I'm trying to follow Feynman's lecture. Unfortunately I'm a bit stuck on a small piece, so if you could show me what I'm doing wrong then I'd greatly appreciate your help. I want to exactly know how Feynman came up with $W_{ac} = \int^c_a \bf{F} \cdot d \bf{s} = F s \cos \theta$ when this implies $\bf{F} = F \cos \theta$. I have not been able to convince myself that this is true using triangles and trigonometry.

The relevant section says,

Let us magnify one of the triangles, as shown in Fig. 13-4. Is the work done in going from $a$ to $b$ and $b$ to $c$ on a triangle the same as the work in going directly from $a$ to $c$? Suppose that the force is acting in a certain direction; let us take the triangle such that the side $bc$ is in this direction, just as example. We suppose that the triangle is so small that the force is essentially constant over the entire triangle. What is the work done in going from $t$ to $c$? It is $W_{ac} = \int^c_a \bf{F} \cdot d \bf{s} = F s \cos \theta$

I'm stuck getting the $F s \cos \theta$. Below you'll see the figure I'm imagining while trying to solve this. The triangle ABC is the original in Feynman's figure; the triangle AED is the triangle representing the forces, where the line segment AD is the "F" in the diagram above. I define line segment AE as b, line segment ED as a, and line segment AD as c.

The law of cosines says that $a^2 = b^2 + c^2 - 2bc \cos \theta $
If we let $a=b$ then,
$b^2 = b^2 + c^2 - 2bc \cos \theta$
iff $c^2 = 2 b c \cos \theta$
iff $c=2b \cos \theta$
which implies
$\frac{c}{2 \cos \theta} = b$

$c$ in this case is in the direction of the applied force but what we need to determine is the work by the actual force along $ae$ which should depend on $c$; yet my answer appears incorrect because according to Feynman the answer is $b=c \cos \theta$. Can you tell me where I went wrong?

Here is where my triangle is located:

  • $\begingroup$ What is the rationale for imposing $a=b$? $\endgroup$ – Jaime Nov 27 '12 at 23:51
  • $\begingroup$ Hi Jaime, I figured that it was arbitrary what $a$ actually is... so as long as $c$ is longer than $a$ or $b$, so to simplify the math I set them equal. $\endgroup$ – sciencenewbie Nov 28 '12 at 0:22

I think your problem is assuming that $a = b$. Your triangle looks isosceles, but it won't usually be. Using your terminology: $\cos \alpha = \overrightarrow{AE} / \overrightarrow{AD}$. If we let $\mathbf{F} = \overrightarrow{AD}$, then $|\overrightarrow{AE}| = F \cos \alpha$, and $|\overrightarrow{AE}|$ is the component of the force in the direction of movement.

The reason you can't just fix $a=b$ is that it makes $\alpha = \frac{\pi}{4}$ automatically, while in reality it can be anything, and it will depend on the relationship between the force and the direcion of movement.

  • $\begingroup$ ugh, I think I made myself stupid while getting caught up in an endless loop. Thanks. $\endgroup$ – sciencenewbie Nov 28 '12 at 3:15

Your first question deals with a comparison of the paths $a$ to $b$ to $c$ with the direct path from $a$ to $c$. The law of conservation of energy deals with this and you can apply it to prove that only the starting and end points matter, the path doesn't.

As for working out the the forces involved, is that part of the problem or just your way of showing that the paths involve the same energy.

One more hint: if there is no heat involved, frictional or otherwise, the energy in going between any two points $E$ is exactly the work. In other words $E = w$.

  • $\begingroup$ I do not have multiple questions, only one: how does Feynman find the force from his own diagram. I tried doing it in a way that makes sense to me, but it has failed no matter what I've tried. $\endgroup$ – sciencenewbie Nov 28 '12 at 1:04

I think Feynman is simply integrating the dot product of the constant force, $\mathbf{F}$ and $d\mathbf{s}$, rather than working this out from first principles per se.

$W_{ac} = \int^c_a \mathbf{F} \cdot d\mathbf{s} = Fs\cos\theta$

This can be derived from $\delta W = \mathbf{F}\cdot\mathbf{v}\delta t$ but I'm not sure that's what you're looking for.

The important point is that the force is constant and always in the same direction in his diagram, thus you can integrate the dot product.

  • $\begingroup$ $W_{ac}=\int_a^c F⋅ds=Fscos \theta$ implies that $F= Fcos \theta$, but I do not see how you get $F cos \theta$ from the diagram using trigonometry. $\endgroup$ – sciencenewbie Nov 28 '12 at 2:16

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