An object weighing 100 lbs is sitting on a frictionless inclined plane that makes a 30° angle with the horizontal. What force, applied parallel to the incline, is needed to keep the object from sliding down?

On a frictionless incline, the component of weight parallel to the incline pulls the object downward along the slope. This component = W × sin(θ), where W = weight and θ = incline angle. For 100 lbs at 30°: parallel force = 100 × sin(30°) = 100 × 0.5 = 50 lbs. To prevent sliding, an equal opposing force is needed (50 lbs up the incline). Incline plane physics: (1) FORCE PARALLEL to incline (driving sliding) = W × sin(θ); (2) FORCE PERPENDICULAR to incline (pressing into surface) = W × cos(θ). At 0° (flat ground): sin=0, parallel force=0, no sliding tendency; perpendicular = full weight. At 90° (vertical): sin=1, parallel = full weight (object in free fall); perpendicular = 0. Common angles to memorize: sin(30°)=0.5, sin(45°)≈0.707, sin(60°)≈0.866, sin(90°)=1; cos same values in reverse order. Inclined plane is a simple machine: mechanical advantage = length of incline / height = 1/sin(θ). A 30° ramp has MA = 1/0.5 = 2 (half the force to lift the same load to the same height, but you push twice as far). Common ASVAB MC inclined plane questions: force to hold/lift on incline, MA of ramp, work done sliding up incline. With friction, additional force = friction coefficient × W × cos(θ).