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Zorluk: OrtaPascal's Principle and Hydraulics

A hydraulic lifting mechanism on a heavy-duty forklift contains an input cylinder and an output cylinder. The output cylinder has 55 times the cross-sectional area of the input cylinder. If the operator applies an input force of 50 lbs50 \text{ lbs} to move the input piston down by 15 inches15 \text{ inches}, what is the resulting lifting force and how far does the output piston move?

  1. 250 lbs250 \text{ lbs} of force and 3 inches3 \text{ inches} of movementCevap
  2. B
    250 lbs250 \text{ lbs} of force and 75 inches75 \text{ inches} of movement
  3. C
    10 lbs10 \text{ lbs} of force and 3 inches3 \text{ inches} of movement
  4. D
    10 lbs10 \text{ lbs} of force and 75 inches75 \text{ inches} of movement

Cevap

The output cylinder produces 250 lbs250 \text{ lbs} of lifting force and moves 3 inches3 \text{ inches}.
The hydraulic system provides a mechanical advantage equal to the ratio of the piston areas. Since the output cylinder has 55 times the area, the output force is 55 times the input force (50×5=250 lbs50 \times 5 = 250 \text{ lbs}). However, to conserve energy and fluid volume, the output piston only moves 1/51/5 of the distance of the input piston (15/5=3 inches15 / 5 = 3 \text{ inches}).

Adım Adım Çözüm

1
Determine the mechanical advantage of the hydraulic system.
Mechanical Advantage = 55.
The output cylinder has 55 times the cross-sectional area of the input cylinder.
2
Calculate the output lifting force.
F2=50×5=250 lbsF_2 = 50 \times 5 = 250 \text{ lbs}.
According to Pascal's Principle, force is directly proportional to the area. The larger piston exerts 55 times the force.
3
Calculate the distance the output piston moves.
d2=15/5=3 inchesd_2 = 15 / 5 = 3 \text{ inches}.
Because the volume of incompressible fluid displaced remains constant, the distance moved is inversely proportional to the area.

Anahtar Kavram

Pascal's Principle and Conservation of Energy in Hydraulics
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Pascal's Principle and Hydraulics

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