+↑ Σ F1_y: F_1N - (m_1*g) = 0
+⟶ Σ F1_x: F - F_t = (m_1*a)
We don't need to solve or substitute with anything in F1_y, so we can just toss it out. The only unknown value in F1_x is that nasty F_t, so let's solve in terms of that.
[code] F_t = F - (m_1 * a) [code]
Sweet, first system down.
Again, we have four separate forces being applied:
+↑ Σ F2_y: F_2N - (m_2*g) = 0
+⟶ Σ F2_x: F_t - f_k = (m_2*a)
Now, of course, f_k can be subbed for (F_2N * u_k), and we can substitute F_t for what we found for it's value in the first system.
[code]+⟶ Σ F2_x: F - (m_1 * a) - (F_2N * u_k) = (m_2*a) [code]
And now we simplify for the value we desire. In this case, u_k:
[code] u_k = (F - (m_2 * a + m_1 * a) / (m_2 * g) [code]
Or, more simply, using F=ma for the whole system:
F - f_k = (m_1 + m_2) * a
F - (m * g * u_k) = (m_1* a + m_2*a)
u_k = ...