Question 12.60: A semicircular slot of 10-in. radius is cut in a flat plate ...

(a) We have \quad\theta=80^{\circ}

Then 

\begin{aligned}N & =(0.8 \mathrm{\ lb})\left[-\cos 80^{\circ}+\frac{1}{32.2 \mathrm{~ft} / \mathrm{s}^{2}} \times \frac{98}{3}\left(13-5 \sin 80^{\circ}\right) \mathrm{\ ft} / \mathrm{s}^{2} \times \sin 80^{\circ}\right] \\& =6.3159 \mathrm{\ lb} \\F & =(0.8 \mathrm{\ lb})\left[\sin 80^{\circ}+\frac{1}{32.2 \mathrm{~ft} / \mathrm{s}^{2}} \times \frac{98}{3}\left(13-5 \sin 80^{\circ}\right) \mathrm{\ ft} / \mathrm{s}^{2} \times \cos 80^{\circ}\right] \\& =1.92601 \mathrm{\ lb}\end{aligned}

Now \quad F_{\max }=\mu_{s} N=0.35(6.3159 \mathrm{\ lb})=2.2106 \mathrm{\ lb}

The block does not slide in the slot, and

\mathbf{F}=1.926 \mathrm{\ lb} \ ⦩ 80^{\circ}\blacktriangleleft

(b) We have \quad\theta=40^{\circ}

Then

\begin{aligned}N & =(0.8 \mathrm{\ lb})\left[-\cos 40^{\circ}+\frac{1}{32.2 \mathrm{\ ft} / \mathrm{s}^{2}} \times \frac{98}{3}\left(13-5 \sin 40^{\circ}\right) \mathrm{\ ft} / \mathrm{s}^{2} \times \sin 40^{\circ}\right] \\& =4.4924 \mathrm{\ lb} \\F & =(0.8 \mathrm{\ lb})\left[\sin 40^{\circ}+\frac{1}{32.2 \mathrm{\ ft} / \mathrm{s}^{2}} \times \frac{98}{3}\left(13-5 \sin 40^{\circ}\right) \mathrm{\ ft} / \mathrm{s}^{2} \times \cos 40^{\circ}\right] \\& =6.5984 \mathrm{\ lb}\end{aligned}

Now 

\begin{gathered}F_{\max }=\mu_{s} N, \text { from which it follows that } \\F>F_{\max }\end{gathered}

Block E will slide in the slot

and

\begin{aligned}\mathbf{a}_{E} & =\mathbf{a}_{n}+\mathbf{a}_{E / \text { plate }} \\& =\mathbf{a}_{n}+\left(\mathbf{a}_{E / \text { plate }}\right)_{t}+\left(\mathbf{a}_{E / \text { plate }}\right)_{n}\end{aligned}

At t=0, the block is at rest relative to the plate, thus \left(\mathbf{a}_{E / p \text { plate }}\right)_{n}=0 at t = 0, so that \mathbf{a}_{E / \text { plate }} must be directed tangentially to the slot.

\begin{aligned}& +\swarrow \Sigma F_{x}=m a_{x}: \quad N+W \cos 40^{\circ} =m \frac{\nu_{E}^{2}}{\rho} \sin 40^{\circ} \\\text{or } & \quad\quad\quad\quad N =W\left\lgroup-\cos 40^{\circ}+\frac{\nu_{E}^{2}}{g \rho} \sin 40^{\circ}\right\rgroup \text { (as above) } \\& \quad\quad\quad\quad\quad =4.4924 \mathrm{\ lb}\end{aligned}

Sliding:

\begin{aligned}F & =\mu_{k} N \\& =0.25(4.4924 \mathrm{\ lb}) \\& =1.123 \mathrm{\ lb}\end{aligned}

Noting that \mathbf{F} and \mathbf{a}_{E / \text { plane }} must be directed as shown (if their directions are reversed, then \Sigma \mathbf{F}_{x} is \searrowwhile m \mathbf{a}_{x} is \nwarrow), we have

the block slides downward in the slot and

\mathbf{F}=1.123\mathrm{\ lb}\ ⦩40^{\circ}\blacktriangleleft

Alternative solutions.

(a) Assume that the block is at rest with respect to the plate.

\Sigma \mathbf{F}=m \mathbf{a}: \quad \mathbf{W}+\mathbf{R}=m \mathbf{a}_{n}

Then 

\begin{aligned}\tan \left(\phi-10^{\circ}\right) & =\frac{W}{m a_{n}}=\frac{W}{\frac{W}{g} \frac{\nu_{E}^{2}}{\rho}}=\frac{g}{\rho\left(\dot{\phi}_{A B C D}\right)^{2}} \\& =\frac{32.2 \mathrm{\ ft} / \mathrm{s}^{2}}{\frac{98}{3}\left(13-5 \sin 80^{\circ}\right) \mathrm{\ ft} / \mathrm{s}^{2}} & \text{(from above)}\end{aligned}

\begin{aligned}& \quad\quad\phi =\phi_{k}, \quad \text { where } \quad \tan \phi_{k}=\mu_{k} \quad \mu_{k}=0.25 \\\text{or} & \quad\quad \phi_{k} =14.0362^{\circ}\end{aligned}