Let's solve the problem step by step: --- ## **Step 1: Draw Free Body Diagrams and Write Force Equilibrium Equations** For **block \( m_1 \) on the incline**: - Weight: \( m_1 g \) (acts vertically downward) - Normal force: \( N \) (perpendicular to the incline) - Friction force: \( f \) (opposes motion, parallel to the incline) - Tension: \( T \) (along the incline, up) Decompose \( m_1 g \): - Parallel to incline: \( m_1 g \sin \theta \) - Perpendicular to incline: \( m_1 g \cos \theta \) **Friction force**: \( f \leq \mu_s N = \mu_s m_1 g \cos \theta \) **Equilibrium condition (along the incline):** \[ T = m_1 g \sin \theta \pm f \] For **hanging mass \( m_2 \)**: \[ T = m_2 g \] --- ### **Explanation (Step 1):** We set up the force balance equations for both the block on the incline and the hanging mass, considering the direction of friction based on the tendency of motion. --- ## **Step 2: Find the Range for \( m_2 \) (Friction Up and Down the Incline)** ### **Case 1: Block tends to move UP the incline** Friction acts DOWN the incline: \[ m_2 g = m_1 g \sin\theta + \mu_s m_1 g \cos\theta \] \[ m_2 = m_1 (\sin\theta + \mu_s \cos\theta) \] ### **Case 2: Block tends to move DOWN the incline** Friction acts UP the incline: \[ m_2 g = m_1 g \sin\theta - \mu_s m_1 g \cos\theta \] \[ m_2 = m_1 (\sin\theta - \mu_s \cos\theta) \] Plug in the values (\(\theta = 26^\circ\), \(\mu_s = .36\)): \[ \sin 26^\circ \approx .4384 \] \[ \cos 26^\circ \approx .8988 \] \[ m_2 = m_1 (.4384 + .36 \times .8988) = m_1 (.4384 + .3236) = m_1 (.762) \] \[ m_2 = m_1 (.4384 - .36 \times .8988) = m_1 (.4384 - .3236) = m_1 (.1148) \] --- ### **Explanation (Step 2):** We analyzed both directions for possible impending motion, applied the friction force accordingly, and calculated the bounds for \( m_2 \). --- ## **Final Answer** \[ \boxed{(.1148\, m_1) \leq m_2 \leq (.762\, m_1)} \] Where: - \( .1148\, m_1 \) is the minimum value - \( .762\, m_1 \) is the maximum value This is the range of \( m_2 \) for which the system is in equilibrium.