To determine the coefficient of friction when an object just begins to slide down an inclined plane, we need to analyze the forces acting on the object. Let's break this down step-by-step. ### Step 1: Understanding the Forces When an object is placed on an inclined plane, two main forces act on it: 1. **Gravitational Force (Weight)**: This force acts vertically downward and can be broken into two components: - **Parallel to the incline**: \( F_{\text{gravity, parallel}} = mg \sin(\theta) \) - **Perpendicular to the incline**: \( F_{\text{gravity, perpendicular}} = mg \cos(\theta) \) Here, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of the incline (30° in this case). 2. **Frictional Force**: The frictional force opposes the motion and is given by: - \( F_{\text{friction}} = \mu F_{\text{normal}} \) - Where \( \mu \) is the coefficient of friction and \( F_{\text{normal}} \) is the normal force acting on the object. ### Step 2: Normal Force Calculation The normal force \( F_{\text{normal}} \) is equal to the perpendicular component of the gravitational force: \[ F_{\text{normal}} = mg \cos(\theta) \] ### Step 3: Setting Up the Equation At the point where the object just begins to slide, the frictional force equals the component of the gravitational force acting down the incline: \[ F_{\text{friction}} = F_{\text{gravity, parallel}} \] Substituting the expressions for these forces, we have: \[ \mu (mg \cos(\theta)) = mg \sin(\theta) \] ### Step 4: Simplifying the Equation We can cancel \( mg \) from both sides (assuming \( m \) is not zero): \[ \mu \cos(\theta) = \sin(\theta) \] Now, we can solve for the coefficient of friction \( \mu \): \[ \mu = \frac{\sin(\theta)}{\cos(\theta)} \] This simplifies to: \[ \mu = \tan(\theta) \] ### Step 5: Calculating for \( \theta = 30^\circ \) Now, we substitute \( \theta = 30^\circ \): \[ \mu = \tan(30^\circ) \] From trigonometric values, we know: \[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \] Thus, the coefficient of friction \( \mu \) is: \[ \mu = \frac{1}{\sqrt{3}} \] ### Conclusion: Correct Option The correct answer is **D. \( \frac{1}{\sqrt{3}} \)**. ### Step 6: Analyzing Other Options - **A. \( \sqrt{3} \)**: This value is much larger than the expected coefficient of friction for a typical incline and does not correspond to any standard trigonometric value for 30°. - **B. \( \frac{\sqrt{3}}{2} \)**: This value is the cosine of 30°, not the tangent. It does not represent the coefficient of friction in this scenario. - **C. \( \frac{1}{\sqrt{2}} \)**: This value corresponds to the tangent of 45°, not 30°. It is incorrect for this problem. ### Revision Summary - The coefficient of friction when an object just begins to slide on an incline can be found using \( \mu = \tan(\theta) \). - For \( \theta = 30^\circ \), \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \). - The correct answer is **D. \( \frac{1}{\sqrt{3}} \)**. - Always remember to analyze the forces acting on the object and use trigonometric identities to find the correct relationships.