On a 60° incline, a 3 kg mass experiences a normal force. What is the normal force?

On a 60° incline, a 3 kg mass experiences a normal force. What is the normal force?

• A. 14.7 N
• B. 29.4 N
• C. 7.35 N
• D. 25.4 N

Answer: (A)

On an incline, the normal force comes from the surface pushing perpendicular to itself, balancing the component of gravity that acts into the plane. The weight mg can be split into a perpendicular component mg cosθ and a parallel component mg sinθ. Since there’s no acceleration into or away from the plane, the normal force equals that perpendicular component: N = mg cosθ. Here, m = 3 kg, g ≈ 9.8 m/s^2, and θ = 60°. So N = 3 × 9.8 × cos60° = 29.4 × 0.5 = 14.7 N. The component of gravity along the plane would be mg sin60° ≈ 29.4 × 0.866 ≈ 25.4 N, which drives motion down the incline if unopposed, but it does not affect the normal force.

On an incline, the normal force comes from the surface pushing perpendicular to itself, balancing the component of gravity that acts into the plane. The weight mg can be split into a perpendicular component mg cosθ and a parallel component mg sinθ. Since there’s no acceleration into or away from the plane, the normal force equals that perpendicular component: N = mg cosθ.

Here, m = 3 kg, g ≈ 9.8 m/s^2, and θ = 60°. So N = 3 × 9.8 × cos60° = 29.4 × 0.5 = 14.7 N.

The component of gravity along the plane would be mg sin60° ≈ 29.4 × 0.866 ≈ 25.4 N, which drives motion down the incline if unopposed, but it does not affect the normal force.