As the strings are weightless and ideal conditions prevail, hence the tensions in the string passing over pulley A will be same. The tensions in the string passing over pulley B will also be same. But the tensions in the strings passing over pulley A and over pulley B will be different as shown in fig.

Let T_{1} = Tension in the string passing over pulley A

T_{2 }= Tension in the string passing over pulley B One end of the string passing over pulley A is connected to a weight 15N, and the other end is connected to pulley B. As the weight 15N is more than the weights (6 + 4 = 10N), hence weight 15N will move downwards, whereas pulley B will move upwards. The acceleration of the weight 15N and of the pulley B will be same.

Let, a = Acceleration of block 15N in downward directiona 1 = Acceleration of 6N downward with respect to pulley B.

Then acceleration of weight of 4N with respect to pulley B = a_{1} in the upward direction.

Absolute acceleration of weight 4N,

= Acceleration of 4N w.r.t. pulley B + Acceleration of pulley B.

= a_{1} + a (upward)

(as both acceleration are in upward direction, total acceleration will be sum of the two accelerations) Absolute acceleration of weight 6N,

= Acceleration of 6 w.r.t. pulley B + Acceleration of pulley

B. = a_{1} – a (downward)

(As a_{1} is acting downward whereas a is acting upward. Hence total acceleration in the downward direction)

Consider the motion of weight 15N

Net downward force = 15 – T_{1}

Using F = ma,

15 – T_{1} = (15/9.81)a ...(1)

Consider the motion of weight 4N

Net downward force = T_{2} - 4

Using F = ma, T_{2} – 4 = (4/9.81)(a + a_{1}) ...(2)

Consider the motion of weight 6N Net downward force = 6 - T_{2}

Using F = ma, 6 – T_{2} = (6/9.81)(a_{1 }– a) ...(3)

Consider the motion of pulley B, T_{1} = 2T_{2} ...(4)

Adding equation (2) and (3)

2 = (4/9.81)(a + a_{1}) + (6/9.81)(a_{1 }– a)

9.81 = 5a_{1} – a ...(5) Multiply equation (2) by 2 and put the value of equation (4), we get

T_{1 }– 8 = (8/9.81)(a_{1} + a) ...(6)

Adding equation (1) and (6), we get

15 – 8 = (15/9.81)a + (8/9.81)(a_{1} + a)

23a + 8a_{1} = 7 X 9.81 ...(7)

Multiply equation (5) by 23 and add with equation (7), we get

a_{1} = 2.39m/sec^{2}

Putting the value of a_{1} in equation (5), we get

a = 2.15m/sec^{2 }

Acceleration of weight 15N = a = 2.15m/sec^{2}

Acceleration of weight 6N = a = 0.24m/sec^{2 }

Acceleration of weight 4N = a = 4.54m/sec^{2}