Question

A system of pulley and block is arranged as shown in the figure. Find the acceleration of the block 5kg, 2kg, 6kg.

Answer 1

Hint: In the above diagram we can see that the block of 5kg is moving downwards. Therefore from that we can imply that the other two blocks will move up as the string across them will get reduced. For the above pulley problem we would use the equation for tension in constrained motion. Further we will determine the net force acting on each of the blocks and then again substituting for acceleration in the equation of tension of the individual blocks will enable us to estimate their respective acceleration.
Formula used:
$\sum Ta=0$
$F_{NET}=ma$

Complete answer:
In the above diagram let us say the block of 5kg, 6kg and 2kg respectively move with acceleration of $a_1\text{, }{a}_{2}and{a}_3$ respectively. Hence we can write the net force on the block of 5kg as difference of force due to gravity and the tension in the rope. Mathematically this can be written as,
$\begin{array}{rl}& F_{NET}=mg-T\\ & \Rightarrow 5a_1=5g-T...(1)\end{array}$
Similarly for a block of 6kg and 2kg we can write, the net force on them as,
$\begin{array}{rl}& \text{For 6kg,}\\ & F_{NET}=2T-mg\\ & \Rightarrow 6a_2=2T-6g...(2)\\ & \text{For 2kg,}\\ & F_{NET}=2T-mg\\ & \Rightarrow 2a_3=2T-2g...(3)\end{array}$
Now as per the equation of constrained motion, it tells us that the sum of the total tension in the rope times the acceleration of each of the block is zero .i.e. mathematically can be written as,
$\sum Ta=0$ where T is the tension in the rope across the individual masses having different acceleration a.
Hence using this above equation in the above figure we get,
$\begin{array}{rl}& \sum Ta=0\\ & \Rightarrow 2T(-a_3)+2T(-a_2)+T{a}_1=0\\ & \Rightarrow 2a_3+2a_2-a_1=0...(4)\end{array}$
Since the tension in the string is the same, from equation 1 and, 2 and 3 and 3 and 1 we get,
From 1 and 2,
$\begin{array}{rl}& T=5g-5a_1=3a_2+3g\\ & \Rightarrow 5a_1+3a_2=2g\end{array}$
From 2 and 3,
$\begin{array}{rl}& T=3a_2+3g=g+a_3\\ & \Rightarrow a_3-3a_2=2g\end{array}$
From 3 and 1,
$\begin{array}{rl}& T=5g-5a_1=a_3+g\\ & \Rightarrow a_3+5a_1=4g\end{array}$
Now let us substitute equation 4 totally in terms of $a_1$,
$\begin{array}{rl}& 2a_3+2a_2-a_1=0\\ & \Rightarrow 2(4g-5a_1)+2({\displaystyle \frac{2g-5a_1}{3}})-a_1=0\\ & \Rightarrow 8g-10a_1+{\displaystyle \frac{4g}{3}}-{\displaystyle \frac{10a_1}{3}}-a_1=0\\ & \Rightarrow -10a_1-{\displaystyle \frac{10a_1}{3}}-a_1=-{\displaystyle \frac{4g}{3}}-8g\\ & \Rightarrow -{\displaystyle \frac{43a_1}{3}}=-{\displaystyle \frac{28g}{3}}\\ & \Rightarrow a_1={\displaystyle \frac{28}{43}}g\text{, }\because g=10m{s}^{-2}\\ & \Rightarrow a_1=6.5m{s}^{-2}\end{array}$
Now let us determine the acceleration using a similar procedure of mass 6kg.
$\begin{array}{rl}& 2a_3+2a_2-a_1=0\\ & \Rightarrow 2(2g+3a_2)+2a_2-({\displaystyle \frac{2g-3a_2}{5}})=0\\ & \Rightarrow 4g+6a_2+2a_2-{\displaystyle \frac{2g}{5}}+{\displaystyle \frac{3a_2}{5}}=0\\ & \Rightarrow 6a_2+2a_2+{\displaystyle \frac{3a_2}{5}}=-4g+{\displaystyle \frac{2g}{5}}\\ & \Rightarrow -{\displaystyle \frac{43a_2}{5}}=-{\displaystyle \frac{18g}{5}}\\ & \Rightarrow a_2=-{\displaystyle \frac{18}{43}}g\text{, }\because g=10m{s}^{-2}\\ & \Rightarrow a_2=-4.2m{s}^{-2}\end{array}$
Now let us determine the acceleration using a similar procedure of mass 2kg.
$\begin{array}{rl}& 2a_3+2a_2-a_1=0\\ & \Rightarrow 2a_3+2({\displaystyle \frac{a_3-2g}{3}})-({\displaystyle \frac{4g-a_3}{5}})=0\\ & \Rightarrow 2a_3+{\displaystyle \frac{2a_3}{3}}-{\displaystyle \frac{4g}{3}}-{\displaystyle \frac{4g}{5}}+{\displaystyle \frac{a_3}{5}}=0\\ & \Rightarrow 2a_3+{\displaystyle \frac{2a_3}{3}}+{\displaystyle \frac{a_3}{5}}={\displaystyle \frac{4g}{3}}+{\displaystyle \frac{4g}{5}}\\ & \Rightarrow -{\displaystyle \frac{43a_3}{15}}={\displaystyle \frac{32g}{15}}\\ & \Rightarrow a_3={\displaystyle \frac{32}{43}}g\text{, }\because g=10m{s}^{-2}\\ & \Rightarrow a_3=7.4m{s}^{-2}\end{array}$
Hence the acceleration of the above blocks i.e. 5kg, 6kg and 2kg are $6.5m{s}^{-2}\text{, 4}\text{.2}m{s}^{-2}and7.4m{s}^{-2}$ respectively.

Note:
It is to be noted that the pulley as well as the string is considered to be massless. Hence We have considered the tension to be the same throughout the string. It is also to be noted that we can take any convention of the masses moving in a particular direction but the final answer would remain the same.