Ch 14: Fluids and Elasticity

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 14: Fluids and Elasticity

Problem 44

Chapter 14, Problem 44

A 55 kg cheerleader uses an oil-filled hydraulic lift to hold four 110 kg football players at a height of 1.0 m. If her piston is 16 cm in diameter, what is the diameter of the football players' piston?

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Understand the principle: This problem involves Pascal's principle, which states that pressure applied to a confined fluid is transmitted undiminished throughout the fluid. The pressure on the cheerleader's piston must equal the pressure on the football players' piston.

Write the formula for pressure: Pressure is defined as force divided by area. For the cheerleader's piston, \( P_{1} = \frac{F_{1}}{A_{1}} \), and for the football players' piston, \( P_{2} = \frac{F_{2}}{A_{2}} \). Since \( P_{1} = P_{2} \), we can equate the two: \( \frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} \).

Express the areas in terms of the diameters: The area of a circle is \( A = \pi \left( \frac{d}{2} \right)^2 \), where \( d \) is the diameter. Substituting this into the equation gives \( \frac{F_{1}}{\pi \left( \frac{d_{1}}{2} \right)^2} = \frac{F_{2}}{\pi \left( \frac{d_{2}}{2} \right)^2} \). The \( \pi \) terms cancel out, simplifying to \( \frac{F_{1}}{\left( \frac{d_{1}}{2} \right)^2} = \frac{F_{2}}{\left( \frac{d_{2}}{2} \right)^2} \).

Rearrange to solve for \( d_{2} \): Multiply through by \( \left( \frac{d_{2}}{2} \right)^2 \) and rearrange to isolate \( d_{2} \): \( d_{2} = d_{1} \sqrt{\frac{F_{2}}{F_{1}}} \).

Substitute the known values: The cheerleader's force is \( F_{1} = m_{1} g = 55 \times 9.8 \), and the football players' force is \( F_{2} = 4 \times 110 \times 9.8 \). The diameter of the cheerleader's piston is \( d_{1} = 16 \ \text{cm} \). Substitute these values into the equation to find \( d_{2} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hydraulic Systems

Hydraulic systems operate based on Pascal's principle, which states that pressure applied to a confined fluid is transmitted undiminished in all directions. This principle allows a small force applied on a small piston to generate a larger force on a larger piston, enabling heavy loads to be lifted with relatively little effort.

Pressure Calculation

Pressure is defined as force per unit area (P = F/A). In hydraulic systems, the pressure exerted by the cheerleader on her piston must equal the pressure exerted by the football players' piston to maintain equilibrium. Understanding how to calculate and equate these pressures is crucial for determining the dimensions of the pistons involved.

Area of a Circle

The area of a circle is calculated using the formula A = π(d/2)², where d is the diameter. This concept is essential for determining the area of both the cheerleader's and the football players' pistons, as the force exerted by each piston is related to its area. By knowing the diameters, one can find the areas and subsequently the pressures involved in the hydraulic lift.

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