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I was asked this:

A force $F = [3,\,-5,\,2]$ in Newtons pulls a sled through a displacement $s = [5,\,-1,\,3]$ in metres.

  • How much work is done on the sled by the force?
  • What is the minimum magnitude of force that could have been applied to the sled to obtain the same displacement? Explain your answer.

For the first one I have no problem, I figured out the work done is $26$ joules.

For the second I'm a bit unsure though. I get the feeling that the answer is the projection of $\vec F$ on $\vec s$. I'm pretty sure that is the answer, but I can't really explain why.

So I'm wondering if I'm right, and if I am, why that is? If I'm not, what is the right answer; and why is it right?

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  • $\begingroup$ Try to define your understanding of projection in mathematical terms, this will lead you to the answer. $\endgroup$ – ccorn Jan 21 '14 at 16:26
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    $\begingroup$ Well, if you draw a line perpendicular to $\vec s$ which connects to the head of $\vec F$, then the point that line intersects $\vec s$ is the location of the head of projection vector. So a projection is the smallest a vector can be while having its head align with a specific point on another vector. For example if a force vector at an angle of 50 degrees from a displacement vector has its head align with the middle point of the displacement vector, then $\vec F$'s projection is the smallest the force can be while still having its head align with the displacement vectors midpoint. $\endgroup$ – Threethumb Jan 21 '14 at 16:35
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    $\begingroup$ That's how I would define it.. but I'm still not entirely sure how to explain exactly why the projection has the minimum magnitude. Particularly because I don't see why the work done has to still be 26 joules for the displacement to be the same. $\endgroup$ – Threethumb Jan 21 '14 at 16:37
  • $\begingroup$ Note that the problem statement does not require you to find the direction of the force that has the minimum magintude. In fact, the question would not provide enough information for that. $\endgroup$ – ccorn Jan 21 '14 at 17:01
  • $\begingroup$ Hmm, that just confuses me more.. $\endgroup$ – Threethumb Jan 21 '14 at 17:10
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The problem statement seems to assume that the displacements are as good as infinitesimal; I will therefore omit some differential symbols and assume that the sled's motion and the ground are approximately uncurved in this problem. "Motion" here means displacement as a function of time.

The second question in the problem statement suggests that different forces $\vec{F}$ can have exactly the same effect on the motion of the sled. In other words, there are nonzero force changes corresponding to zero motion changes. This implies that the system is constrained.

The constraint that comes to mind for a typical sled on the ground is that it only follows the direction of its skids. Since we are assuming that the geometry here is as good as infinitesimal, this can be simplified to mean that the sled moves only along a fixed axis. Obviously, the axis determines the direction (up to sign) of $\vec{s}$.

It will be helpful to decompose $$\vec{F} = \vec{F}_s + \vec{F}_o$$ where $\vec{F}_s$ is collinear with $s$ and $\vec{F}_o$ is orthogonal to $\vec{s}$. Thus, $\vec{F}_s$ is your projection of $\vec{F}$ on $\vec{s}$. You can achieve this with the following matrix $\mathbf{S}$: $$\begin{align} \text{Let}\quad \mathbf{S} &= \frac{1}{\vec{s}\cdot\vec{s}}\,\vec{s}\circ\vec{s} = \frac{1}{35}\begin{pmatrix}5\\-1\\3\end{pmatrix} \begin{pmatrix}5&-1&3\end{pmatrix} \\\text{Then}\quad \vec{F}_s &= \mathbf{S}\cdot\vec{F} = \frac{\vec{F}\cdot\vec{s}}{\vec{s}\cdot\vec{s}}\,\vec{s} ,\quad\vec{F}_o = \vec{F}-\vec{F}_s = (\mathbf{I}-\mathbf{S})\cdot\vec{F} \end{align}$$ where $\cdot$ denotes the dot product, $\circ$ denotes the outer product, and $\mathbf{I}$ is the $3\times3$ identity matrix. Note that $\mathbf{S}$ is symmetric and positive semidefinite. Furthermore, $\mathbf{S}$ is idempotent: $\mathbf{S}\cdot\mathbf{S}=\mathbf{S}$. This allows you to verify that $\vec{F}_s\cdot\vec{F}_o=0$.

Back to the mechanics. The balance of forces requires that all force components orthogonal to the sled's acceleration sum to zero. Due to the constraint imposed, acceleration can happen only along the axis determined by $\vec{s}$, therefore $\vec{F}_o$ must be counteracted by a constraint force $$\vec{F}_c=-\vec{F}_o$$

The fact that $\vec{F}_o\cdot\vec{s}=0$ also means that $\vec{F}_o$ does not contribute to the work done. The same applies to $\vec{F}_c$ and is consistent with the general fact that constraint forces do not do work: $$W = \vec{F}\cdot\vec{s} = \vec{F}_s\cdot\vec{s} + \underbrace{\vec{F}_o\cdot\vec{s}}_0 = \vec{F}_s\cdot\vec{s}$$ In fact, the sled's motion can only be influenced by the sum $$\vec{F}+\vec{F}_c = \vec{F}_s+ \underbrace{\vec{F}_o+\vec{F}_c}_0 = \vec{F}_s$$ which again points to the conclusion that preserving motion requires preserving $\vec{F}_s$, whereas $\vec{F}_o$ can be varied arbitrarily without effect on the sled's motion.

Now, since $\vec{F}_s$ and $\vec{F}_o$ are orthogonal by design, we can apply the pythagorean theorem: $$\|\vec{F}\|^2 = \|\vec{F}_s\|^2+\|\vec{F}_o\|^2$$

And this means that minimization of $\|\vec{F}\|$ while preserving $\vec{F}_s$ means minimizing $\|\vec{F}_o\|$. This is achieved with $\vec{F}_o=\vec{0}$. Thus, the minimal required force is $$\vec{F} = \vec{F}_s = \frac{\vec{F}\cdot\vec{s}}{\vec{s}\cdot\vec{s}}\,\vec{s} = \frac{26\,\mathrm{Nm}}{35\,\mathrm{m}^2}\vec{s} \quad\text{with}\quad \|\vec{F}\| = \|\vec{F}_s\| = \frac{26}{\sqrt{35}}\mathrm{N}$$

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  • $\begingroup$ Oh wow, this is way beyond the scope of what I'm learning at the moment. I just someone changed my tags to say "classical mechanics", which is absolutely the wrong tag for this. I'm only learning about basic vectors, not mechanics. I'll change the tag back to "vectors". $\endgroup$ – Threethumb Jan 22 '14 at 21:44
  • $\begingroup$ Ignore the virtual work thing, then it's just about linear equality constraints. Are there specific pieces where you cannot follow? $\endgroup$ – ccorn Jan 22 '14 at 21:47
  • $\begingroup$ The mechanical considerations are there to prove what you intuitively suspected to be the solution. Along the way, I have noted how to express projection formally. TL;DR: Your guess is right, and it can be proven right because the sled is essentially constrained to move in one direction only. (Without that constraint, there would not be enough information to predict what would happen when $\vec{F}$ changes.) $\endgroup$ – ccorn Jan 22 '14 at 23:23
  • $\begingroup$ I see.. Well at the moment the pieces I can't follow would be most of it. Constraint forces and "outer product" are concepts completely alien to me. $\frac {26}{\sqrt(35)}$ is what I found too, though, so while my answer seems to be right, I'm still not entirely certain how I would simply explain why this is the right answer. If I'm getting you right, it as to do with the fact that the displacement has to stick to the same direction. $\endgroup$ – Threethumb Jan 23 '14 at 9:24
  • $\begingroup$ Hmm, I just thought of this: When the sled is supposed to move in the same direction and the same length, that ultimately means the work done has to be the same. Whenever the vector is at an angle, it has to have a large magnitude to make up for the angle, since the force isn't applied directly. So when at an angle of zero, the force requires the least magnitude to achieve the same displacement. And a vectors projection represents a zero angle vector that has the same amount of work done, thus also the least magnitude force vector. Would this be a valid explanation? $\endgroup$ – Threethumb Jan 23 '14 at 9:30

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