Consider the situation shown in figure (a). Both the pulleys and the strings are light and all the surface are frictionless. Calculate.
tension in the siring PQ.

Explanation

As pulley Q is not fixed, so if.it moves a distance d, the length of siring between P and Q will change be 2d(d from the above and d from below), i.e., M will move 2d. This in turn implies that if `a(to 2d)` is the acceleration of M, the acceleration of Q and of 2 M will be `(a/2)(tod)`
· Now.if we consider the motion of mass M, it is accelerated downward, so
T = M (g- a) --------- (i)
and for the motion of Q,
`2T-T.=0xx(a//2)=0,i.e., T.=2T.....(ii)`
and for the motion of mass 2M,
T. = 2M( a/ 2), i.e., T. =Ma-------- (iii)
From Eqns. (ii) and (iii) as`T=(1/2)`Ma, so Eqn (i) reduces to
`(1/2)Ma=M(g-a)` or `a=(2/3)g`
So, the acceleration of mass M is `(2/3)`g while tension in tl)e siring PQ .from-Eqn. (i) will be,
`T=M[g-(2/3g)]=(1/3)Mg`