This problem is solved without any numbers provided. It is an algebraic rearrangement of the forces interacting on mass 1 and mass 2. ![[Unbalanced_Systems_on_Incline-removebg-preview.png]] Find equation for $m_1$ $\Sigma \vec{F} = \vec{F}_T + \vec{F}_f + \vec{F}_{g1}$ $m_1 a$ = ${F}_T$ - $F_f$ - $F_{g1}$ $m_1a$ = $F_T - \mu mg \cdot cos\theta - mg \cdot sin\theta$ Find equation for $m_2$ $\Sigma \vec{F} = \vec{F}_T + \vec{F}_{g2}$ $m_2a$ = $F_T + F_{g2}$ $m_2a = -F_T + m_{2} g$ $F_T = m_2g - m_2a$ To Find Acceleration insert $m_2$ equation into equation 1 $m_1a$ = $(m_2g - m_2a) - \mu m_1g \cdot cos\theta - m_1g \cdot sin\theta$ $m_1a + m_2a$ = $m_2g - \mu m_1g \cdot cos\theta - m_1g \cdot sin\theta$ $a(m_1 + m_2)$ = $g(m_2 - \mu m_1 \cdot cos\theta - m_1 \cdot sin\theta)$ $a$ = $\frac{g(m_2 - \mu m_1 \cdot cos\theta - m_1 \cdot sin\theta)}{(m_1 + m_2)}$ To Find $m_2$ mass insert $m_2 $ equation into equation 1 $m_1a$ = $(m_2g - m_2a) - \mu m_1g \cdot cos\theta - m_1g \cdot sin\theta$ $m_2a - m_2g$ = $- m_1a - \mu m_1g \cdot cos\theta - m_1g \cdot sin\theta$ $m_2(a - g)$ = $- m_1a - \mu m_1g \cdot cos\theta - m_1g \cdot sin\theta$ $m_2$ = $\frac{- m_1a - \mu m_1g \cdot cos\theta - m_1g \cdot sin\theta}{a - g}$ Once the acceleration is known, then solve for the tension force $\vec{F}_T$ by substituting in the acceleration value in equation 2.