Question

Problema este anexata.

Answer 1

Asupra corpului de masa m:

$F_f=\mu mg\cos\alpha$

Asupra corpului de masa M actioneaza: $G_M, F_{f \text{ reactiune}}=F_f, G_m_n$. Avem pe cele doua axe de coordonate relatiile vectoriale:

$\vec{F}_f_y+\vec{N}+\vec{G}_M+\vec{G}_n_m_y=0\\\\\implies \boxed{N=Mg+mg\cos^2\alpha-\mu m g\cos\alpha\sin\alpha}$

$F_f'=\mu'N=\mu'(Mg+mg\cos^2\alpha-\mu m g\cos\alpha\sin\alpha)$

$\vec{F'}_f+\vec{G}_m_n_x+\vec{F}_f_x=0\\\\\implies F_f'=\mu mg\cos^2\alpha +mg\cos\alpha\sin\alpha$

Cum $\alpha =45^\circ\implies x=\sin\alpha=\cos\alpha\:\:\:(\text{Notam cu }x)$:

$\mu'(Mg+mgx^2-\mu m gx^2)=\mu mgx^2 +mgx^2\\\\\implies \mu'=\dfrac{mx^2+\mu mx^2}{M+mx^2-\mu mx^2}=\dfrac{mx^2(1+\mu)}{M+mx^2(1-\mu)}$

Numeric: $\boxed{\mu'_{\min}\approx 0,16279}$