ASVAB (Armed Services Vocational Aptitude Battery)Mechanical Comprehension (MC)Medium
A worker needs to move a 500 N barrel up a ramp onto a platform. The ramp is 5 meters long and the platform is 1.5 meters high. Ignoring friction, what is the minimum force the worker must apply parallel to the ramp to move the barrel?
- A250 N
- B75 N
- C150 N
- D500 N
Show answer & explanationAnswer & explanation
Correct answer: C. 150 N
For an inclined plane, the ideal mechanical advantage (IMA) is length/height. IMA = 5m / 1.5m = 3.33. The force required is Load / IMA = 500 N / 3.33 = 150 N. Alternatively, Force = Load * (height / length) = 500 N * (1.5 m / 5 m) = 500 N * 0.3 = 150 N.
Why the other options are wrong
- A. This force is too high; it would be the force needed if the ramp were shorter or steeper.
- B. This force is too low; it would not overcome the gravitational component along the incline.
- D. This is the full weight of the barrel; less force is needed due to the mechanical advantage of the ramp.
Force on Inclined Plane (Ideal)
The minimum force required to move an object up an inclined plane, ignoring friction. It is calculated by multiplying the object's weight by the ratio of the ramp's height to its length.
- Provides mechanical advantage by reducing the required force.
- Ideal force = Weight * (Height / Length).
- Actual force is always higher due to friction.
Memory trick: Ramps 'lend' a hand: the 'longer' the ramp, the 'less' force you need for the 'height'.