The small piston of a hydraulic lift has a cross-sectional area of 3.00 cm^2 and its large pisto

Name: Problem 8, Chapter 14: Physics for Scientists and Engineers with Modern Physics
Uploaded: 2020-07-29T10:52:56-08:00
Duration: 57 s
Channel: Surjit Tewari
Description: The small piston of a hydraulic lift has a cross-sectional area of 3.00 cm^2 and its large piston has a cross-sectional area of 200 cm^2 (Fig. 14.4 a ) . What force must be applied to the small piston for the lift to raise a load of 15.0 kN ? (In service stations, this force is usually exerted by compressed air.)

step 1 Question number 8, according to Pascal law, the force at the small piston upon the area of the s

Question

The small piston of a hydraulic lift has a cross-sectional area of 3.00 $\mathrm{cm}^{2}$ and its large piston has a cross-sectional area of 200 $\mathrm{cm}^{2}$ (Fig. $14.4 \mathrm{a} ) .$ What force must be applied to the small piston for the lift to raise a load of 15.0 $\mathrm{kN}$ ? (In service stations, this force is usually exerted by compressed air.)

   The small piston of a hydraulic lift has a cross-sectional area of 3.00 $\mathrm{cm}^{2}$ and its large piston has a cross-sectional area of 200 $\mathrm{cm}^{2}$ (Fig. $14.4 \mathrm{a} ) .$ What force must be applied to the small piston for the lift to raise a load of 15.0 $\mathrm{kN}$ ? (In service stations, this force is usually exerted by compressed air.)

Physics for Scientists and Engineers with Modern Physics

Raymond A. Serway,… 7th Edition

Chapter 14, Problem 8

Instant Answer

Solved by Expert Surjit Tewari

07/29/2020

Step 1

This means that the force per unit area (pressure) on the small piston is equal to the force per unit area on the large piston. Mathematically, this can be written as: \[ \frac{F_{s}}{A_{s}} = \frac{F_{L}}{A_{L}} \] where $F_{s}$ is the force on the small Show more…

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The small piston of a hydraulic lift has a cross-sectional area of 3.00 $\mathrm{cm}^{2}$ and its large piston has a cross-sectional area of 200 $\mathrm{cm}^{2}$ (Fig. $14.4 \mathrm{a} ) .$ What force must be applied to the small piston for the lift to raise a load of 15.0 $\mathrm{kN}$ ? (In service stations, this force is usually exerted by compressed air.)

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Key Concepts

Mechanical Advantage in Hydraulics

The concept of mechanical advantage in hydraulic systems arises from the difference in areas between two connected pistons. A small force applied over a small area can, through the transmission of pressure, generate a much larger force over a larger area, allowing heavy loads to be lifted with relatively little input force.

Pascal's Principle

Pascal's principle states that any change in pressure applied to an enclosed, incompressible fluid is transmitted equally in all directions throughout the fluid. This concept is fundamental in understanding how forces can be amplified in systems like hydraulic lifts.

Pressure as Force per Unit Area

Pressure is defined as the force applied per unit area (P = F/A). Recognizing this relationship is key to analyzing how different areas in a hydraulic system can produce different forces by maintaining equal pressure throughout.

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