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Zorluk: OrtaPascal's Principle and Hydraulics

A factory uses a closed hydraulic press system to stamp metal parts. The system features an input piston with a cross-sectional area of 3 in23 \text{ in}^2 and an output stamping piston with a cross-sectional area of 15 in215 \text{ in}^2. If the output stamping piston must travel a distance of 4 inches4 \text{ inches} to complete the stamping operation, how far must the input piston be depressed?

  1. 20 inches20 \text{ inches}Cevap
  2. B
    60 inches60 \text{ inches}
  3. C
    5 inches5 \text{ inches}
  4. D
    0.8 inches0.8 \text{ inches}

Cevap

20 inches20 \text{ inches}
In a closed hydraulic system, the volume of fluid displaced by the input piston must equal the volume gained at the output piston (V=A×dV = A \times d). Because the output piston has 55 times the area of the input piston (15/3=515 / 3 = 5), the input piston must travel 55 times as far to move the required volume of fluid. Therefore, multiplying the area ratio (55) by the output distance (4 inches4 \text{ inches}) gives an input travel distance of 20 inches20 \text{ inches}.

Adım Adım Çözüm

1
Calculate the mechanical advantage (area ratio) between the output and input pistons.
15 in2/3 in2=515 \text{ in}^2 / 3 \text{ in}^2 = 5
The area ratio dictates the proportional difference in how far each piston must travel.
2
Apply the principle of conservation of fluid volume (A1×d1=A2×d2A_1 \times d_1 = A_2 \times d_2) to find the input distance.
din=5×4 inches=20 inchesd_{in} = 5 \times 4 \text{ inches} = 20 \text{ inches}
Because the input piston is 5 times smaller in area, it must travel 5 times the distance of the output piston to displace the same volume of fluid.

Anahtar Kavram

Conservation of fluid volume in hydraulics (A1×d1=A2×d2A_1 \times d_1 = A_2 \times d_2)
Tahmini Süre:1m 0s
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Konular

Pascal's Principle and Hydraulics

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