Q. 283.6( 15 Votes )

# Let m_{1} = 1 kg, m_{2} = 2 kg and m_{3} = 3 kg in figure (5-E12). Find the accelerations of m_{1}, m_{2} and m_{3}. The string from the upper pulley to m_{1} is 20 cm when the system is released from rest. How long will it take before m_{1} strikes the pulley?

Answer :

Given: Let m_{1} = 1 kg, m_{2} = 2 kg and m_{3} = 3 kg

Suppose the block m_{1} moves upward with acceleration a_{1} and the blocks m_{2} and m_{3} have relative acceleration a_{2} due to the difference of weight between them.

Actual acceleration of the blocks m_{1}, m_{2} and m_{3} will be: a1, (a1 − a2) and (a1 + a2)

From figure 2, T − g – a_{1} = 0 …(i)

From figure 3, -2g-2(a_{1}-a_{2})=0 …(ii)

From figure 4, -3g-3(a_{1}+a_{2})=0 …(iii)

From equations (i) and (ii), eliminating T, we get:

g + a_{2} = 4g + 4 (a_{1} + a_{2})

5a_{2} − 4a_{1} = 3g …(iv)

From equations (ii) and (iii), we get:

2g + 2(a_{1} – a_{2}) = 3g − 3 (a_{1} – a_{2})

5a_{1} + a_{2}= g …(v)

Solving equations (iv) and (v), we get: a_{1}=

a_{2}=g-5a_{1}

Then

And,

So, accelerations of m_{1}, m_{2} and m_{3} are up, down and down, respectively.

Now, u = 0, s = 20 cm = 0.2 m

∴

⇒ t=0.25 s

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