On a ramp with angle θ, the normal force is N = m g cos θ. Which expression is the normal force?

On a ramp with angle θ, the normal force is N = m g cos θ. Which expression is the normal force?

• A. N = m g sin θ
• B. N = m g cos θ
• C. N = m g
• D. N = m g / cos θ

Answer: (B)

The normal force on an incline comes from balancing the part of gravity that pushes perpendicularly into the surface. Gravity mg can be split into a component perpendicular to the plane and a component parallel to the plane. The perpendicular component has magnitude mg cos θ, because the angle between gravity (straight down) and the plane’s normal is θ. If the block stays in contact and doesn’t accelerate into or away from the plane, the forces perpendicular to the surface must cancel, so the normal force equals that perpendicular component: N = mg cos θ. The component along the plane, mg sin θ, would drive motion along the incline (if friction is negligible). The full weight mg isn’t the normal force, and mg/ cos θ isn’t the correct perpendicular component.

The normal force on an incline comes from balancing the part of gravity that pushes perpendicularly into the surface. Gravity mg can be split into a component perpendicular to the plane and a component parallel to the plane. The perpendicular component has magnitude mg cos θ, because the angle between gravity (straight down) and the plane’s normal is θ. If the block stays in contact and doesn’t accelerate into or away from the plane, the forces perpendicular to the surface must cancel, so the normal force equals that perpendicular component: N = mg cos θ. The component along the plane, mg sin θ, would drive motion along the incline (if friction is negligible). The full weight mg isn’t the normal force, and mg/ cos θ isn’t the correct perpendicular component.