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Solution:

Solution:

At the initial moment, obviously the tension in the thread connecting m₁ and m₂ equals the weight of m₂.

(a) For the block m₂ to come down or the block m₁ to go up, the conditions is

 \( {{m}_{2}}g-T\ge 0 \) and  \( T-{{m}_{1}}g\sin \alpha -fr\ge 0 \)

Where T is tension and fr is friction which in the limiting case equals  \( k{{m}_{1}}g\cos \alpha \) . Then

Or  \( {{m}_{2}}g-{{m}_{1}}\sin \alpha >k{{m}_{1}}g\cos \alpha \)  or  \( \frac{{{m}_{2}}}{{{m}_{1}}}>k\cos \alpha +\sin \alpha \) .

(b) Similarly in the case

 \( {{m}_{1}}g\sin \alpha -{{m}_{2}}g>f{{r}_{\lim }} \) or,  \( {{m}_{1}}g\sin \alpha -{{m}_{2}}g>k{{m}_{1}}g\cos \alpha \)  or,  \( \frac{{{m}_{2}}}{{{m}_{1}}}<\sin \alpha -k\cos \alpha \) .

(c) For this case, neither kind of motion is possible, and fr need not be limiting.

Hence,  \( k\cos \alpha +\sin \alpha >\frac{{{m}_{2}}}{{{m}_{1}}}>\sin \alpha -k\cos \alpha  \).