Answer
(a) $m_2 = m_1~sin(\alpha) + m_1~cos(\alpha)~\mu_k$
(b) $m_2 = m_1~sin(\alpha) - m_1~cos(\alpha)~\mu_k$
(c) The maximum value of $m_2 $ is $m_2 = m_1~sin(\alpha) + m_1~cos(\alpha)~\mu_s$.
The minimum value of $m_2 $ is $m_2 = m_1~sin(\alpha) - m_1~cos(\alpha)~\mu_s$.
When $m_2$ is between this range of values, the system will remain at rest.
Work Step by Step
(a) Let's consider the system of $m_1$. Since the system is moving at constant speed, the acceleration is zero. Then the tension $T$ is equal in magnitude to the component of weight directed down the slope plus the force of kinetic friction directed down the slope.
$T = m_1~g~sin(\alpha) + m_1~g~cos(\alpha)~\mu_k$
Let's consider the system of $m_2$.
$m_2~g = T$
$m_2~g = m_1~g~sin(\alpha) + m_1~g~cos(\alpha)~\mu_k$
$m_2 = m_1~sin(\alpha) + m_1~cos(\alpha)~\mu_k$
(b) Let's consider the system of $m_1$. Since the system is moving at constant speed, the acceleration is zero. Then the tension $T$ plus the force of kinetic friction directed up the slope is equal in magnitude to the component of weight directed down the slope.
$T + m_1~g~cos(\alpha)~\mu_k= m_1~g~sin(\alpha)$
$T = m_1~g~sin(\alpha) - m_1~g~cos(\alpha)~\mu_k$
Let's consider the system of $m_2$.
$m_2~g = T$
$m_2~g = m_1~g~sin(\alpha) - m_1~g~cos(\alpha)~\mu_k$
$m_2 = m_1~sin(\alpha) - m_1~cos(\alpha)~\mu_k$
(c) When $m_2$ is at its maximum value, the force of static friction exerted on $m_1$ is directed down the slope.
$m_2 = m_1~sin(\alpha) + m_1~cos(\alpha)~\mu_s$
When $m_2$ is at its minimum value, the force of static friction exerted on $m_1$ is directed up the slope.
$m_2 = m_1~sin(\alpha) - m_1~cos(\alpha)~\mu_s$
When the $m_2$ is between these two values, the force of static friction exerted on $m_1$ will be able to hold the system at rest.
