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.

Explanation

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=0*048g~~0*05g`
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`.