In the following shown a; = 6m/s²
(downwards) and a: = 4m/s2 The acceleration of 1 is (upwards).
Answers
Note: Look at the ATTACHMENT given below to find the figure.
Given data:
Acceleration shown by block 3 is: a₃ = 6m/s²
Acceleration shown by block 3 is: a₂ = 4m/s²
To find:
Acceleration shown by block 1 is: a₁ = ? m/s²
Solution:
Here, the observer is watching the moveable system. The relative displacements can be seen.
So, at pulley P₂, block 2 moves with an acceleration of 'b' upwards and so block 3 in downwards.
Coming to the pulley P₁, block 1 moves with an acceleration of 'a' upward, and on the other side, at the pulley 2 moves with an acceleration of 'a' downwards.
Now, by applying the given data upon the blocks, we get:
Acceleration at block 3: a + b = 6m/s² -------> 1 (Same directions)
Acceleration at block 2: a - b = 4m/s² ------> 2 (Opposite directions)
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By solving eqns 1 and 2: 2a = 10m/s²
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a = 10/2 = 5m/s²
Calculating the acceleration of Block 1:
The acceleration at block 1 is 'b', as 'a' acceleration is balanced on both the sides from the values a and a+b on either side.
To find the acceleration 'b', substitute 'a' value in any of the equations 1 and 2.
a+b = 6m/s²
5m/s² + b = 6m/s²
b = 6m/s² - 5m/s²
b = 1m/s² upwards.
Therefore, the acceleration of block 1 is 1m/s² upwards.
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