Question

Let m₁ = 1 kg, m₂ = 2 kg and m₃ = 3 kg in the following figure. 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

 The free-body diagram for mass m₁ is shown below:
a
The free-body diagram for mass m₂ is shown below:

The free-body diagram for mass m₃ is shown below:

Suppose the block m₁ moves upward with acceleration a₁ and the blocks m₂ and m₃have relative acceleration a₂ due to the difference of weight between them.
So, the actual acceleration of the blocks m₁, m₂ and m₃ will be a₁, (a₁ − a₂) and (a₁ + a₂), as shown.

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

From figure 3,\[\frac{T}{2} - 2g - 2\left( a_1 - a_2 \right) = 0 . . . \left( ii \right)\]

From figure 4,\[\frac{T}{2} - 3g - 3\left( a_1 + a_2 \right) = 0 . . . \left( iii \right)\]

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:

\[a_1 = \frac{2g}{29}\]

\[\text{ and }a_2 = g - 5 a_1 \]

\[ \Rightarrow a_2 = g - \frac{10g}{29} = \frac{19g}{29}\]

\[\text{So}, a_1 - a_2 = \frac{2g}{29} - \frac{19g}{29} = - \frac{17g}{29}\]

So, accelerations of m₁, m₂ and m₃ are \[\frac{19g}{29}\left( up \right),

\frac{17g}{29} \left(\text{ down }\right)\text{ and }\frac{21g}{29}\left(\text{ down }\right)\]

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

\[a_2 = \frac{19g}{29} \]

\[ \therefore s = ut + \frac{1}{2}a t^2 \]

\[ \Rightarrow 0 . 2 = \frac{1}{2} \times \frac{19}{29}g t^2 \]

\[ \Rightarrow t = 0 . 25 s\]