely determined at any point in space. You want the work of this force, not of another one that will occur in different conditions.

MatinSAR
nasu said:
How would the work depend on mass for a given, non-conservative force?
A simple example is where the force is time-dependent.

Ok, this makes sense. I did not push the non-conservative part far enough. :)
Just trying to imagine a constant non-conservative force whose work depends on mass. For a time dependent force the work will also depend upon when the work was done.

## 1. Is a vector a function of its position?

Yes, a vector is a mathematical object that is defined by both magnitude and direction, and its position or location is an essential component in determining its value.

## 2. Can a vector exist without a position?

No, a vector needs to have a position in order to have a magnitude and direction. Without a position, there is no way to determine the value of the vector.

## 3. How is the position of a vector represented?

The position of a vector is typically represented by its initial and terminal points, or by a coordinate system such as Cartesian coordinates.

## 4. Is the position of a vector constant?

No, the position of a vector can change depending on its movement or displacement. For example, a vector representing the position of a moving object will have a different position at different points in time.

## 5. Can a vector have a negative position?

Yes, a vector can have a negative position if it is measured relative to a reference point or origin. In a coordinate system, a vector with a negative position would be located in the opposite direction of the positive position.

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