/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 In a hydraulic lift, the radii o... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 10

In a hydraulic lift, the radii of the pistons are \(2.50 \mathrm{cm}\) and $10.0 \mathrm{cm} .\( A car weighing \)W=10.0 \mathrm{kN}$ is to be lifted by the force of the large piston. (a) What force \(F_{\mathrm{a}}\) must be applied to the small piston? (b) When the small piston is pushed in by $10.0 \mathrm{cm},$ how far is the car lifted?(c) Find the mechanical advantage of the lift, which is the ratio \(W / F_{\mathrm{a}}\).

Short Answer

Expert verified
Given: - Radii of the pistons: \(r_1 = 2.50\,\text{cm}\) and \(r_2 = 10.0\,\text{cm}\) - Weight of the car: \(W = 10.0\,\text{kN}\) - Distance the small piston is pushed in: \(d_1 = 10.0\,\text{cm}\) Answer: - The force applied to the small piston (\(F_{\mathrm{a}}\)) is \(0.625\,\text{kN}\). - When the small piston is pushed in by \(10.0\,\text{cm}\), the car is lifted by \(0.625\,\text{cm}\). - The mechanical advantage of the hydraulic lift is \(16\).

Step by step solution

01

Write down the given quantities

The given quantities are: - Radii of the pistons: \(r_1 = 2.50\,\text{cm}\) and \(r_2 = 10.0\,\text{cm}\) - Weight of the car: \(W = 10.0\,\text{kN}\) - Distance the small piston is pushed in: \(d_1 = 10.0\, \text{cm}\)
02

Find the areas of the pistons

Using the formula for the area of a circle, we can find the areas of both pistons: \(A_1 = \pi r_1^2\): Area of the small piston. \(A_2 = \pi r_2^2\): Area of the large piston. \(A_1 = \pi (2.50\,\text{cm})^2 = 19.63\,\text{cm}^2\) \(A_2 = \pi (10.0\,\text{cm})^2 = 314.16\,\text{cm}^2\)
03

Calculate the force on the small piston (Part a)

According to Pascal's law, the pressure exerted on both pistons is the same: \(P_1 = P_2\) We can write the pressures in terms of forces and areas: \(\frac{F_1}{A_1} = \frac{W}{A_2}\) We can solve for the force on the small piston, \(F_1\): \(F_1 = \frac{W \times A_1}{A_2}\) \(F_1 = \frac{10.0\,\text{kN} \times 19.63\,\text{cm}^2}{314.16\,\text{cm}^2} = 0.625\,\text{kN}\) So, the force that must be applied to the small piston is \(F_{\mathrm{a}} = 0.625\,\text{kN}\).
04

Calculate the distance the car is lifted (Part b)

Using the principle of conservation of volume, we can write the equation for the distances the pistons move: \(V_1 = V_2\) Since the volume of a cylinder is \(V = A \times d\), we can write the equation in terms of areas and distances: \(A_1 \times d_1 = A_2 \times d_2\) We can solve for the distance the car is lifted, \(d_2\): \(d_2 = \frac{A_1 \times d_1}{A_2}\) \(d_2 = \frac{19.63\,\text{cm}^2 \times 10.0\,\text{cm}}{314.16\,\text{cm}^2} = 0.625\,\text{cm}\) So, when the small piston is pushed in by \(10.0\,\text{cm}\), the car is lifted by \(0.625\,\text{cm}\).
05

Calculate the mechanical advantage (Part c)

Mechanical advantage is the ratio of output force (or weight of the car) to input force (or force on the small piston): \(MA = \frac{W}{F_{\mathrm{a}}}\) \(MA = \frac{10.0\,\text{kN}}{0.625\,\text{kN}} = 16\) The mechanical advantage of the hydraulic lift is \(16\).

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