Let's solve the problem step by step: --- ## **Step 1: Resolve Forces and Find the Normal Reaction (N)** **Given:** - Weight of block, \( W = 800\,N \) - Angle of incline, \( \theta = 25^\circ \) - Applied force, \( P = 400\,N \) at \( 40^\circ \) to the horizontal - Coefficient of static friction, \( \mu_s = .20 \) - Coefficient of kinetic friction, \( \mu_k = .15 \) **Resolving forces perpendicular to the incline:** - Normal force (\( N \)) acts perpendicular to the incline. - Components perpendicular to the incline: - Weight: \( W \cos25^\circ \) (down into the plane) - Applied force, \( P \): \( P \sin(40^\circ - 25^\circ) = P \sin(15^\circ) \) (up away from the plane) **Equation:** \[ N = W \cos25^\circ - P \sin(15^\circ) \] **Calculate:** - \( W \cos25^\circ = 800 \times .9063 = 725.04\,N \) - \( P \sin15^\circ = 400 \times .2588 = 103.52\,N \) \[ N = 725.04 - 103.52 = 621.52\,N \] **End of Step 1 Explanation:** The normal reaction force is \( 621.52\,N \). (Note: The negative value in your attempt suggests an error in sign or direction. Here, it is positive, which is physically correct.) --- ## **Step 2: Find the Friction Force (F)** **Static friction maximum:** \[ F_{s,\text{max}} = \mu_s N = .20 \times 621.52 = 124.30\,N \] **Now, resolve forces parallel to the incline to check equilibrium:** - Down the plane: \( W \sin25^\circ = 800 \times .4226 = 338.08\,N \) - Up the plane: \( P \cos(15^\circ) = 400 \times .9659 = 386.36\,N \) - Friction force (\( F \)), direction opposes motion tendency **Sum of forces along incline for equilibrium:** \[ P \cos(15^\circ) + F = W \sin25^\circ \] \[ 386.36 + F = 338.08 \] \[ F = 338.08 - 386.36 = -48.28\,N \] A negative value means friction acts **up the plane** (opposite to assumed direction), and its magnitude is \( 48.28\,N \). Since \( |F| < F_{s,\text{max}} \), the block is in equilibrium. --- ## **Final Answers** 1. **Normal component (N) of reaction:** **\( N = 621.52\,N \)** 2. **Friction force (F):** **\( F = 48.28\,N \) (up the plane)** --- ### **Summary** - **Step 1:** Calculate normal force: \( N = 621.52\,N \) - **Step 2:** Calculate friction force: \( F = 48.28\,N \) (acting up the plane for equilibrium; within static limit) - **Final Answers:** - Normal Force: \( \boxed{621.52\,N} \) - Friction Force: \( \boxed{48.28\,N} \)