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I just don't understand how to get to the normal force from the P.E. given; specially, when its circular motion. I'm stuck on this one question. Thanks in advance for helping me out!

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  • $\begingroup$ After solving for external force on the bead, you'll realize that there isn't any circular motion. The bead is stationary. $\endgroup$
    – dnaik
    Oct 29, 2020 at 2:22

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The following may be useful. In particular, the force can be obtained from the potential using:

$$ \mathbf{F}=-\nabla U = -\frac{\partial}{\partial x}U\hat{x} - \frac{\partial}{\partial y}U\hat{y} = -2x \hat{x} - 4y \hat{y}. $$

Since the bead is located at $x=2, y=0$, this force only has an $x$-component (in the negative $x$ direction). Now, there is also the force due to gravity which acts vertically. If you take both of these forces into account, I believe you should be able to determine the normal force of the wire frame acting on the bead, since the bead is not moving.

I hope this helps.

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