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Zorluk: OrtaInclined Planes

A combat engineer squad needs to push a 60 kg60\text{ kg} supply crate up a smooth, frictionless ramp onto a platform. The platform is 2 meters2\text{ meters} above the ground, and the ramp is 6 meters6\text{ meters} long. Assuming the acceleration due to gravity is 10 m/s210\text{ m/s}^2, what is the ideal effort force required to push the crate up the ramp?

  1. A
    20 N20\text{ N}
  2. 200 N200\text{ N}Cevap
  3. C
    600 N600\text{ N}
  4. D
    1200 N1200\text{ N}

Cevap

The required effort force is 200 N200\text{ N}.
The ideal mechanical advantage (IMA) of an inclined plane is calculated by dividing its length by its height (6 m / 2 m = 3). The crate's weight is its mass multiplied by gravity (60 kg × 10 m/s² = 600 N). The effort force required is the weight divided by the IMA (600 N / 3 = 200 N).

Adım Adım Çözüm

1
Calculate the weight (gravitational force) of the crate.
Weight=60 kg×10 m/s2=600 NWeight = 60\text{ kg} \times 10\text{ m/s}^2 = 600\text{ N}
Effort force must oppose the weight (force of gravity) pulling the object down, not just the mass.
2
Calculate the ideal mechanical advantage (IMA) of the ramp.
IMA=6 m2 m=3IMA = \frac{6\text{ m}}{2\text{ m}} = 3
IMA compares the length of the ramp (distance the effort is applied) to the vertical height the load is lifted.
3
Calculate the effort force.
Effort=600 N3=200 NEffort = \frac{600\text{ N}}{3} = 200\text{ N}
The inclined plane reduces the required force by a factor equal to its mechanical advantage.

Anahtar Kavram

Ideal Mechanical Advantage of Inclined Planes
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Konular

Inclined Planes

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