In the pulley system shown in figure. P and Q are fixed pulleys while A, B and C are movable pulleys each of mass 1kg. The strings are vertical and inextensible. Find the tension in the string and acceleration of frictionless pulleys A, B and C

Explanation

A single string whose ends are tied to centres of A and B, passes over all the pulleys, so tension at each point of string is same equal to T. Weight of each pulley A, B and C is
mg = 1 g Newton.
Let `y_A ,y_B and y_C` be the distances of centres of pulleys A, B and C from fixed pulleys at any time t. Following the string starting from end A and reaching upto end B, we have
`(y_B - y_A) + y_B + 2y_A + y_C + y_C- y_B = L`= constant
i.e.,` y_A + y_B + 2y_C = L = ` contant
DIfferentitating twice with respect to t , we get
` (d^(2)y_A)/(dt^(2)) + (d^(2)y_B)/(dt^(2)) + 2(d^(2)y_C)/(dt^(2)) =0`
i.e., `a_A + a_B + 2a_C =0` ...... (1)
where ` a_A , a_B and a_C` are acceleration of pulley A,B,C respectively .
Now equations of motion of pulleys A,B and C are
`mg + T-2T = ma_A implies mg-T =ma_A`............ (2)
`mg + T-2T = ma_B implies mg -T = ma_B`......... (3) and
`mg-2T = ma_C`
From (2) and (3) it is obvious that
and ` a_A = a_B = (g-(T)/(M)) `............ (5) and from (4) , `a_C = g-(2T)/(m)`............. (6)
susbtituting `a_A , a_B and a_C` in (1) , we get
`(g-(T)/(m)) + (g-(T)/(m)) + 2(g-(2T)/(m)) =0`
` 4g- (6T)/(m) =0 implies T = (2)/(3) mg = (2)/(3) xx 1 xx 9.8 = 6.5 N`
`:. a_A =a_B = (g-(T)/(m)) = 9.8 -(6.5)/(1) = 3.3 m//s^(2)`
From (1) , `ac=-a_A = -3.3 m//s^(2)`