Two blocks of masses `m_(1)` and `m_(2)` are attached at the ends of an inextensible string which passes over a smooth massless pulley. If `m_(1)gtm_(2)`, find :
(i) the acceleration of each block
(ii) the tension in the string.

Explanation

The block `m_(1)` is assumed to be moving downward and the block `m_(2)` is assumed to be moving upward, It is merely an assumption and it does not imply the real direction. If the values of `a_(1)` and `a_(2)` come out to be positive then body the assumed directions are correct, oherwise the body moves in the opposite direction. Since the pulley is smoth and massless, therefore, the tension on each side of the pulley is same. The free body diagram of each block is shown in the figure.

Fig. 1.1
Applying Newton's second Law on blocks `m_(1)` and `m_(2)`
Block `m_(1)" "m_(1)g-T=m_(1)a . . . . .. . . . (1)`
Block `m_(2)" "-m_(2)g+T=m_(2)a_(2) . . . . . . . . . . . (2)`
Number of unknowns : T, `a_(1)` and `a_(2)` (three)
Number of equations : only two
Obviously, we require one more equation to solve the problem Note that whenever one finds the number of equations less than the number of unknowns, one must think about the constraint relation. Now we are going to explain the mathematical procedure for this.
How to determine Constraint Relation ?
(1) Assume the direction of acceleration of each block e.g. `a_(1)` (downward) and `a_(2)` (upward) in this case.
(2) Locate the position of each block from a fixed point (depending on convenience), e.g. centre of the pulley in this case.
(3) Identify the constraint and write down the equation of constraint in terms of the distance assumed. For example, in the chosen problem, the length of string remains constant is the constraint or restriction.
Thus, `x_(1)+x_(2)=` constant
we get `(dx_(1))/(dt)+(dx_(2))/(dt)=0`
Each term on the left side represents the velocity of the blocks. Since we have to find a relation between acceletations, therefore we differntiate it once again w.r.t. time.

Thus `(d^(2)x_(1))/(dt^(2))+(d^(2)x_(2))/(dt^(2))=0`
Since, the block `m_(1)` is assumed to be moving downward (`x_(1)` is increasing with timw)
`:.(d^(2)x_(1))/(dt^(2))=+a_(1)`
and block `m_(2)` is assumed to be moving upward (`x_(2)` is decreasing with time)
`:.(d^(2)x_(2))/(dt^(2))=-a_(2)`
Thus `a_(1)-a_(2)=0`
or `a_(1)=a_(2)=a` (say) is the required constraint relation. Substituting `a_(1)=a_(2)=a` in equations (1) and (2) and solving them, we get
`(i)a=[(m_(1)-m_(2))/(m_(1)+m_(2))]g` (ii) `T=[(2m_(1)m_(2))/(m_(1)+m_(2))]g`