# Finding friction forces of stacked boxes on a table

Consider the following system. The given friction coefficients are for static. Let $g=10$ (meter per square second).

If the $F(t)=10t$, for example, determine the friction forces $f_{AB}(t)$ (between the boxes $A$ and $B$) and $f_{BF}(t)$ (between the box $B$ and the floor).

I am confused how the friction forces grow reacting the external force $F(t)$. Could you explain the concept to solve this kind of problem? It is not a homework for sure.

$f_{AB}(t)$ will be on the left side and $f_{BF}(t)$ on the right.

Maximum $f_{BF}(t)=\mu (m_A+m_B)g=(0.6)(30)(10)=180N$

So, $F(t)$ will have to be greater than $180N$ so that $B$ can move. When it does, $A$ will experience a pseudo force say $F'(t)$ in the left direction. If $B$ accelerates at $a_B$, then $$F'(t)=m_Aa_B=10(180/30)=60N$$

But, Maximum $f_{AB}(t)=\mu m_Ag=(0.3)(10)(10)=30N$

Thus, $F'(t)$>$f_{AB}(t)$ and hence $A$ will appear to slide towards the left if $B$ moves and will eventually fall.

• A will slide towards the left if... er, wouldn't it be better to phrase it as "A will resist sliding to the right along with B ..." because it definitely wouldn't slide left. – user80551 Apr 6 '14 at 11:54
• @user80551 A will appear to slide towards the left because B is moving forward. – user42733 Apr 6 '14 at 12:09
• "A will appear to slide towards the left because B is moving forward" - exactly, appear, not actually move to the left. – user80551 Apr 6 '14 at 12:25