The inclined plane of figure forms an angle `alpha=30^@` with the horizontal. The mass ratio `m_2//m_1=eta=2//3`. The coefficients of friction between the body `m_1` and the inclined plane is equal to `k=0.10`. The masses of the pulley and the threads are negligible. Find the magnitude and the direction of acceleration of the body `m_2` when the formely stationary system of masses starts moving.

The inclined plane of figure forms an angle `alpha=30^@` with the hori

1.	The inclined plane of figure forms an angle `alpha=30^@` with the horizontal. The mass ratio `m_2//m_1=eta=2//3`. The coefficients of friction between the body `m_1` and the inclined plane is equal to `k=0.10`. The masses of the pulley and the threads are negligible. Find the magnitude and the direction of acceleration of the body `m_2` when the formely stationary system of masses starts moving.
Answer» Form the conditions, obtained in the previous problem, first we will check whether the mass `m_2` goes up or down. Here, `m_2//m_1=etagtsin alpha+kcos alpha`, (substituting the values). Hence the mass `m_2` will come down with an acceleration (say w). From the free body diagram of previous problem, `m_2-g-T=m_2w` (1) and `T-m_1gsin alpha-km_1gcosalpha=m_1w` (2) Adding (1) and (2), we get, `m_2g-m_1g sin alpha-km_1g cos alpha=(m_1+m_2)w` `w=((m_2//m_1-sin alpha-kcos alpha)g)/((1+m_2//m_1))=((eta-sin alpha-k cos alpha)g)/(1+eta)` Substituting all the values, `w=0048g~~005g` As `m_2` moves down with acceleration of magnitude `w=0*05ggt0`, thus in vector form acceleration of `m_2`: `vecw_2=((eta-sin alpha-k cos alpha)vecg)/(1+eta)=0.05vecg`.

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