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Zorluk: OrtaWheel and Axle

A maintenance worker operates a hand-cranked rotary fluid pump. The pump's crank handle rotates in a circle with a radius of 18 inches18 \text{ inches}. It turns a central drive shaft with a diameter of 6 inches6 \text{ inches}. If the worker applies a steady perpendicular force of 60 pounds60 \text{ pounds} to the handle, what is the output force exerted at the outer edge of the drive shaft? (Assume an ideal simple machine with no friction losses.)

  1. A
    180 pounds180 \text{ pounds}
  2. B
    10 pounds10 \text{ pounds}
  3. 360 pounds360 \text{ pounds}Cevap
  4. D
    20 pounds20 \text{ pounds}

Cevap

The output force is 360 pounds360 \text{ pounds}.
The correct answer is derived by first converting the axle's diameter to a radius (3 inches3 \text{ inches}). The system's mechanical advantage is the ratio of the wheel's radius to the axle's radius (18 / 3 = 6). Multiplying the input force of 60 pounds60 \text{ pounds} by the mechanical advantage of 6 yields an output force of 360 pounds360 \text{ pounds}.

Adım Adım Çözüm

1
Determine the radius of both the wheel (crank path) and the axle (drive shaft).
Wheel radius (RR) is given as 18 inches18 \text{ inches}. Axle diameter is 6 inches6 \text{ inches}, so axle radius (rr) is 3 inches3 \text{ inches}.
Mechanical advantage formulas require the radii of the rotating components, not the diameter.
2
Calculate the Ideal Mechanical Advantage (IMA) of the wheel and axle.
IMA=Rr=183=6\text{IMA} = \frac{R}{r} = \frac{18}{3} = 6.
The IMA indicates how many times the machine multiplies the input force.
3
Calculate the output force by multiplying the input force by the IMA.
Output Force=60 pounds×6=360 pounds\text{Output Force} = 60 \text{ pounds} \times 6 = 360 \text{ pounds}.
In an ideal simple machine, the input force multiplied by the mechanical advantage equals the output load force.

Anahtar Kavram

Mechanical Advantage of a Wheel and Axle
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Konular

Wheel and Axle

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