+↑ Σ F1_y: F_1N - (m_1*g) = 0 +⟶ Σ F1_x: F - F_t = (m_1*a) [code] 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. [big]System 2[big] Again, we have four separate forces being applied: [code] +↑ Σ F2_y: F_2N - (m_2*g) = 0 +⟶ Σ F2_x: F_t - f_k = (m_2*a) [code] 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] [big]Fast Method[big] Or, more simply, using F=ma for the whole system: [code] F - f_k = (m_1 + m_2) * a F - (m * g * u_k) = (m_1* a + m_2*a) u_k = ...