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?

Figure

Answer 1

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

(Figure 1)
The free-body diagram for mass m₂ is shown below:


(Figure 2)
The free-body diagram for mass m₃ is shown below:


(Figure 3)

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.

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

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:
$$\begin{gathered} a_1=\frac{2g}{29} \\ \mathrm{and}{a}_2=g-5a_1 \\ \Rightarrow a_2=g-\frac{10g}{29}=\frac{19g}{29} \\ \mathrm{So},a_1-a_2=\frac{2g}{29}-\frac{19g}{29}=-\frac{17g}{29} \\ \mathrm{and}{a}_1+a_2=\frac{2g}{29}+\frac{19g}{29}=\frac{21g}{29} \end{gathered}$$
So, accelerations of m₁, m₂ and m₃ are $\frac{19g}{29}\left(\mathrm{up}\right),\frac{17g}{29}\left(\mathrm{down}\right)\mathrm{and}\frac{21g}{29}\left(\mathrm{down}\right)$ , respectively.
Now, u = 0, s = 20 cm = 0.2 m
$$\begin{gathered} a_2=\frac{19g}{29} \\ \therefore s=ut+\frac{1}{2}a{t}^2 \\ \Rightarrow 0.2=\frac{1}{2}\times \frac{19}{29}g{t}^2 \\ \Rightarrow t=0.25\mathrm{s} \end{gathered}$$