On an incline plane with angle θ, which statement correctly identifies the normal force and the component of gravity along the plane?

On an incline plane with angle θ, which statement correctly identifies the normal force and the component of gravity along the plane?

• A. N = mg; along plane = mg sin θ.
• B. N = mg cos θ; along plane = mg sin θ.
• C. N = mg cos θ; along plane = mg cos θ.
• D. N = mg; along plane = mg cos θ.

Answer: (B)

On an incline, gravity can be split into two parts relative to the surface: a perpendicular component mg cos θ pulling into the plane, and a component parallel to the plane mg sin θ pulling down the slope. The normal force acts perpendicular to the surface and, when there’s no acceleration into the plane, it balances the perpendicular component of weight. So the normal force has magnitude mg cos θ, and the gravity component along the plane is mg sin θ directed down the slope. This matches the statement with N = mg cos θ and along-plane = mg sin θ. The other options mix up these components (for example, the normal force would be mg only on a horizontal plane, and the along-plane component would not be mg cos θ).

On an incline, gravity can be split into two parts relative to the surface: a perpendicular component mg cos θ pulling into the plane, and a component parallel to the plane mg sin θ pulling down the slope. The normal force acts perpendicular to the surface and, when there’s no acceleration into the plane, it balances the perpendicular component of weight. So the normal force has magnitude mg cos θ, and the gravity component along the plane is mg sin θ directed down the slope. This matches the statement with N = mg cos θ and along-plane = mg sin θ. The other options mix up these components (for example, the normal force would be mg only on a horizontal plane, and the along-plane component would not be mg cos θ).