Determining Whether Solution is Vector vs Scalar?

[Ascendant0]

Ok, so currently, I'm working on problems involving work done by a general variable force. I had a question as far as solving these that ties into other problems as well...

I see in some of the worked problems in this textbook, when they take the integral of a force equation to determine the amount of total work done, it converts it from a vector equation to a scalar equation. Based on what I had learned previously, my first thoughts prior to seeing how they worked it out was to solve for the x direction, then solve for the y direction, then do what I'd normally do to determine the magnitude of something, which would be the sqrt(x^2 + y^2) you would use in problems prior to this to determine overall distance, overall velocity, etc.

Is it because you integrate, and that's what removes the vectors? Is there some way in future problem-solving (with things other than force or work) to see a tell-tale sign that something is to be converted from a vector, to a scalar in the end? I want to make sure I know the difference in future problems as to when I'm converting a vector to a scalar, and when it remains a scalar?

 

[phinds]

Not sure what you mean "removes the vectors". Scalar things like work and speed are scalar and vector things like force and velocity are vectors. How you get from one to the other is not the issue.

It should be easy to understand what is a vector and what is a scalar. Is that what you have a problem with?

 

[fresh_42]

Work done is the product of two vectors: force and path. The result of that multiplication is a scalar, the work. If we have a linear path and a uniform force $\vec{F}$ along that path $\vec{s}$ then we get
$$
W=\vec{F} \cdot \vec{s} = |\vec{F}| \cdot |\vec{s}|\cdot \cos \sphericalangle\left(\vec{F} , \vec{s}\right)
$$
without any integration. The integral formula
$$
W=\int_P^Q \vec{F}\left(\vec{s}\right) \cdot\,d\vec{s}
$$
of a path from point $P$ to point $Q$ comes into play if the force varies along the path or the path isn't linear anymore. In this case, we parameterize the path $\vec{s}\, : \,[0,1] \longrightarrow \mathbb{R}^3$ as if we would walk along and solve the integral by
$$
W=\int_0^1 \vec{F}\left(\vec{s}(t)\right) \cdot \dfrac{d\vec{s}(t)}{dt}\,dt\,.
$$

 

[Ascendant0]

Not sure what you mean "removes the vectors". Scalar things like work and speed are scalar and vector things like force and velocity are vectors. How you get from one to the other is not the issue.

It should be easy to understand what is a vector and what is a scalar. Is that what you have a problem with?

I probably didn't explain it well. Apologies for the poorly-edited image above, but that is the problem I am referring to. It starts out with an i and j direction for the forces, but then after integrating, everything just gets merged together as a single scalar value.

On this problem, initially before seeing their answer, I took the answer I got from the i direction (19), and the answer I got from the j direction (-12), and thought to find the magnitude of the force, take:

sqrt (19^2 + (-12)^2) = 22.5J

Hopefully, that makes more sense of my thought process and where I'm a bit confused as to why i and j disappear, and they are added together regardless of direction.

 

[pasmith]

Work done is a line integral; the result is a scalar. In general it is $$W = \int_0^S \mathbf{F}(\mathbf{x}(s)) \cdot \mathbf{x}'(s)\,ds$$ where $s$ is a parameter along the path (which need not be equal to time $t$). Here the force is conservative so the result is independent of the path taken between the initial and final positions, and the text has implicitly taken $$\begin{split} \mathbf{x}(s) &= \begin{cases} (2 + 2s)\mathbf{i} + 3\mathbf{j} & 0 \leq s \leq \frac 12 \\ 3\mathbf{i} + (6 - 6s)\mathbf{j} & \frac12 < s \leq 1\end{cases} \\ \mathbf{x}'(s) &= \begin{cases} 2\mathbf{i} & 0 \leq s \leq \frac12 \\ -6\mathbf{j} & \frac12 < s \leq 1. \end{cases} \end{split}$$ Thus only the $\mathbf{i}$ component appears in the integral over $0 \leq s \leq \frac12$ (and we can make the substitution $x = 2(1 + s)$) and only the $\mathbf{j}$ component appears in the integral over $\frac12 \leq s \leq 1$ (and we can make the substitution $y = 6(1 - s)$), resulting in the calculation given in the text.

 

[Lnewqban]

On this problem, initially before seeing their answer, I took the answer I got from the i direction (19), and the answer I got from the j direction (-12), and thought to find the magnitude of the force, take:

sqrt (19^2 + (-12)^2) = 22.5J

As described in the problem, we start with two orthogonal components (12 N and 4 N) of a force vector, which has magnitude 12.65 N and direction 18° from horizontal.
That is the actual initial force acting on the particle.

At the time when the trajectory ends, we have two orthogonal components (27 N and 4 N) of a new force vector, 27.3 N and which has magnitude 12.65 N and direction 8° from horizontal.
That is the actual final force acting on the particle.

That force, that is constantly increasing in magnitude and rotating clockwise, is the one doing the work of increasing the kinetic energy of the particle.
Within that range, any individual force vector that we choose will give us an incorrect result.

Note how the second member of the equation is negative because the 4 N j component of the forces points in opposite direction to the displacement of the particle.
The particle is forced to move in that direction by the component (in the direction of the trajectory) of the increasing and rotating actual force.

 

[Ascendant0]

Here the force is conservative so the result is independent of the path taken between the initial and final positions

That right there is what I was missing. Now I'm seeing it how I should, thank you!