Question

O sanie de greuate G coboara fara viteza initiala pe o panta AB de lungime x inclinata cu unghul alpha, dupa care isi continua drumul pe orizntala pana in C unde se opreste. Cunoscand lungimea x a pantei AB si distanta de oprire BC, sa se determine coeficientul de frecare la alunecare dintre sanie si zapada.

Answer 1

[tex]\displaystyle Se~da:\\ \\ x,\alpha,s\\ \\ \mu=?\\ \\ \\ Formule:\\ \\ Pe~panta:\\ \\ x=\frac{\Delta v^2}{2\times a_1}\\ \\ x=\frac{v_1^2-v_0^2}{2\times a_1},~unde~v_0=0\\ \\ x=\frac{v_1^2}{2\times a_1}\\ \\ v_1^2=2\times x\times a_1\\ \\ \\[/tex]


[tex]\displaystyle Pe~O_y:\\ \\ N-G_y=0\\ \\ N=G\times \cos\alpha\\ \\ N=m\times g\times\cos\alpha \\ \\ \\ Pe~O_x:\\ \\ G_x-F_{fr}=m\times a_1\\ \\ G\times\sin\alpha-\mu\times N=m\times a_1\\ \\ m\times g\times\sin\alpha-\mu\times m\times g\times \cos\alpha=m\times a_1\\ \\ a_1=g\times(\sin\alpha-\mu\times\cos\alpha)\\ \\ \\[/tex]


[tex]\displaystyle v_1^2=2\times x\times g\times(\sin\alpha-\mu\times\cos\alpha)\\ \\ \\ Pe~orizontala:\\ \\ \\ s=\frac{\Delta v^2}{2\times a_2}\\ \\ s=\frac{v_2^2-v_1^2}{2\times a_2},~unde~v_2=0\\ \\ s=\frac{-v_1^2}{2\times a_2}\\ \\ v_1^2=-2\times s\times a_2\\ \\ \\[/tex]


[tex]\displaystyle Pe~O_y:\\ \\ N-G=0\\ \\ N=m\times g\\ \\ \\ Pe~O_x:\\ \\ -F_{fr}=m\times a_2\\ \\ -\mu\times N=m\times a_2\\ \\ -\mu\times m\times g=m\times a_2\\ \\ a_2=-\mu\times g\\ \\ \\ v_1^2=2\times s\times (-\mu\times g)\\ \\ \\[/tex]


[tex]\displaystyle Egalam:\\ \\ 2\times x\times g\times(\sin\alpha-\mu\times\cos\alpha)=2\times s\times \mu\times g\\ \\ x\times(\sin\alpha-\mu\times\cos\alpha)=s\times\mu\\ \\ x\times \sin\alpha-x\times\mu\times\cos\alpha=s\times\mu\\ \\ s\times\mu+x\times\mu\times\cos\alpha=x\times\sin\alpha\\ \\ \mu\times(s+x\times\cos\alpha)=x\times\sin\alpha\\ \\ \\ \\ \\ \mu=\frac{x\times\sin\alpha}{(s+x\times\cos\alpha)}[/tex]