ASVAB (Armed Services Vocational Aptitude Battery)Mechanical Comprehension (MC)Medium
A 10 kg object is placed on an inclined plane that makes an angle of 30° with the horizontal. Assuming no friction, what is the magnitude of the force required to prevent the object from sliding down the plane? (Use g = 10 m/s²)
- A86.6 N
- B100 N
- C50 N
- D173.2 N
Show answer & explanationAnswer & explanation
Correct answer: C. 50 N
The force required to prevent the object from sliding down is equal to the component of gravity acting parallel to the inclined plane. This component is calculated as Weight × sin(angle). First, calculate weight: Weight = mass × g = 10 kg × 10 m/s² = 100 N. Then, Force = 100 N × sin(30°). Since sin(30°) = 0.5, Force = 100 N × 0.5 = 50 N.
Why the other options are wrong
- A. This is incorrect. This would be 100 N * cos(30°), which is the normal force component, not the parallel force.
- B. This is incorrect. This is the total weight, not the component acting down the slope.
- D. This is incorrect. This would be 2 * Weight * sin(30°) or 100 N * tan(60°).
Inclined Plane Forces
Forces acting on an object on an inclined plane include gravity (weight), normal force, and potentially friction or an applied force.
- Weight (W) acts vertically downwards.
- Normal force (N) acts perpendicular to the surface.
- Component of weight parallel to plane = W × sin(θ).
- Component of weight perpendicular to plane = W × cos(θ).
Memory trick: Slope's Sine Slips, Cosine Holds Firm!