If the 10-lb block passes point A on the smooth track with a speed of vA = 5 ft/s, determi...

Consider the relation for the radius of curvature and calculate the radius of curvature of the point \(B\).

$$ \rho_{B}=\frac{\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}} \mid \ldots $$

Consider the equation of curve and differentiate with respect to \(x\) to obtain the following relation:

$$ \begin{aligned} &y=\frac{1}{32} x^{2} \\ &y=\frac{1}{32} x^{2} \\ &d y=\frac{1}{32} \times 2 x \times d x \ldots \ldots \text { (2) } \\ &\frac{d y}{d x}=\frac{1}{16} x \end{aligned} $$

Differentiate the equation (2) with respect to \(x\) to obtain the following relation:

$$ \frac{d^{2} y}{d x^{2}}=\frac{1}{16} \ldots \ldots $$

Substitute the values of equation (2) and (3) in the equation (1).

$$ \begin{aligned} \rho_{B} &=\frac{\left(1+\left(\frac{1}{16} x\right)^{2}\right)^{\frac{3}{2}}}{\left(\frac{1}{16}\right)} \\ &=\frac{\left(1+\left(\frac{1}{16}(0)\right)^{2}\right)^{\frac{3}{2}}}{\left(\frac{1}{16}\right)} \\ &=16 \mathrm{ft} \end{aligned} $$

Calculate normal acceleration.

$$ a_{n}=\frac{v_{B}^{2}}{\rho_{B}} $$

Substitute 16 ft for \(\rho_{B}\) and \(23.24 \mathrm{ft} / \mathrm{s}\) for \(v_{B}\).

$$ \begin{aligned} a_{n} &=\frac{23.24^{2}}{16} \\ &=33.75 \mathrm{ft} / \mathrm{s}^{2} \end{aligned} $$

Calculate the slope at position \(B\).

$$ \tan \theta=\left.\left(\frac{d y}{d x}\right)\right|_{x=0} $$

Substitute equation (2) for \(\left(\frac{d y}{d x}\right)\).

$$ \begin{aligned} &\tan \theta=\left.\left(\frac{1}{16} x\right)\right|_{x=0} \\ &\tan \theta=\left(\frac{1}{16} \times 0\right) \\ &\theta=\tan ^{-1}\left(\frac{1}{16} \times 0\right) \\ &\theta=0^{\circ} \end{aligned} $$

Calculate the normal reaction at the point \(B\) by applying newton's law of equation as follows:

$$ \begin{aligned} &\sum F_{n}=m a_{n} \\ &N-W \cos \theta=m a_{n} \\ &N-W \cos \theta=\frac{W}{g} a_{n} \end{aligned} $$

Substitute \(0^{\circ}\) for \(\theta, 10 \mathrm{lb}\) for \(W, 32.2 \mathrm{ft} / \mathrm{s}^{2}\) for \(g, 33.75 \mathrm{ft} / \mathrm{s}^{2}\) for \(a_{n}\).

$$ \begin{aligned} &N-10 \cos 0^{\circ}=\frac{10}{32.2} \times 33.75 \\ &N=20.48 \mathrm{lb} \end{aligned} $$

Therefore, the normal reaction at point \(B, N_{B}\) is \(20.48 \mathrm{lb}\).