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Problema este anexata.

Problema Este Anexata class=

Răspuns :

Asupra corpului de masa m:

[tex] F_f=\mu mg\cos\alpha [/tex]

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

[tex] \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} [/tex]

[tex] F_f'=\mu'N=\mu'(Mg+mg\cos^2\alpha-\mu m g\cos\alpha\sin\alpha) [/tex]

[tex] \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 [/tex]

Cum [tex] \alpha =45^\circ\implies x=\sin\alpha=\cos\alpha\:\:\:(\text{Notam cu }x) [/tex]:

[tex] \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)} [/tex]

Numeric: [tex] \boxed{\mu'_{\min}\approx 0,16279} [/tex]