物理

求助提问贴

$\therefore F = 0 时, mg\sin\theta \gt \mu mg\cos\theta$

$现在分两种情况讨论，f向下和f向上$

$\blue{(1)} f向下$

$\begin{cases} mg\sin\theta = F_{min}\cos\theta + f_{静max} \\ F_N = mg\cos\theta + F_{min}\sin\theta \\ f_{静max} = \mu F_N \end{cases}$

$\Rightarrow F_{min} = \dfrac{\sin\theta - \mu\cos\theta}{\mu\sin\theta + \cos\theta} \cdot mg$

$\blue{(2)} f向上$

$\begin{cases} mg\sin\theta - F_{max}\cos\theta = - f'_{静max} \\ F'_N = mg\cos\theta + F_{max}\sin\theta \\ f'_{静max} = \mu F'_N\end{cases}$

$\Rightarrow F_{max} = \dfrac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta} \cdot mg$

$综上，F \in [\dfrac{\sin\theta - \mu\cos\theta}{\mu\sin\theta + \cos\theta}mg , \dfrac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}mg]$