# A system of weight connected by string passing over pulleys A and B is shown in fig.

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A system of weight connected by string passing over pulleys A and B is shown in fig. Find the acceleration of the three weights. Assume weightless string and ideal condition for pulleys. +1 vote
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selected by 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 T1 = Tension in the string passing over pulley A

T2 = 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 = a1 in the upward direction.

Absolute acceleration of weight 4N,

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

= a1 + 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. = a1 – a (downward)

(As a1 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 – T1

Using F = ma,

15 – T1 = (15/9.81)a ...(1)

Consider the motion of weight 4N

Net downward force = T2 - 4

Using F = ma, T2 – 4 = (4/9.81)(a + a1) ...(2)

Consider the motion of weight 6N Net downward force = 6 - T2

Using F = ma, 6 – T2 = (6/9.81)(a1 – a) ...(3)

Consider the motion of pulley B, T1 = 2T2 ...(4)

2 = (4/9.81)(a + a1) + (6/9.81)(a1 – a)

9.81 = 5a1 – a ...(5) Multiply equation (2) by 2 and put the value of equation (4), we get

T1 – 8 = (8/9.81)(a1 + a) ...(6)

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

15 – 8 = (15/9.81)a + (8/9.81)(a1 + a)

23a + 8a1 = 7 X 9.81 ...(7)

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

a1 = 2.39m/sec2

Putting the value of a1 in equation (5), we get

a = 2.15m/sec

Acceleration of weight 15N = a = 2.15m/sec2

Acceleration of weight 6N = a = 0.24m/sec

Acceleration of weight 4N = a = 4.54m/sec2