1. ## physics problem

OK so you have the diagram here, with only a bit of the left side cut off, but it's just the wall. the coefficient of static friction between B and the table is .25, B weighs 711 N. find the maximum weight of block A, such that the system is stationary.

I got this:
the rope attatched to B exerts a horizontal force on B, so for B to just barely not move ther force must be 177.75 N, and this is the tension in the rope. Therefore, that is the force exerted on the knot by the wall, so the vertical force must be $177.75*\sin 30^o$, and this is the weight of the block. The book says im wrong, but i dont see where i erred.

2. Hi

Equilibrium of B
$T_B - F = 0$
$- m_B g + N = 0$

Just at the rupture of equilibrium $F = fN$

Therefore $T_B = F = fN = f m_B g$

Equilibrium of the knot
$T_B - T \cos \alpha = 0$

$T_A - T \sin \alpha = 0$

Therefore $T_A = T \sin \alpha = T_B \tan \alpha = f m_B g \tan \alpha$

Equilibrium of A
$- m_A g + T_A = 0$

$m_A = \frac{T_A}{g} = f m_B \tan \alpha$

3. okay, thanks. My main problem was that I haven't worked with more complex rope systems, and i wasn't sure how the force on block B worked. I was thinking that maybe, because gravity was pulling block A down, the force on block B may have been partially downwards.