Question

21. For the system shown in figure. The acceleration of centre of mass is (g = 10 m/s2).
(2) -0.4j+0.81
(1) -0.49-0.8î
(3) -0.89+0.4
(4) None​

Answer 1

Given:

A two pulley system is given.

To find:

Acceleration of centre of mass ?

Calculation:

• First , let's calculate the acceleration of each of the objects.

According to FBD , we can say :

$1) \: 8g - T1 = 8a$

$2) \: T2 - 4g= 4a$

$3) \: T1 - T2= 8a$

• Now, addition of the three equations:

$\implies \: 8g - 4g = 20a$

$\implies \: 20a = 4g$

$\implies \: a = 2 \: m {s}^{ - 2}$

Now, let's find out the acceleration of centre of mass:

$a_{x} = \dfrac{ \sum(m_{i}a_{i})}{ \sum( m_{i})}$

$\implies a_{x} = \dfrac{ (8 \times 2) +(8 \times 0) + (4 \times 0) }{ 8 + 4 + 8}$

$\implies a_{x} = 0.8 \: \hat{i}$

Now, in y axis :

$a_{y} = \dfrac{ \sum(m_{i}a_{i})}{ \sum( m_{i})}$

$\implies a_{y} = \dfrac{ (8 \times 0) + \{8 \times (- 2 )\} + (4 \times 2) }{ 8 + 4 + 8}$

$\implies a_{y} = - 0.4 \hat{j}$

So, final answer is :

$\boxed{ \bf a_{net} = - 0.4 \hat{j} + 0.8 \hat{i} }$

OPTION 2) IS CORRECT ✔️