Question

In the arrangement show, find the acceleration of each block, tension in the string and reaction in the pully. The string and pully are light (massless)(m1>m2).neglect the friction in pully

Answer 1

Answer:

Since m2>m1, the block of mass m1 will move downward and the block of mass m2, upward.

↓:m1g−T=m1a (i)

↑:T−m2g=m2a (ii)

Solving (i) and (ii), we get

a=(m1−m2)g(m1+m2)

T=2m1m2gm1+m2

R−2T=0 [since the pulley is massless]

where R is the reaction in the pulley.

Answer 2

the answer is R-2T=0

Explanation:

since the pulley is massless