Two touching bars 1 and 2 are placed on an inclined plane forming an angle `alpha` with the horizontal (figure). The masses of the bars are equal to `m_1` and `m_2`, and the coefficients of friction between the inclined plane and these bars are equal to `k_1` and `k_2` respectively, with `k_1gtk_2`. Find:
(a) the force of interaction of the bars in the process of motion,
(b) the minimum value of the angle `alpha` at which at bars start sliding down.

Explanation

Let us indicate the positive direction of x-axis along the incline (figure). Figures show the force diagram for the blocks.
Let, R be the force of interaction between the bars and they are obviously sliding down with the same constant acceleration w.


Newton's second law of motion in projection from long x-axis for the blocks gives:
`m_1g sin alpha-k_1m_1gcos alpha+R=m_1w` (1)
`m_2g sin alpha-R-k_2m_2g cos alpha=m_2w` (2)
Solving Eqs. (1) and (2) simultaneously, we get
`w=g sin alpha-g cos alpha (k_1m_1+k_2m_2)/(m_1+m_2)` and
`R=(m_1m_2(k_1-k_2)gcos alpha)/(m_1+m_2)` (3)
(b) when the blocks just slide down the plane, `w=0`, so from Eqn. (3)
`gsin alpha-gcos alpha(k_1m_1+k_2m_2)/(m_1+m_2)=0`
or, `(m_1+m_2)sin alpha=(k_1m_1+k_2m_2)cosalpha`
Hence `tan alpha=((k_1m_1+k_2m_2))/(m_1+m_2)`