The free-body diagram for the box on the ramp shows the normal force perpendicular to the surface in the +y-direction. The weight is downward, and the static friction force is up (parallel to) the incline. The +x-direction is down the incline. The weight makes an angle of theta relative to the normal vector (surface perpendicular) direction. Describe the components of the forces acting on the box based on this coordinate system.

Answer

Based on the description provided for the free-body diagram, we can decompose the forces into x and y components using the coordinates where +x is down the incline and +y is perpendicular to the incline. 1. Normal Force (N): Since the normal force is exactly in the +y-direction, its components are: x-component = 0, y-component = N. 2. Static Friction Force (fs): Since it acts straight up the incline (opposite to +x), its components are: x-component = -fs, y-component = 0. 3. Weight (W): The weight acts vertically downward. Given that the weight vector makes an angle theta with the negative y-axis (the normal vector direction), we can use trigonometry to find its components. The component down the incline (x-direction) is W*sin(theta), and the component perpendicular to the incline (into the surface, or negative y-direction) is W*cos(theta). Thus: x-component = W*sin(theta), y-component = -W*cos(theta). In summary, the total force in the x-direction is W*sin(theta) - fs, and the total force in the y-direction is N - W*cos(theta).