Please provide the Attached questions.This is from NLM and friction

m_A = 6 kg, m_B = 3 kg, m_C = 6 kg, m_D = 1 kg,

₁ = 0.7 = friction coefficient between A and B, ₂ = 0.2 = friction coefficient between surface and B.

Let us assume that the system has a common acceleration ‘a’ i.e. C is moving downwards, A and B are moving together to the left and D is moving upwards with same acceleration a.

 

Let T₁ and T₂ be the tensions in the strings connecting blocks A and C and blocks B and D, N₁ be the normal reaction between A and B, N₂ be the normal reaction between surface and B, f₁ be the force of friction between A and B, f₂ be the frictional force between the surface and B.

 

FBDs (free body diagrams) of the blocks are shown below.

                    T₁                                     N₁                                              N₂

                                                                                                         

Block C :   |  6 kg  |   ,  Block A :  T₁ |  6 kg  | f₁  ,  Block B : f₁ | 3 kg | T₂ + f₂

                                                                                                             

             m_c g = 60 N                      m_A g = 60 N                                     N₁   m_B g = 30 N

            

                  T₂

                  

Block D :  | 1 kg |

                  

              m_D g = 10 N

There isn’t vertical motion of A and B i.e. net force acting on them in vertical direction is zero.

N₁ = m_A g = 60 N  and  N₂ = N₁ + m_B g  N₂ = (60 + 30) N = 90 N.

Limiting static friction or kinetic friction between A and B is

f_1max = ₁ N₁ = (0.7)(60) = 42 N.

Similarly, maximum static friction or kinetic friction between B and the surface is

f_2max = (0.2)(90) = 18 N.

Since we have assumed that A and B are moving together, B is sliding on the surface and the associated friction is kinetic. So, f₂ = 18 N.

Applying F_net = m.a for each mass, we get

For C,  60 – T₁ = 6 a                             ...(1)

For A,  T₁ – f₁  = 6 a                             ...(2)

For B,  f₁ – T₂ – f₂ = 3 a    f₁ – T₂ – 18 = 3 a                      ...(3)

For D,  T₂ – 10 = 1 a                             ...(4)

Add (1), (2), (3) & (4) 60 – 18 – 10 = (6+6+3+1) a  32 = 16a or a = 2 m/s²

Put the value of a in (1), (2), (3), (4) to get

T₁ = 48 N , T₂ = 12 N , f₁ = 36 N    f_1max.

f_{1 1max} , our assumption is correct.

Thus, options ‘A’ and ‘D’ are correct.