A small block of mass m rests on the rough, sloping side of a triangular block of mass M which itself rests on a horizontal frictionless table as shown in the figure c Eigure 1) If the coefficient of static ic friction is it determine the minimum horizontal force F applied to M that will cause the small block m to start moving up the in incline Express your answer in terms of the given quantities and appropriate constants. F_(min)=

To determine the minimum horizontal force \( F_{\text{min}} \) required to cause the small block \( m \) to start moving up the incline, we need to consider the forces acting on both the small block and the larger block.<br /><br />### Step-by-Step Solution:<br /><br />1. **Forces on the Small Block \( m \):**<br /> - Gravitational force acting downward: \( mg \)<br /> - Normal force perpendicular to the incline: \( N \)<br /> - Frictional force opposing the motion: \( f_{\text{friction}} = \mu N \)<br /><br />2. **Forces on the Large Block \( M \):**<br /> - Gravitational force acting downward: \( Mg \)<br /> - Normal force from the table: \( N_{\text{table}} \)<br /> - Applied horizontal force: \( F \)<br /> - Frictional force from the small block: \( f_{\text{friction}} = \mu N \)<br /><br />### Equilibrium Conditions:<br /><br />- The system is in equilibrium, so the sum of forces in both the vertical and horizontal directions must be zero.<br /><br />#### Vertical Forces:<br />- For the small block:<br /> \[<br /> N = mg \cos(\theta)<br /> \]<br /> where \( \theta \) is the angle of the incline.<br /><br />- For the large block:<br /> \[<br /> N_{\text{table}} = Mg<br /> \]<br /><br />#### Horizontal Forces:<br />- For the small block:<br /> \[<br /> f_{\text{friction}} = \mu mg \cos(\theta)<br /> \]<br /><br />- For the large block:<br /> \[<br /> F - f_{\text{friction}} = 0<br /> \]<br /><br />### Minimum Force Calculation:<br /><br />To cause the small block to start moving, the applied force \( F \) must overcome the maximum static friction force. The maximum static friction force is given by:<br />\[<br />f_{\text{max}} = \mu mg \cos(\theta)<br />\]<br /><br />Since the large block is in equilibrium, the frictional force exerted by the small block on the large block is equal and opposite to the frictional force exerted by the table on the large block. Therefore, the applied force \( F \) must be equal to the maximum static friction force to start the motion.<br /><br />Thus, the minimum horizontal force \( F_{\text{min}} \) required is:<br />\[<br />F_{\text{min}} = \mu mg \cos(\theta)<br />\]<br /><br />### Final Answer:<br />\[<br />F_{\text{min}} = \mu mg \cos(\theta)<br />\]