Not slip sliding away

Let's denote masses by M = 50 kg, and m = 30 kg; and let the coefficient of friction be k. Considering the free body diagram of the heavier mass, the maximum friction on either side and the tension in the rope must at least balance down-the-slope component of its weight. $M g \sin \theta = T + k (M+m) g \cos \theta + k m g \cos \theta$ . Similarly from the free body diagram of the lighter mass: $T = m g \sin \theta + k m g \cos \theta$ . Eliminating T from the above equations, we have: $k = \frac{(M - m) tan \theta} {M + 3m}$ . Substituting the values, we have k = 0.082.

Since both blocks are at rest, the net force on each block must be zero. Forces on the top block, parallel to the incline (taking up the incline as positive): $\\T-W_{\parallel }- f_{bottom} = 0 \\T-m_{top}gsin\theta - \mu m_{top}gcos\theta = 0 \\T-30gsin30^{\circ} - \mu 30gcos30^{\circ} = 0$

Forces on the bottom block, parallel to the incline (taking up the incline as positive): $\\T+f_{floor}+f_{top}-W_{\parallel} = 0 \\T+\mu (m_{top}+m_{bottom})gcos\theta+\mu m_{top}gcos\theta-m_{bottom}gsin\theta=0 \\T+\mu 80gcos30^{\circ}+\mu 30gcos30^{\circ}-50gsin30^{\circ}=0 \\T+\mu 110gcos30^{\circ}-50gsin30^{\circ}=0$

The tension in the rope connecting the blocks must be the same everywhere. Solving for tension using the equation for the top block:

$\\T=30gsin30^{\circ} + \mu 30gcos30^{\circ}$

then substituting for tension into the equation for the bottom block:

$\\30gsin30^{\circ} + \mu 30gcos30^{\circ}+\mu 110gcos30^{\circ}-50gsin30^{\circ}=0$

Combining like terms and solving for $\mu$ : $\\\mu 140gcos30^{\circ}-20gsin30^{\circ}=0 \\\mu=\frac{tan30^{\circ}}{7} \\\mu=0.0825$

First, choose coordinate axes such that the x-axis is parallel to the surface of the plane and the y-axis is perpendicular. Next, realize that if the blocks slip, the bottom one will slide down the slope (this helps you get the direction of the frictional forces right). We can then write the following force balances for the x and y components of the forces on the top and bottom block:

$F_t,x=-m_tgsin\theta + T - \mu N_{b~on~t}=0$

$F_t,y=-m_t g cos\theta + N_{b~on~t}=0$

$F_b,x=-m_b g sin\theta + \mu N_{t~on~b} + \mu N_{plane~on~b} + T=0$

$F_b,y=-m_b g cos\theta + N_{plane~on~b}-N_{t~on~b}=0$ .

All the net forces above are zero because if $\mu$ is the minimum coefficient necessary to keep the blocks from slipping then, well, before the blocks slip they are not moving.

We can solve this system of linear equations for $\mu$ ,

$\mu=\frac {m_b-m_t} {3 m_t+m_b} tan\theta$ .

Entering in numbers gives $\mu=0.0824$ .