# three blocks in contract with each other are pushed across

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First, treat the system as one unit with a total mass $M=2+3.1+2.6 =7.7kg$ and find its acceleration. The total normal force is $N=Mg$. Then the total frictional force is $f=\mu N$. The net force is $110-f=Ma$, which gives $a$ and all blocks accelerate with the same $a$.
From Newton's Third Law, the force that the 2.6kg block exerts on the 3.1 kg block is equal and opposite to the force the 3.1kg block applies on the 2.6kg block. This latter force (call it $F_3$) makes the 2.6kg block accelerate at the $a$ found above. Then we have $F_3-f=2.6a$. Here $f=\mu 2.6g$.  $F_3$ can now be found.