Question

Question is given below ?

The question is the following arrangement the system is initially at rest. The 5 kg block is now released. Assuming the pulleys and strings to be massless and smooth, the acceleration of block 'C' is , the diagram is given

now the solution of this question is also give but how can this possible since B is grounded .

Solution which is given :-

Answer 1

See below.

Explanation:

Considering

`m_1 = 5` [Kg]
`m_2=10` [Kg]
`m_3 = 8` [Kg]

The movement for `m_1` is

`m_1 alpha_1 = T-m_1 g`

The movement for `m_3` is

`m_3 alpha_3 = 2T-m_3 g`

Now due to the inextensible strings

`alpha_1=-2alpha_3`

Now solving

`{(m_1 alpha_1 = T-m_1 g),(m_3 alpha_3 = 2T-m_3 g),(alpha_1=-2alpha_3):}`

gives

`{(alpha_1 = 2(m_3-2m_1)/(4m_1+m_3)g),(alpha_2 = (m_3-2m_1)/(4m_1+m_3)g),(T=(m_1m_3)/(4m_1+m_3)g):}`

and finally

`alpha_3 = g/14`

NOTE

`T = 30/7 g` so `2T = 60/7 g > 8 g` hence `m_3` elevates and `m_2` remains in the ground because `T < m_2g`

Answer 2

Only point of contention appears to be acceleration `a/2` of mass `8kg` shown in its FBD.

it is difficult to understand this with the help of Newtonian Mechanics. It is clearer if we use Euler-Lagrange equations and use Lagrangian mechanics

However, to appreciate this in terms of Newtonian mechanics,
if both masses `Aand B` are moving with constant acceleration,
it is clear that

`veca_A=-veca_B` .....(1)

This is due to the fact that string joining the masses is inextensible and therefore, its length is always constant. (1) is purely an outcome of conservation of length even though `Aand B` may be have different masses. Rate of change of length of one side of string must be equal and opposite to the rate of change of length of the other.

By the same conservation of length, if all three masses are moving then

`veca_C=-(veca_A+veca_B)/2` ......(2)

In our problem, `B` does not move. Hence, `veca_B=0`. From (2) we get

`veca_C=-veca_A/2`

This explains the FBD of block `C`