Question

 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.

Answer 1

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) 

Adding equation (2) and (3) 

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/sec² 

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

Acceleration of weight 6N = a = 0.24m/sec² 

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