Geometry for Feynman’s Lectures Exercises 2.24

Solving this exercise via force components is straightforward, leading to the answer $\Delta x \cos 30^\circ \sin 30^\circ$. However, I struggled with solving it using the principle of virtual work.

The hardest part of this exercise was calculating how the weight moves when we displace the cart. For simplicity, I reflected the diagram horizontally, aligning the movement with the standard x-axis.

One important point is that we cannot simply multiply the displacement by $\tan 30^\circ$, as this does not account for the fact that the weight moves along a circular path due to the rope.

I arrived at the following diagram, and the solution became clear from its geometry. In this diagram, the card moves to the right and $AB = \Delta x$.

We can approximate the path of the weight by a straight line, as the angle of the rope will be small. Thus, we obtain:

$$\angle ACB = 90^\circ \implies AC = AB \cdot \cos 30^\circ = \Delta x \cdot \cos 30^\circ$$

Finally, we arrive at the correct answer:

$$\angle AOC = 90^\circ \implies OC = AC \cdot \sin 30^\circ = \Delta x \cdot \cos 30^\circ \cdot \sin 30^\circ$$

I would be interested in solving this analytically by equating the equation for a circle with that of the plane to find the coordinates of the point. Ultimately, for the principle of virtual work, we only need to determine $\frac{\delta y}{\delta x}$.

If anyone stumbles upon this post and finds such a solution, please feel free to send a copy to $\text{simon.uvarov}@\text{icloud.com}$.