공간에서의 운동

We don't have enough information to gauge the size of the vectors, but we can figure out their direction. Gravity pulls straight down, so the vector $-mg\hat{\boldsymbol{\jmath}}$ points straight down. The normal force will be normal to the curve.

This equation stems from $\mathbf{F}=m\mathbf{a}$. In that equation, $\mathbf{a}$ is acceleration — the second derivative of position with respect to time. So, $\mathbf{v}$ is the derivative of position with respect to time.

We previously used $\mathbf{v}$ as the derivative of position with respect to the parameter we use to define our position — which was often called $t$, but was not the necessarily time. So this is a good point to keep straight.

Solution 1:

For large, negative values of $x$, the wire is closer and closer to a vertical line. If the bead were sliding down a vertical wire, it could do so without even touching the wire, so the force exerted on the bead would be zero. As $x$ approaches 0 from the left, the wire approximates a horizontal line. If the bead were sitting on a horizontal line, the wire would be pushing up to counter gravity. So, we imagine the magnitude of the force exerted by the wire might increase as $x$ increases. That is, $\frac{\mathrm{d}W}{\mathrm{d}x}>0$.

Solution 2:

The net force exerted on the bead is

$$\begin{aligned} F=m\mathbf{a}&= W\hat{\mathbf{N}} -mg\hat{\boldsymbol{\jmath}} \\ \text{We dot both sides with $\hat{\mathbf{N}}$.} \\ W\hat{\mathbf{N}} \cdot \hat{\mathbf{N}} - mg\hat{\boldsymbol{\jmath}}\cdot\hat{\mathbf{N}}&=m\mathbf{a}\cdot\hat{\mathbf{N}} \\ \text{Using the equation $\mathbf{a}(t) = \frac{\mathrm{d}^2s}{\mathrm{d}{t}^2}\hat{\mathbf{T}} + \kappa \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2\hat{\mathbf{N}}$,} \\ W-mg\hat{\boldsymbol{\jmath}}\cdot\hat{\mathbf{N}}&=m\kappa \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2\\ W&=mg\hat{\mathbf{n}}\cdot\hat{\mathbf{N}}+m\kappa \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2\\ &=mg\cos\theta+m\kappa \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2 \end{aligned}$$

where $\theta$ is the angle between $\hat{\boldsymbol{\jmath}}$ and $\hat{\mathbf{N}}$.

As $x$ moves from a highly negative number to zero, $\theta$ moves from nearly $\pi/2$ to nearly $0$. Therefore $\cos \theta$ increases from nearly zero to nearly one. Then $mg\cos\theta$ is increasing.

Furthermore, as $x$ increases, we see from the picture that the curvature $\kappa$ increases, and speed $\frac{\mathrm{d}s}{\mathrm{d}t}$ increases as well (kinetic energy is increasing as potential energy decreases).

So, $\frac{\mathrm{d}W}{\mathrm{d}x}>0$.