Evaluate the acceleration of the three weights, Mechanical Engineering

Evaluate the acceleration of the three weights:

A system of weight connected by the string passing over pulleys A and B is shown in figure given below. Find out acceleration of the three weights. Assume the weightless string and ideal condition for the pulleys.

Sol: As strings are weightless and ideal conditions prevail, thus the tensions in string passing over pulley A will be same. The tensions in string passing over pulley B will be same. But tensions in the strings passing over pulley A and over pulley B will be different as shown in the given figure.

Let T1  = Tension in string passing over pulley A
T2  = Tension in string passing over pulley B

One end of string passing over pulley A is connected to the weight 15N, and other end is connected to pulley B. As weight 15N is more than weights (6 + 4 = 10N), thus weight 15N will move downwards, while pulley B will move upwards. The acceleration of weight 15N and of pulley B will be same.

Let, a  = Acceleration of block 15N in the downward direction1  = Acceleration of 6N downward with respect to the pulley B.

Then acceleration of weight 4N with respect to the pulley B = a1  in upward direction.

1385_Evaluate the acceleration of the three weights.png

The absolute acceleration of weight 4N,

= Acceleration of 4N with respect pulley B + Acceleration of pulley B. = a1 + a (upward)

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

Absolute acceleration of weight 6N,

= Acceleration of 6 with respect to pulley B + Acceleration of pulley B.

= a1 - a (downward)

(As a1 is acting downward while a is acting upward. Thus total acceleration in downward direction)

Consider motion of weight 15N Net downward force = 15 - T1

Using F = ma,

15 - T1 = (15/9.81)a                                                                                                                         ...(1)

Consider 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 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 value of equation (4),

T1  - 8 = (8/9.81)(a1  + a)                                                                                                                ...(6)

Add 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),

a1 = 2.39m/sec2                                                                                                                         .....ANS

Putting value of a1 in equation (5),

a = 2.15m/sec2                                                                                                                            ....ANS

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

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

Accelerationofweight  4N = a = 4.54m/sec2                                                                      .......ANS

Posted Date: 10/20/2012 12:47:51 AM | Location : United States

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