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Zorluk: OrtaInclined Planes

A construction team uses various temporary ramps to roll heavy equipment platforms to different elevated levels. Assuming friction is negligible, arrange the following ramp configurations in order from the LEAST effort force required to the GREATEST effort force required.

  1. 1A 25 ft25\text{ ft} long ramp reaching a height of 5 ft5\text{ ft}
  2. 2A 20 ft20\text{ ft} long ramp reaching a height of 5 ft5\text{ ft}
  3. 3A 18 ft18\text{ ft} long ramp reaching a height of 6 ft6\text{ ft}
  4. 4A 14 ft14\text{ ft} long ramp reaching a height of 7 ft7\text{ ft}

Cevap

The correct order from least effort force to greatest effort force is: the 25 ft ramp, followed by the 20 ft ramp, then the 18 ft ramp, and finally the 14 ft ramp.
The effort force required to move an object up a frictionless inclined plane is determined by the ratio of the height to the length of the plane. A lower height-to-length ratio means a higher mechanical advantage and less required effort force. By calculating this ratio for each ramp, we find the 25 ft ramp requires 15\frac{1}{5} (20%) of the load's weight, the 20 ft ramp requires 14\frac{1}{4} (25%), the 18 ft ramp requires 13\frac{1}{3} (33.3%), and the 14 ft ramp requires 12\frac{1}{2} (50%). Therefore, the sequence is 25 ft ramp, 20 ft ramp, 18 ft ramp, and 14 ft ramp.

Adım Adım Çözüm

1
Identify the relationship between effort force, load, ramp length, and height.
Effort Force=Load×(heightlength)\text{Effort Force} = \text{Load} \times \left(\frac{\text{height}}{\text{length}}\right).
The effort force is inversely proportional to the ideal mechanical advantage (IMA=lengthheight\text{IMA} = \frac{\text{length}}{\text{height}}).
2
Calculate the effort force fraction for each ramp configuration.
25 ft ramp: 525=15\frac{5}{25} = \frac{1}{5}; 20 ft ramp: 520=14\frac{5}{20} = \frac{1}{4}; 18 ft ramp: 618=13\frac{6}{18} = \frac{1}{3}; 14 ft ramp: 714=12\frac{7}{14} = \frac{1}{2}.
Comparing these fractions allows us to determine the relative force required for an identical load.
3
Order the calculated fractions from lowest to highest.
15<14<13<12\frac{1}{5} < \frac{1}{4} < \frac{1}{3} < \frac{1}{2}.
The prompt asks to arrange the setups from the least effort force to the greatest effort force.

Anahtar Kavram

Ideal Mechanical Advantage and Effort Force on an Inclined Plane
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Konular

Inclined Planes

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