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Zorluk: KolayInclined Planes

A worker needs to push a supply cart weighing 150 lbs150\text{ lbs} up to a loading dock. The dock is 3 feet3\text{ feet} above the ground, and the straight ramp leading up to it is 30 feet30\text{ feet} long. Assuming friction is negligible, what is the effort force required to push the cart up the ramp?

  1. A
    150 lbs150\text{ lbs}
  2. B
    450 lbs450\text{ lbs}
  3. 15 lbs15\text{ lbs}Cevap
  4. D
    1,500 lbs1,500\text{ lbs}

Cevap

The correct effort force required is 15 lbs15\text{ lbs}.
The effort force is found by dividing the total load by the Ideal Mechanical Advantage (IMA). The IMA of an inclined plane is the ratio of its length to its vertical height (30 feet/3 feet=1030\text{ feet} / 3\text{ feet} = 10). Therefore, the required effort force is the load divided by the IMA: 150 lbs/10=15 lbs150\text{ lbs} / 10 = 15\text{ lbs}.

Adım Adım Çözüm

1
Identify the known variables from the problem.
Load = 150 lbs150\text{ lbs}, Ramp Length = 30 feet30\text{ feet}, Height = 3 feet3\text{ feet}.
These values are required to calculate the mechanical advantage and the resulting effort force.
2
Calculate the Ideal Mechanical Advantage (IMA) of the ramp.
IMA=LengthHeight=303=10\text{IMA} = \frac{\text{Length}}{\text{Height}} = \frac{30}{3} = 10
The IMA tells us by what factor the machine multiplies the input force (or divides the required effort).
3
Calculate the required effort force.
Effort Force=LoadIMA=15010=15 lbs\text{Effort Force} = \frac{\text{Load}}{\text{IMA}} = \frac{150}{10} = 15\text{ lbs}
Dividing the total weight of the object by the mechanical advantage gives the theoretical force needed to move it up the incline.

Anahtar Kavram

Ideal Mechanical Advantage (IMA) and Effort Force on an Inclined Plane
Tahmini Süre:45s
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Konular

Inclined Planes

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