Friction Force on a System at Rest

1. Oct 23, 2007

odie5533

1. The problem statement, all variables and given/known data
http://img337.imageshack.us/img337/2514/diagramfn4.png [Broken]

3. The attempt at a solution
$$\sum F_{x} = w_{x} - T = m_{x}a$$
$$T = (6.9)g - (6.9)a$$

$$\sum F_{a} = T - w_{a} - f_{s} = m_{a}a$$
$$T - m_{a}g - m_{a}a = f_{s}$$
$$(6.9g - 6.9a) - m_{a}g - m_{a}a = f_{s}$$
I don't see a 5kg block with friction, but I do see a 3kg block with friction:
$$6.9g - 6.9a - 3g - 3a = f_{s}$$
$$3.9g - 9.9a = f_{s}$$

Since the system is at rest, $$a = 0$$
$$38N = f_{s}$$

But I have two problems with my answer:
1) $$f_{s-max} = (0.40)(50) = 20N < 38N$$ in my answer
2) Using the "5kg" block, $$f_{s} = 19N$$

Am I missing something?

Last edited by a moderator: May 3, 2017
2. Oct 23, 2007

Staff: Mentor

Looks to me like the 3 kg was a typo and they meant to write 5 kg. (Otherwise the problem doesn't work, as you point out.)