Question 12.48: A 250-g block fits inside a small cavity cut in arm OA, whic...

\text { +} \swarrow \Sigma F_{n}=m a_{n}: \quad P+m g \sin \theta-Q=m \frac{\nu^{2}}{\rho}

To have contact with the specified surface, we need Q ≥ 0,

or \quad \quad Q=P+mg\;\mathrm{sin}\,\theta-\frac{m\nu^{2}}{\rho}\gt 0

\sin\theta\gt \frac{1}{g}\left\lgroup\frac{\nu^{2}}{\rho}-\frac{P}{m}\right\rgroup    (1)

Data:\quad \quad m=0.250 \mathrm{~kg}, \quad \nu=3 \mathrm{~m} / \mathrm{s}, \quad P=1.5 \mathrm{~N}, \quad \rho=0.9 \mathrm{~m} 

Substituting into (1):

\begin{aligned}& \sin \theta>\frac{1}{9.81}\left[\frac{(3)^{2}}{0.9}-\frac{1.5}{0.25}\right] \\& \sin \theta>0.40775\end{aligned}