Question

Let m1 = 1 kg, m2 = 2 kg and m3 = 3 kg in figure (5−E12). Find the accelerations of m1, m2 and m3. The string from the upper pulley to m1 is 20 cm when the system is released from rest. How long will it take before m1 strikes the pulley?

Answer 1

Given :

Let m1 = 1 kg

m2 = 2 kg and

m3 = 3 kg

Suppose the block m1 moves upward with acceleration a1 and the blocks m2 and m3 have relative acceleration a2 due to the difference of weight between them.

Actual acceleration of the blocks m1, m2 and m3 will be

a1, (a1 − a2) and (a1 + a2)

From figure 2, T − 1g − 1a1 = 0    …(i)

From figure 3,

T/2-2g-2(a1-a2)=0    …ii

From figure 4,

T/2-3g-3(a1+a2)=0    …iii

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

1g + 1a2 = 4g + 4 (a1 + a2)

5a2 − 4a1 = 3g    …(iv)

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

2g + 2(a1 − a2) = 3g − 3 (a1 − a2)

5a1 + a2 = g    …(v)

Solving equations (iv) and (v), we get:

a1=2g/29

a2=g-5a1

a2=g-10g/29

=19g/29

Then  a1-a2=2g/29-19g/29

= -17g/ 29

and a1+a2=2g/29+19g/29

=21g/29

So, accelerations of m1, m2 and m3 are

19g/29up, 17g/29 down and 21g/29down, respectively.

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

a2=19g/29        

 ∴ s=ut+1/2at

⇒0.2=1/2×19/29gt²

⇒t=0.25 s

Answer 2

Answer:

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