# Stack of Blocks

A stack of Block A, B of mass 2.0 kg and 4.0 kg respectively are pushed along a smooth horizontal surface by an external force of 12 N. Both blocks move as a stack with the same acceleration. Calculate frictional force between Block A and B.

Solution

Fxy denote the contact force that block X exerts on Y.

Block B is accelerated rightward because of the 12 N external force. In the absence of friction, block A would have slid leftward relative to B. This explains why block B exerts a rightward frictional force FBA on block A. Basically it is block B’s attempt to get block A to “stick” with it.

Considering all both blocks as a combined 6 kg mass,

\displaystyle \begin{aligned}({{F}_{net}}&=ma)\\12&=(4+2)a\\a&=2.0\text{ m }{{\text{s}}^{-2}}\text{ }\end{aligned}

Considering block A by itself as a 2 kg mass,

\displaystyle \begin{aligned}({{F}_{net}}&=ma)\\{{F}_{BA}}&=(2)(2.0)\\&=4.0\text{ N }\end{aligned}

Alternatively, we can consider blocks B by itself as a 4 kg mass.

\displaystyle \begin{aligned}{{F}_{net}}&=ma\\12-{{F}_{AB}}&=(4)(2.0)\text{ }\\{{F}_{AB}}&=4.0\text{ N}\end{aligned}

In summary,