# Frictional force acting between 2 stacked blocks

1. Sep 7, 2015

### Rococo

1. The problem statement, all variables and given/known data

In the figure above, block A has mass $m_A=25kg$ and block B has mass $m_B=10kg$. Both blocks move with constant acceleration $a=2m/s^2$ to the right, and the coefficient of static friction between the two blocks is $\mu_s = 0.8$. The static frictional force acting between the blocks is:
A) 20 N
B) 50 N
C) 78 N
D) 196 N
E) 274 N

2. Relevant equations

$F_{net} = ma$
$f_s = \mu_s N$ (if blocks are at the point of slipping)

The correct answer is supposed to be A) 20N.

3. The attempt at a solution

Looking at the system as a whole the overall force F acting on the blocks is:

$F = (m_A + m_B)a = (25+10)(2) N = 70N$

I was confused about this question because it doesn't say if this force is applied to the top or bottom block. For example if it was applied to the bottom block A you have:

Block A: $F - f_s = m_A a$ giving $f_s = 70 - (25)(2) = 20N$

Block B: $f_s = m_B a$ giving $f_s = (10)(2) = 20N$

Which gives the frictional force as 20N, as given in the answer key.

However if the force F was applied to the top block B you have:

Block A: $f_s = m_A a$ giving $f_s = (25)(2) = 50N$

Block B: $F - f_s = m_B a$ giving $f_s = 70 - (10)(2) = 50N$

But this gives the frictional force between the blocks as 50N.

So I was wondering how this problem is supposed to be solved, given only the information in the question.

Last edited: Sep 7, 2015
2. Sep 7, 2015

### Staff: Mentor

Bad problem statement, but applying it to the upper one would be very unusual, it would also mean there is information missing (the coefficient of friction with the surface).

3. Sep 7, 2015

### Rococo

Thank you, I think I see now, that I should have said the 70N is actually the net force $F_{net}$ acting on the block system, which will be equal to $F - f_g$ where $F$ is the 'applied force' and $f_g$ is the friction between the ground and the block system.

Then if the force F was applied to the top block B you get:

Block A: $f_s - f_g = m_A a$
Block B: $F - f_s = m_B a$

From which it's impossible to find the frictional force between the blocks, $f_s$, since as you said the coefficient of friction with the ground is unknown, and we only know $F_{net} = F - f_g$.

So the force F must be applied to the bottom block A giving:

Block A: $F - f_g - f_s = m_A a$ i.e. $F_{net} - f_s = m_A a$
Block B: $f_s = m_B a$

Which gives $f_s = 20N$ as given in the answer key.