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Zorluk: OrtaInclined Planes

A logistics team needs to roll a heavy generator weighing 900 lbs900\text{ lbs} from the ground onto a raised platform that is 5 feet5\text{ feet} high. If the team can apply a maximum combined push force of 150 lbs150\text{ lbs} parallel to the ramp surface, what is the minimum ramp length required to accomplish this task? Assume ideal conditions with zero friction.

Cevap: 30 feet

Cevap

The minimum ramp length required is 30 feet30\text{ feet}.
In an ideal frictionless system, the ratio of the load force to the effort force equals the ratio of the ramp length to its height. A load of 900 lbs900\text{ lbs} requiring only 150 lbs150\text{ lbs} of effort means the system needs a mechanical advantage of 66. Since the height is 5 feet5\text{ feet}, the ramp must be 66 times longer than the height, which is 30 feet30\text{ feet}.

Adım Adım Çözüm

1
Calculate the required Ideal Mechanical Advantage (IMA).
IMA=LoadEffort=900 lbs150 lbs=6IMA = \frac{\text{Load}}{\text{Effort}} = \frac{900\text{ lbs}}{150\text{ lbs}} = 6
The mechanical advantage indicates how many times the machine multiplies the effort force. We need a mechanical advantage of 6 to move the 900 lbs900\text{ lbs} load with only 150 lbs150\text{ lbs} of force.
2
Calculate the required ramp length using the IMA and the given height.
IMA=LengthHeight⇒6=Length5 feet⇒Length=6×5=30 feetIMA = \frac{\text{Length}}{\text{Height}} \Rightarrow 6 = \frac{\text{Length}}{5\text{ feet}} \Rightarrow \text{Length} = 6 \times 5 = 30\text{ feet}
For an inclined plane in ideal conditions, the mechanical advantage is also the ratio of the ramp's length to its vertical height.

Anahtar Kavram

The relationship between effort force, load, ramp length, and vertical height on an ideal inclined plane (Ideal Mechanical Advantage).
Tahmini Süre:1m 30s
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