Answer
(a) $F_f = 12.0~N$
(b) The maximum weight $w$ is 15.0 N.
Work Step by Step
(a) We can find the tension $T$ in the wire attached to the hook.
$T~sin(\theta) = w$
$T = \frac{w}{sin(\theta)}$
The force of static friction exerted on block A is equal in magnitude to the horizontal component of $T$.
$F_f = T~cos(\theta) = w~cot(\theta)$
$F_f = (12.0~N)~cot(45.0^{\circ})$
$F_f = 12.0~N$
(b) From part (a), we can see that for the system to remain in equilibrium, $w = F_f$. We can use $\mu_s$ to find the maximum possible force of friction $F_f$.
$w = F_f = (m_A)~g~\mu_s = (60.0~N)(0.25)$
$w = 15.0~N$
The maximum weight $w$ is 15.0 N.
