Question

Let $m_1=1$kg $m_2=2$kg and $m_3=3$ kg in figure.

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 1

Let the block $m_1$ moves upward with acceleration $a_1$ and the two blocks $m_2$ and $m_3$ have relative acceleration $a_2$ due to the difference of weight between them.

So, the actual acceleration of the blocks $m_1$, $m_2$, and $m_3$, will be $a_1,(a_1-a_2)$ and $(a_1+a_2)$ as shown

From figure 2, $T-1g-1a_2=0$ ...(i)

From Figure 3,

$\frac{T}{2}-2g-2(a_1-a_2)=0$ ...(ii)

From figure 4,

$\frac{T}{2}-3g-3(a_1+a_2)=0$ ...(iii)

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

$1g+1a_2=4g+4(a_1+a_2)$

$\Rightarrow 5a_2-4a_1=3g$ ....(iv)

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

$2g+2(a_1-a_2)=3g-3(a_1-a_2)$

$\Rightarrow 5a_1+a_2=g$

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

$a_1=\frac{2g}{29}$

and $a_2=g-5a_1$

=$g-\frac{10g}{29}=\frac{19g}{29}$

So, $a_1-a_2=\frac{2g}{29}-\frac{19g}{29}=-\frac{17g}{29}$

and $a_1+a_2=\frac{2g}{29}+\frac{19g}{29}=\frac{21g}{29}$

So, acceleration of $m_1,m_2$ and $m_3$ are $\frac{19g}{29}$ (up) , $\frac{17g}{29}$ (down) and $\frac{21g}{29}$ (down), respectively,

Again, from u = 0, S = 20 cm = 0.2 m and

$a_2=\frac{19g}{29}$ [g = $10m/s^2$]

$\therefore$ 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 sec